An Invitation to Galois Theory and Galois Representations Clifton Cunningham April 6, 2007

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Course Information Sheet: PMAT 613, Winter Term 2007

Monday

Tuesday

Wednesday

Thursday 12:30–14:00 in CHE202 15:00-16:00 in MS 452

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• Professor: Clifton Cunningham • Course website: go to http://blackboard.ucalgary.ca • Recommended Reading: Field and Galois Theory by Patrick Morandi. • Pre-requisite: PMAT431 or approval from the Division • Evaluation: – take-home test: 30% – presentation and oral exam: 20% – take-home exam: 50% All grading will use the GPA (grade point average) system, following official Faculty of Science guidelines.

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Date 2007.01.11 2007.01.18 2007.01.25 2007.02.01 2007.02.08 2006.02.15 2007.02.22 2007.03.01 2007.03.08 2007.03.15 2007.03.22 2007.04.05 2007.03.29 2007.04.12 2007.04.12

Topic categories: fields, subfields, subgroups no lecture the Galois adjunction; algebraic extensions finite extensions; finite subgroups finite Galois extensions; extension theorems The Galois is Normal and Separable Theorem Reading Week/Take-home test: due 2007.03.01 the Krull topology The Fundamental Theorem no lecture some infinite abelian extensions of Fp and Q ¯ Decomposition and inertia subgroups of Gal(Q/Q) ¯ Some `-adic representations of Gal(Q/Q) no lecture Take-home exam: due 2007.04.26

Comments Chapter 1: The Galois Adjunction Chapter 2: Galois Extensions Chapter 3: FTGT Chapter 4: Galois groups Chapter 4: Galois Rep’ns

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Preface to the course notes

These notes correspond roughly to the lectures for PMAT613 offered in the Winter Term of 2007 at the University of Calgary. These notes are still in very rough form: some proofs are missing, the notation is not yet completely uniform, exercises are almost entirely missing, not all internal references are working, and it is no-doubt riddled with typographic errors. I decided to make these notes available since I was not able to find a text for the course which reflected the perspective I have taken on Galois theory (categorical, topological and representation theoretic), but I imagine the present state of these notes must be very frustrating, and I apologize wholeheartedly for distributing such unfinished work. I can only hope that, despite its flaws, these notes help you understand the material presented in the course. I thank you all for bearing with me and with this work-in-progress. *** Number theorists used to boast that their subject was safe from the vagaries of mathematical fashion, since no applications of the subject could possibly be found. Of course, this absurd claim was always meant to be taken with a grain of salt, but it was proved completely false in the most spectacular manner when the work on Fermat’s Last Theorem lead directly to new cryptosystems which are now, only a few years later, ubiquitous in information security. As cryptography research groups have sprung up over the country, many graduate mathematics programs now attract significant numbers of talented students who are interested in applications of algebraic number theory to information security. On the other hand, many students of pure mathematics have been electrified by the spectacular breakthroughs in arithmetic geometry in recent years, including, but not limited to, the proof of the Shimura-Tanyama-Weil conjecture by Christophe Breuil, Brian Conrad, Fred Diamond, and Richard Taylor, and building on the work of Andrew Wiles. Many of these students have sensed - correctly - that the Langlands Program provides a unifying framework with which to understand this work and from which to attack open problems. It is natural to approach the Langlands Programme through Galois theory, and specifically, through the study of Galois representations and corresponding L-functions. The usual pedagogical approach to this field begins with a course in finite or classical Galois theory, and proceeds to a course in class field theory, which is a huge leap. This approach is completely inappropriate for students interested in applications of number theory and also rather heavy handed for students in pure mathematics who would like a more direct route through Galois theory to Galois representations. Tragically, it is difficult for many students to understand the goals and significance of class field theory from the perspective of a standard course in galois theory. One goal of this course is to produce students motivated and prepared to take a course in class field theory. There are three defining features to the treatment of Galois theory in this course, and each reflects a desire to reach both modern students interested in applications of algebraic number theory and also those students setting out toward the Langlands Program. The first is that we take the point of view that

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Galois theory is an example of a subject which straddles two categories and therefore, at its heart, the subject concerns functors and natural transformations. In fact, Galois theory is probably the first deeply categorical subject, both historically and pedagogically. In this course, the Galois Correspondence is an adjunction between categories, which is a weak form of an equivalence of categories, and the term ‘Galois’ is defined in terms of the adjoint pair of functors which form the Galois Equivalence. The course does not, however, assume any great familiarity with category theory apart from the very basic definitions. Consequently, this course provides an introduction to category theory, by way of a highly important example of an adjoint pair of functors. The second defining feature of our treatment of Galois theory is that, almost from the beginning, we remind the reader that a Galois group is a topological group, not just a group. The major results are stated and proved for arbitrary Galois extensions, not just finite Galois extensions. Of course, there are many excellent elementary treatments of Galois theory which include infinite Galois extensions, but for the most part they begin by studying the finite theory and then treating infinite Galois extensions as a sort of add-on. By contrast, we incorporate arbitrary Galois extensions into the discussion from the very beginning, and the big theorems in this course (such as the ‘Galois is Normal and Separable’ Theorem, and the Fundamental Theorem of Galois Theory) are stated and proved in that context. This is made possible by extensive use of (direct and inverse) limits throughout the course, in various categories, which simplifies many arguments, such as the existence of an algebraic closure. Although this course is meant to serve as an introduction to the subject, our ultimate goal is to provide students, as efficiently as possible, with the machinery necessary to study Galois representations, which are basic objects in modern number theory. This is the third and most important defining feature of our treatment of Galois theory. Consequently, this course introduces the inertia and decomposition subgroups of the absolute Galois group of the Rational numbers, and also provides some simple examples of `-adic Tate modules for curves, together with the action of the absolute Galois group on these modules. In this way we produce explicit examples of one- and two-dimensional `-adic ¯ representations of Gal(Q/Q). Galois Theory is an old subject, with a dramatic history reaching back into antiquity and forward to the most active branches of modern mathematics. History tells us that in order to understand the present it is necessary to study the past, and it is not surprising, in light of this maxim, that there are many textbooks available today which emphasize the classical roots of Galois Theory. However, the classical aspect of Galois theory is altogether absent from the lectures in this course. Here the reader will find no treatment of the hallowed topics of Galois theory, such as soluble polynomials and ruler-and-straightedge constructions. Instead, we go directly to the central concepts and principles necessary to understand how Galois theory plays a central role in modern number theory through Galois representations. The classical topics are well treated in existing introductory textbooks and make excellent material for student presentations.

Contents 0.1 0.2 1 The 1.1 1.2 1.3 1.4 1.5 1.6

Course Information Sheet: PMAT 613, Winter Term 2007 . . . . Preface to the course notes . . . . . . . . . . . . . . . . . . . . . Galois Adjunction A field guide to fields . . . The Galois functors . . . The Galois adjunction . . Pull-back and push-out . Products and co-products Limits and co-limits . . .

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2 Galois Extensions 2.1 Algebraic extensions . . . . . . . . . . . . . . 2.2 Galois extensions and subgroups . . . . . . . 2.3 Finite extensions . . . . . . . . . . . . . . . . 2.4 Finitely generated extensions . . . . . . . . . 2.5 Algebraic extensions and profinite extensions 2.6 Finite subgroups are Galois . . . . . . . . . . 2.7 Splitting extensions . . . . . . . . . . . . . . . 2.8 Algebraic closure . . . . . . . . . . . . . . . . 2.9 Normal extensions . . . . . . . . . . . . . . . 2.10 Restriction . . . . . . . . . . . . . . . . . . . 2.11 Separable extensions . . . . . . . . . . . . . . 2.12 Galois extensions . . . . . . . . . . . . . . . .

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3 The 3.1 3.2 3.3 3.4 3.5 3.6

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Fundamental Theorem Galois groups are profinite . The Krull topology . . . . . Galois (topological) groups Closed subgroups . . . . . . The Fundamental Theorem The Galois Equivalence . .

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6 4 Some important Galois groups 4.1 The Pruffer ring . . . . . . . . . . . . ¯ p /Fp ) . . . . 4.2 Some subgroups of Gal(F ¯ 4.3 Some subgroups of Gal(Q/Q) . . . . . 4.4 p-adic numbers . . . . . . . . . . . . . ¯ 4.5 Decomposition subgroups of Gal(Q/Q) ¯ 4.6 Inertia subgroups of Gal(Q/Q) . . . .

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5 Galois Representations 5.1 Ramification . . . . . . . . . . . . . . . . . . 5.2 `-adic cyclotomic characters . . . . . . . . . . 5.3 Tate module of the algebraic group GL(1) . . 5.4 Tate module of an elliptic curve . . . . . . . . 5.5 Complex Galois representations . . . . . . . . 5.6 Weil-Deligne group . . . . . . . . . . . . . . . 5.7 Weil-Deligne representations . . . . . . . . . . 5.8 An introduction to the Langlands Programme

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6 Student Presentations 6.1 Damian Brotbek: L¨ uroth’s Theorem, 2007.03.28, 08:45-10:00 in MS452 . . . . . . 6.2 Kelly Rose: Cubics and Quartics, 2007.03.30, 08:45-10:00 in MS452 . . . . . . 6.3 Ryan Stratford: Ruler and Compass, 2007.03.30, 11:00-12:15 in MS528 . . . . . . 6.4 Shuai Zhang: Solution by Radicals, 2007.04.04, 08:45-10:00 in MS452 . . . . . . 6.5 Robert Parkinson: Galois Cohomology, 2007.04.11, 08:45-10:00 in MS452 . . . . . . 6.6 Thomas Kuwahara: Inverse Galois Problem, 2007.04.13, 14:00-14:50 in ICT516 . . . . . . 6.7 Matthew Musson: Kummer Theory, 2007.04.20, 14:00-14:50 in ICT516 . . . . . . 6.8 Safa Ismail: Primitive Element Theorem, 2007.04.27, 08:45-10:00 in MS452 . . . . . .

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7 Tests and Exams 75 7.1 Take-home Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 7.2 Take-home Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Chapter 1

The Galois Adjunction 1.1

A field guide to fields

First things first: What is a field? Definition 1 A field is a non-zero commutative ring with identity such that every non-zero element is a unit. In this section we look at two equivalent definitions which provide some insight into fields. The first of these, Proposition 2, will not be used very much later, but it does makes clear just how thoroughly we adopt Zorn’s lemma (or equivalently, the Axiom of Choice) in the course. The second, Proposition 3, goes straight to the heart of the matter, and shows one of the very special features of the category of fields. Throughout this course, we refer to commutative ring with identity as a cring. For an arbitrary cring, let Spec(A) denote the set of prime ideals of A. Proposition 2 Let A be a non-zero cring; then A is a field if and only if Spec(A) = {(0)}. Proof. Suppose A is a field and let I be a prime ideal in A. If I is not the ideal (0), then I contains a non-zero element, say a. Since A is a field and a is non-zero, a is a unit. Thus, I contains a unit, so I = A. Thus, A has exactly two ideals, (0) and A itself; equivalently, (0) is maximal, thus prime so Specm(A) = {(0)} and Spec(A) = {(0)}. Conversely, suppose A is a non-zero cring and Spec(A) = {(0)}. Then Specm(A) = {(0)} or Specm(A) empty. In the first case, A has no zero divisors, which in itself is not enough to conclude that A is a field. However, the second case contradicts Zorn, which we accept in this course, so we are done. From the proof above we see that if A is a non-zero cring then A is a field if and only (0) is a maximal ideal, in which case Specm(A) (the set of maximal ideals in A) is the singleton {(0)}. Of course, we do not need Zorn to make this observation. 7

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Our next observation concerning fields comes from thinking about homomorphisms. In this book we use the term cring homomorphism for a homomorphism of crings which maps the identity of the domain to the identity of the codomain. Proposition 3 Let A be a non-zero cring. Then A is a field if and only if, for every non-zero cring B, Hom(A, B) consists entirely of monomorphisms (of crings). Proof. Suppose A is a field. Let B be a non-zero cring. Suppose φ ∈ Hom(A, B). The First Isomorphism Thoerem (FIT) gives the following commutative diagramme. A  A/ ker φ

φ

∼ =

/B O O / imφ

Now ker φ is an ideal of A, and A is a field, so ker φ = (0) or ker φ = A. Suppose, for a contradiction, that ker φ = A. Then the ring A/ ker φ is the zero ring, so imφ is (0). In particular, φ(1A ) = 0B . From the definition of maps in cring we see that this is possible only if 0B = 1B , in which case B is the zero ring. This is the desired contradiction, showing that ker φ = (0). It follows immediately that φ is injective. Conversely, suppose A is a non-zero cring such that Hom(A, B) consists entirely of injections for every non-zero cring B. Suppose, for a contradiction, that A is not a field. Then, by Proposition 2 A has a non-zero proper ideal I. Let B = A/I. Since I is proper, this is a non-zero cring. Consider the quotient map φ : A → B. Since ker φ = I is non-zero, φ is not a injective. This is the desired contradiction, showing that A is a field. In the category of fields, objects are field, maps are cring homomorphisms between fields, composition is given by function composition, and the identities are identity functions. Corollary 4 The category of fields is a full subcategory of the category of crings. Moreover, in the category of fields every homomorphism is a monomorphism! Proof. We have seen above that if K and L are fields, then Homfields (K, L) = Homcrings (K, L). Thus, the category of fields is a full subcategory of the category of crings. From Proposition 3 it follows immediately that if K and L are fields then Homcrings (K, L) consists entirely of monomorphisms. Consequently, the study of maps of fields may be divided into two parts: the study of inclusions and the study of isomorphisms. It is common in the literature to regard any map of fields K → L as an inclusion. We will not proceed that way in this course. In particular, we use the term extension to refer to any homomorphism of fields, not only inclusions; in this course, the terms ‘extension’ and ‘field homomorphism’ are synonymous.

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The Galois functors

Let Z be an object in an arbitrary category. Let sub(Z) denote the category of subobjects of X; this, an object is a monic with codomain Z (i.e., a monic map α : X → Z in the ambient category), a map from α : X → Z to β : Y → Z is a commutative triangle ? Z ?_ ?  ??β   ??  ?   / Y, X

(1.1)

α

composition is indicated by the diagramme, ? ZO `@@  @@  @@   @  /X /7 Y, S

(1.2)

and the identity idα:X→Z is the triangle Z ~> `AAA AAα α ~~ AA ~~ ~ A ~ o X. X id

(1.3)

X

Notice that there is an obvious forgetful functor from sub(Z) to the ambient category, taking α : X → Z to X and taking the ‘triangle’ X → Y → Z to X → Y . We will often make implicit use of this functor. Given a field L, the Galois correspondence is a pair of contravariant functors GALL and FIXL , the former from the category sub(L) of subfields of L to the category subAut(L) of subgroups of Aut(L), and the latter in the opposite direction. While the composition of these two functors is not an identity functor, there is a natural transformation from the identity functor to the composition FIXL ◦ GALL and likewise from the identity functor to the composition GALL ◦ FIXL . In this section we define these functors; in the next section we show that (GALL , FIXL ) is an adjoint pair of contravariant functors. Let L be a field and let Aut(L) denote the group of automorphisms of L (in the category of fields, of course). Let K be a subfield of L. The Galois group of L over K is the group AutK (L) of field automorphisms σ of L such that the following diagramme commutes. L `@ @@ @@ α @@

σ

K

/L; ~? ~ ~ ~~α ~ ~

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thus, AutK (L) = {σ ∈ Aut(L) σ ◦ α = α}. We write GALL (K) for the group AutL (K) considered as a subgroup of Aut(L); in other words, GALL (K) is the object AutL (K) → Aut(L) in sub(Aut(L)). Next, let γ be a map in the category of subfields of L; thus, γ is a commuting triangle (1.4) > L _@@ }} @@ } } @@ }} @ }} /N M as in Diagramme 1.1. Let GALL (γ) be the commuting triangle Aut(L) eLLL s9 s LLL s s s LLL s s s L ss / AutM (L) AutN (L)

(1.5)

where each arrow is the map ϕ 7→ ϕ, which is clearly a group monomorphism. The verification that GALL really is a functor is left as an exercise. We can summarize these rules by the following short-hand: GALL (K) = AutK (L)

and

GALL (M → N ) = (AutM (L) ,→ AutN (L)).

Lemma 5 If α and β are subfields of L and Hom(α, β) is not empty, then GALL (β) is a subgroup of GALL (α). Proof. GALL is contravariant, so GALL : Hom(α, β) → Hom(GALL (β), GALK (α)). If the former is non-empty, then the later is non-empty. We now define a functor FIXL from the category of subgroups of Aut(L) to the category of subfields of L. Let H → Aut(L) be an object in the category of subgroups of Aut(L); thus, H → Aut(L) is a group monomorphism. Let LH denote the field of elements of L fixed by the image of each element of H under f : H → Aut(L); thus, LH = {u ∈ L ∀h ∈ H, f (h)(u) = u}. Let FIXL (H → Aut(L)) denote the obvious field monomorphism LH → L; thus, FIXL (H → Aut(L)) is an object in the category of subfields of L. Next, consider the map φ in the category of subgroup of Aut(L) given by the following triangle Aut(L) ; cGG GG ww w GG w w GG w w G ww / H2 H1

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let FIXL (φ) denote the map in the category of subfields of L given by the triangle > L aC || CCC | CC | CC || || / LH1 , LH2 where the group monomorphisms are all inclusions. Note that we can summarize these rules by the following shorthand: for any subgroup H of Aut(L), FIXL (H) = LH

1.3

and

FIXL (H1 → H2 ) = (LH2 ,→ LH1 ).

The Galois adjunction

As mentioned in the preceding section, the functors GALL and FIXL are not equivalences. However, they are closely related, as we shall now see. We begin by defining a natural transformation F L : idL → FIXL ◦ GALL , where idL denotes the identity functor on the category of subfields of L. Let α : K → L be a subfield of L. (FIXL ◦ GALL )(K → L)

= FIXL (AutK (L) → Aut(L)) = (LAutK (L) → L).

Compare this with idL (K → L) = (K → L). Recall that σ ∈ AutK (L) implies σ ◦ α = α, so σ(α(u)) = α(u) for each u ∈ K. Thus, the image of α : K → L is actually contained in LAutK (L) . With this in mind, let F L (K → L) be the map of subfields of L given by the triangle ? L dIII II   II  II  I   α / LAutK (L) . K α

Thus, F L (α) is α with codomain restricted to LAutK (L) . We can summarize the definition of the natural transformation F L by the following shorthand: for any subfield α : K → L, α

α

F L (K → L) = (K → LAutK (L) ). Using this shorthand, F L is given by the following square. K1

α1

/ LAutK1 (L)

γ

 K2

 / LAutK2 (L) α2

(1.6)

12

DRAFT : April 6, 2007 Next, we define a natural transformation G L : idAut(L) → GALL ◦ FIXL ,

where idAut(L) denotes the identity functor in sub(Aut(L)). Let τ : H → Aut(L) be a subgroup. Then (GALL ◦ FIXL )(H)

= GALL (LH ) = AutLH (L).

Now, G L (H) is the map (in the category of subgroups of Aut(L)) from H → Aut(L) to AutLH (L) → Aut(L) given by the triangle Aut(L) fLLL < xx LLL x x LLL xx x L xx / AutLH (L) H To define the bottom field homomorphism simply note that the image of H → Aut(L) is contained in AutLH (L). Note also that LAutLH (L) = AutLH (L). It follows from the discussion above that (GALL , FIXL ) is an adjoint pair of contravariant functors. This observation has important consequences, as any adjoint pair establishes an equivalence of certain closely associated categories. To see what this means in this case we make the following definition. Definition 6 Let L be a field. A subfield K → L is Kaplansky if F L (K → L) is an isomorphism; a subgroup H of Aut(L) is Kaplansky if G L (H → Aut(L)) is an isomorphism. Equivalently, K → L is Kaplansky if the image of K → L is LAutK (L) , and H → Aut(L) is Kaplansky if the image of H → Aut(L) is AutLH (L). These notions define two full subcategories and an equivalence between them. Let the category of Kaplansky subfields of L be the full subcategory of the category of subfields of L such that each object is a Kaplansky subfield; likewise, let the category of Kaplansky subgroups of Aut(L) be the full subcategory of the category of subgroups of Aut(L) such that each object is a Kaplansky subgroup. Then GALL and FIXL are equivalences between these categories. Before concluding this section, we make one further observation: the natural transformation G L induces an isomorphism of functors FIXL ◦ GALL ◦ FIXL ∼ = FIXL , and the natural transformation F L induces an isomorphism of functors GALL ◦ FIXL ◦ GALL ∼ = GALL , as summarized in the following diagrammes.

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subfields of L

subgroups of Aut(L)

LO

Aut(L) O

LAutO K (L) i

K



GALL

/ Aut AutK (L) (L) L

FIXL

 GAL

L

 / AutK (L)

subfields of L

subgroups of Aut(L)

LO

Aut(L) O

LAutLH (L) o + LH o

FIXL

 Aut H (L) L 5 O

GALL FIXL



H

Although the concepts introduced in the remainder of this chapter are important to this course, the reader pressed for time may skip them on a first reading, and go directly to Chapter 2, referring back to Sections 1.4 and 1.6 as necessary.

1.4

Pull-back and push-out

One important property that the category of rings enjoys is the existence of pull-backs and push-outs. To explain what this means, suppose for the moment that α : A → C and β : B → C be maps. A pull-back of the pair (α, β) is a pair of maps (α0 , β 0 ) such that the square below is commutative and also enjoys the following property: if α00 : D0 → B and β 00 : D0 → A are any maps making the outer square commute, then there is a unique map θ : D0 → D making the

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triangles commute. > C `AA AA β }} } AA } AA }} } } AO `A B AA }> N AA }} } A }} 0 β 0 AA }} α DO α

β 00

α00

∃! θ

D0

Proposition 7 Pull-backs exist in the category of rings, crings, fields and groups.

Proof. Let A, B, C and α, β be as above. Let A ×C B denote the subset of (a, b) ∈ A × B such that α(a) = β(b). To see that A ×C B is a ring it suffices to show that it is a subring of A×B with pairwise addition and multiplication (left for the reader). Next, define α0 : A×C B → B by α0 (a, b) = b and β 0 : A×C B → A by β 0 (a, b) = a. These are clearly ring homomorphisms, and if (a, b) ∈ A×C B then β ◦ α0 (a, b) = β(b) = α(a) = α ◦ β 0 (a, b), so the square appearing in the definition above commutes. In order to show that (α0 , β 0 ) is a pull-back, suppose α00 : D0 → A ×C B → B and β 00 : D0 → A ×C B → A are ring homomorphisms such that α00 ◦ β = β 00 ◦ α (so the outer square commutes in the diagramme above). Define θ : D0 → A ×C B by θ(d) = (β 00 (d), α00 (d)). The codomain of θ is A ×C B precisely because the outer square commutes. Moreover, the triangles in the diagramme above commute because β 0 ◦ θ(d) = β 0 (β 00 (d), α00 (d)) = β 00 (d) and α0 ◦ θ(d) = α0 (β 00 (d), α00 (d)) = α00 (d) for all d ∈ D0 . It is also clear from this calculation that θ is the only function with this property, thus completing the proof that pull-backs exists in the category of rings. To see that pull-backs also exist in the category of fields it is sufficient to suppose that A, B and C are fields and show that A ×C B is a field. It is clear that A ×C B is a cring. If (a, b) ∈ A ×C B and (a, b) 6= (0, 0), then a 6= 0 and b 6= 0, so a and b are units. Since α(a−1 ) = α(a)−1 = β(b)−1 = β(b−1 ), (a−1 , b−1 ) ∈ A ×C B. Moreover, (a, b) × (a−1 , b−1 ) = (1, 1), so (a, b) is a unit in A ×C B. This completes the proof that A ×C B is a field, and therefore the proof that pull-backs exist in the category of fields. The dual notion also exists in the category of rings and fields. Again, suppose for the moment that α : C → A and β : C → B are maps in an arbitrary category. A push-out of α and β is a pair of maps (α0 , β 0 ) such that the square below is commutative and also enjoys the following property: if (α00 , β 00 ) is any other map making the outer square commute, then there is a unique map

DRAFT: April 6, 2007

15

θ : D → D0 making the triangles commute. 0

< DO b ∃! θ β 00

α00

> D `AA AA }} } AA }} 0 } α0 AA }} β A `A >B AA }} AA } } α AAA }} }} β C In the category of rings, push-outs are given by the tensor product. However, the existence of push-outs in the category of fields is a much thornier issue, which we do not need in this course. Instead, we have the following weaker result. Proposition 8 The push-out of a pull-back exists in the category of rings, fields and groups. Proof. Let α : M → L and β : N → L be field homomorphisms. Let R denote the intersection of all subfields (in the literal sense) of L which contain α(M ) ∪ β(N ). We claim that R is a push-out of a pull-back of (α, β). B LO \ 0

α

; RO c

β

∃! θ

R |> `AAA AA || | AA | 00 00 | A α β ||

M aB BB BB B β 0 BB

K

N }> }} } }} 0 }} α

[under construction]

1.5

Products and co-products

In any category, the product of objects X1 and X2 is an object X equipped with maps π2 π1 / X2 X1 o X

16

DRAFT : April 6, 2007

with the following universal property: for any X 0 and any π10 : X 0 → X1 and π20 : X 0 → X2 there is a unique map θ : X 0 → X such that X1 a o

π1 π10

XO

/ X2 =

π2

π20

∃! θ

X0 Q

commutes. Products are usually denoted X1

X2 .

Proposition 9 Products exists in the category of subfields and in the category of subgroups. Proof. Let α : M → L and β : N → L be objects in the category of subfields of L. Let (α0 , β 0 ) be a pull-back of (α, β) in the category of fields, as pictured below. L {= O `BBB BBβ α {{{ BB { { B {{ γ N MP aC CC β 0 |> N CC α ||| CC | C ||| KO 00 00 β

α

∃! θ

K0 We claim that γ : K → L is a product in the category of subfields of L [under construction]. In any category, the co-product of objects X1 and X2 is an object X equipped with maps /Xo X1 X2 f1

f2

with the following universal property: for any X 0 and any f10 : X1 → X 0 and f20 : X2 → X 0 there is a unique map θ : X → X 0 such that 0

f10

X1

f1

= XO a ∃! θ

/Xo

f20

f2

commutes. Co-products are often denoted X1

X2 `

X2 .

Proposition 10 Co-products exist in the category of subfields and in the category of subgroups. Proof. Let α : M → L and β : N → L be objects in the category of subfields of L. The coproduct of these two objects is given by considering the push-out of the pull-back of (α, β) in the category of fields. [details left to the reader]

DRAFT: April 6, 2007

1.6

17

Limits and co-limits

In this section we present two constructions which will be very important later in the course. Let I be a partially ordered set. Suppose I is directed, which is to say that it satisfied the following property: ∀i ∈ I, ∀j ∈ I, ∃k ∈ I,

i ≤ k and j ≤ k.

Let ext(K) denote the the category of extensions of K: in this category, objects are field homomorphisms α : K → N or β : K → M and Hom(α, β) is the set of commuting triangles as below, which may also denoted HomK (N, M ). γ

N `A AA AA α AA

/M |> | || || || β

K

Suppose we are given, for each ordered pair i ≤ j in I, a map αi≤j : Ki → Kj in ext(K). This family of maps (αi≤j : Ki ≤ Kj )i,j∈I is a direct system in ext(L) if it satisfies the following two conditions: ∀i, j ∈ I, i = j =⇒ αi≤j = idKi ∀i, j, k ∈ I i ≤ j ≤ k =⇒ αi≤k = αj≤k ◦ αi≤j Proposition 11 If I is directed and (αi≤j : Ki → Kj )i,j∈I is a direct system in ext(L), then there is a field extension αI : K → KI equipped with a family of homomorphisms (αi : Ki → KI )i∈I such that: αi = αj ◦αi≤j for each i ≤ j; and if (β i : Ki → KI0 )i∈I is another such family, then there is a unique θ : KI → KI0 such that β i = θ ◦ αi for each i ∈ I.

∃!θ

4 KI0 > L

βi

n6 > K F I nnn|||

|| nn nnn |||

n n n Ki aBαi≤j / Kj

αI O BB

BB αj

αi BBB

K nj αi nnα

βj

` Proof. Let KI denote the set of equivalence classes of elements from i∈I Ki (coproduct in the category of sets, i.e., disjoint union) with respect to the equivalence relation defined by ui ≡ uj ⇐⇒ ∃k ∈ I, αi≤k (ui ) = αj≤k (uj ).

18

DRAFT : April 6, 2007

We begin by showing that this set is canonically equipped with addition and multiplication rules making it into a field. We will then show that the field K enjoys the universal property promised above. Suppose x, y ∈ K. Then x = [ui ] and y = [uj ] for some ui ∈ Ki and uj ∈ Kj . Pick k greater than both i and j in I and define [ui ]+[uj ] := [αi≤k (ui )+ αj≤k (uj )] and [ui ] × [uj ] := [αi≤k (ui ) × αj≤k (uj )]. (Exercise: Show that this is well-defined...) It is now clear that K is a cring with 0K = [0] and 1K = [1]. (Exercise: Do this.) To see that K is a field, suppose x ∈ K × , so x = [ui ] −1 for some ui ∈ Ki× , and observe that [ui ] × [u−1 i ] = [ui × ui ] = [1] = 1K , so −1 −1 x = [ui ]. It remains to show that K has the universal property promised above. (exercise) Definition 12 If I is a directed system and (αi≤j : Ki → Kj )i,j∈I is a direct system in ext(L) then the extension K → KI constructed in Proposition 11 is denoted lim Ki . − − → i∈I

In Section 3.6, we also need the direct limit in the category of topological groups and subgroups The dual notion will play an important role in the category of groups. With I as above, suppose we are given, for each ordered pair i ≤ j in I, a group homomorphism ϕi≤j : Gj → Gi . Suppose that this family of group homomorphisms (ϕi≤j : Gj ≤ Gi )i,j∈I is an inverse system, which is to say, it satisfies the conditions: ∀i, j ∈ I, i = j =⇒ ϕi≤j = idGj ∀i, j, k ∈ I i ≤ j ≤ k =⇒ ϕi≤k = ϕi≤j ◦ ϕj≤k . Proposition 13 If I is directed and (ϕi≤j : Gj → Gi )i,j∈I is an inverse system, then there is a group GI equipped with a family of group homomorphisms (ϕi : GI → Gi )i∈I such that: ϕi = ϕi≤j ◦ αj for each i ≤ j; and if (ϕ0i : G0 → G0i )i∈I is another such family, then there is a unique θ : G0 → GI such that ϕ0i = ϕi ◦ θ for each i ∈ I.

G0 ϕ0i

GI nnn} ϕi nnn }} nnn }} nnn ~}}} ϕj n n vn Gi o ϕi≤j Gj q

∃!θ

~ ϕ0j

DRAFT: April 6, 2007

19

Proof. Let GI denote the set of x = (xi )i∈I ∈ of sets) such that ∀i, j ∈ I,

Q

i∈I

Gi (product in the category

i ≤ j =⇒ xi = ϕi≤j (xj ).

We begin by showing that this set is canonically equipped with a multiplication rule making it into a group. We will then show that the group G enjoys the universal property promised above..... [under construction] Definition 14 If I is a directed system and (ϕi≤j : Gj → Gi )i,j∈I is an inverse system of groups then the group GI constructed in Proposition 13 is denoted lim G. ← − − i i∈I

Definition 15 A group G is said to be profinite if there is a directed partially ordered set I and an inverse system of groups (ϕi≤j : Gj → Gi )i,j∈I such that G∼ G and each Gi is finite. = lim ← − − i i∈I

Profinite groups play an extremely important role in Galois theory; in particular, we will see later that every Galois group is profinite. In Chapter 3 we will see that this endows every Galois group with a topology, with important consequences.

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Chapter 2

Galois Extensions 2.1

Algebraic extensions

The study of algebraic extensions of a field K is inextricably linked to the ring K[x], so we begin this section by recalling that K[x] is a principal ideal cring and that Spec(K[x]) = Specm(K[x]) ∪ {(0)}. Let us also take this moment to fix some notation: for each u ∈ K, we write u : K[x] → K for the unique splitting of ι : K → K[x] such that u (x) = u; we refer to u as evaluation at u. Now, let α : K → L be a fixed extension of fields and consider the set HomK (K[x], L) of cring homomorphisms φ : K[x] → L such that the triangle K[x] aDD DD D ι DD D

φ

K

/L ?    α  

is commutative. Our first goal is to understand this set. If we write αx : K[x] → L[x] for the obvious extension of α : K → L, then it is clear that u ◦ αx is an element of HomK (K[x], L), for each u ∈ L. A moments reflection shows that every element of HomK (K[x], L) takes this form, since if φ ∈ HomK (K[x], L) then X X φ( ai xi ) = φ(ai )φ(x)i i

i

=

X

=

X

φ(ι(ai ))φ(x)i

i

α(ai )φ(x)i

i

X = φ(x) ( α(ai )xi ) i

X = φ(x) ◦ αx ( ai xi ). i

21

22

DRAFT : April 6, 2007

In otherwords, the function HomK (K[x], L) → L φ 7→ φ(x) is a bijection. Observe next that if φ ∈ HomK (K[x], L) then ker φ is a prime ideal in K[x], since if f1 f2 ∈ ker φ then φ(f1 f2 ) = 0 so φ(f1 )φ(f2 ) = 0, in which case φ(f1 ) = 0 or φ(f2 ) = 0, so f1 ∈ ker φ or f2 ∈ ker φ. Definition 16 Let α : K → L be an extension of fields. We say that u ∈ L is algebraic over K if ker(u ◦ αx ) is a maximal ideal. In this case we write fu,K ∈ K[x] for the unique monic polynomial generator for ker(u ◦ αx ); the irreducible polynomial fu,K ∈ K[x] is called the minimal polynomial for u over K. Otherwise, u ∈ L is transcendental over K. The extension α : K → L is algebraic if the image of the map HomK (K[x], L) → Spec(K[x]) φ 7→ ker φ is contained in Specm(K[x]); otherwise, the extension is said to be transcendental. Some of you may recognize the geometric nature of this definition. The set Spec(K[x]) is the set underlying the affine line A1K as a K-scheme and HomK (K[x], L) is precisely the set of L-valued points in A1K as a K-scheme. Thus, Definition 16 may be paraphrased as follows: K → L is algebraic if and only if every L-valued point on A1K is closed. Lemma 17 Every isomorphism of fields is an algebraic extension. Proof. Let α : K → L be an isomorphism. Pick u ∈ L. Then the minimal polynomial for u over K is fu,K = x − α−1 (u), since fu,K is irreducible, monic, has coefficients in K and αx (fu,K )(u) = u − u = 0. Thus, α : K → L is algebraic. Lemma 18 Suppose α = β ◦ γ. If α is algebraic then β and γ are algebraic.

? L `A ~~ AAAAβ ~ AA ~~ A ~~ /M K α

γ

Proof. Suppose α is algebraic. Pick u ∈ L. Consider fu,K ∈ K[x]× , which exists since α : K → L is algebraic, and γx (fu,K ) ∈ M [x]× . Now, βx (γx (fu,K ))(u) = αx (fu,K )(u) = 0, so ker(u ◦ βx ) is maximal, which shows that u is algebraic

DRAFT: April 6, 2007

23

over M . Since u ∈ L was arbitrary, it follows that β : M → L is algebraic. Now, pick v ∈ M . Consider X fβ(v),K = bi xi ∈ K[x]× , i

which exists since α : K → L is algebraic and β(v) ∈ L. Since αx (fβ(v),K )(β(v)) = 0, we have αx (fβ(v),K )(β(v)) = 0 X αx ( bi xi )(β(v)) = 0 i

X

α(bi )β(v)i = 0

i

X

β ◦ γ(bi )β(v i ) = 0

α=β◦γ

i

X β( γ(bi )v i ) = 0 i

X

γ(bi )v i = 0 β is monic

i

X γx ( bi xi )(v) = 0 i

γx (fβ(v),K )(v) = 0 Thus, ker(v ◦ γx ) is maximal, which shows that v is algebraic over K. Since v ∈ M was arbitrary, it follows that γ : K → M is algebraic. In due course we will see that the converse to Lemma 18 is also true (see Proposition 36).

2.2

Galois extensions and subgroups

Definition 19 Let L be a field. (a) A field homomorphism K → L (i.e., an object in the category of subfields of L) is Galois if it is algebraic and Kaplansky. (b) A subgroup H → Aut(L) is Galois if it is an object in the subcategory of subAut(L) which is equivalent to the category of Galois subfields of L; equivalently, a subgroup H of Aut(L) is Galois if H = GALL (K) for some Galois subfield K of L. In Chapter 2 we assemble various important properties of Galois extensions; in the next chapter we study the category of Galois subgroups. Here we make a very simple observations concerning Galois extensions. Lemma 20 Every isomorphism of fields is a Galois extension of fields.

24

DRAFT : April 6, 2007

Proof. Let α : K → L be an isomorphism. If σ ∈ AutK (L) then α = σ ◦ α. Since α is an isomorphism, α ◦ α−1 = σ ◦ α ◦ α−1 so idL = σ. Now, AutK (L) = {idL } so LAutK (L) = L{idL } = L. Recalling the definition of the natural transformation F L from Equation 1.6, we see that F L (α) = α. Since α is an isomorphism it follows from Definition 6 that α is Kaplansky. We already saw (Lemma 17 that α is algebraic, so α is Galois, by Definition 19.

2.3

Finite extensions

Let α : K → L be a field homomorphism. Then L is a vector space over K with the following definition: K ×L→L (k, u) 7→ α(k)u. (We will often write k ·u for α(k)u, below.) Thus, we can apply a basic invariant from the theory of vector spaces to field extensions. Definition 21 Let K → L be an extension. The degree of L over K is the dimension of L as a vector space over K. This number, which may be infinite, is denoted dimK (L) or [L : K]. If the degree of L over K is finite, then we say K → L is a finite; otherwise, K → L is a infinite. Proposition 22 (Tower Law) Let K → M and M → L be field homomorphisms; then dimK (L) = dimK (M ) dimM (L). Proof. Let {ui i ∈ I} be a basis for L over M and let {vj j ∈ J} be a basis for M over K. Then {ui vj (i, j) ∈ I × J} is a basis for L over K. Corollary 23 Suppose α = β ◦ γ. Then α is a finite extension if and only if β and γ are both finite extensions. Proposition 24 If K → L is finite then K → L is algebraic. Proof. Pick u ∈ L. SincePdimK (L) = n, the set {1, u, u2 , . . . un } is linearly dependent over K.PThus, i ai · ui = 0 for some ai ∈ K not all zero. Define f ∈ K[x]× by f = i ai xi . Since αx (f )(u) = 0, it follows that ker(u ◦ αx ) is a non-zero ideal, whence u ∈ L is algebraic over K by Definition 16.

2.4

Finitely generated extensions

Definition 25 Let α : K → L be a field extension and let u be an element of L. Let K(u) denote the intersection (in L) of all subsets of L which contain u, α(u) and are fields. Define αu : K → K(u) by αu (x) = α(x); thus, αu is just α with restricted codomain. Then αu : K → K(u) is a simple extension. In fact, K(u) admits another description which is very helpful.

DRAFT: April 6, 2007

25

Lemma 26 Let α : K → L be a field homomorphism. Let u : L[x] → L denote the cring homomorphism defined by u (f ) = f (u). Let K[u] denote the image of u ◦ αx , where αx : K[x] → L[x] is the obvious cring homomorphism. Then K(u) is the quotient field of the integral domain K[u]. Proof. (exercise) Proposition 27 Let α : K → L be a field homomorphism. Then αu : K → K(u) is finite (and thus algebraic) if and only if u is algebraic over K. Proof. The FIT gives us the following commuting diagramme. K[x]  K[x]/ ker u

u

¯u ∼ =

/L O O / imu

Proposition 28 Let K → L be a field homomorphism. Suppose u ∈ L is algebraic over K. Let n = deg(mK,u ). Then {1, u, u2 , . . . , un−1 } is a basis for K(u) over K. Thus, K → K(u) is finite and dimK (K(u)) = degK (u). Proof. Pick a ∈ L. By Definition 31 and Lemma [inset reference] there is some f ∈ K[x] such that a = u (f ). By the Euclidean Division Theorem, f = qfu,K + r for some q, r ∈ K[x] with r = 0 or deg r < n. Now u (f ) = u (q)u (fu,K ) + u (r) so u (f ) = u (r). Thus, a = u (r). Since r ∈ K[x] and r = 0 or deg(r) < n it follows that a is a linear combination of {1, u, u2 , . . . , un−1 } over K. In other words, {1, u, u2 , . . . , un−1 } spans L over K. To see that Pn−1 {1, u, u2 , . . . , un−1 } is linearly independent over K, suppose i=0 ai · ui = 0 Pn−1 with ai ∈ K. Let g = i=0 ai xi . Then u (g) = 0, so g = hmK,u for some h ∈ K[x]. If g 6= 0 then n − 1 ≥ deg(g) = deg(h) + deg(fu,K ) ≥ n. Since this is clearly impossible, g = 0. This concludes the proof that {1, u, u2 , . . . , un−1 } is a basis for L over K. It follows immediately that n = dimK (K(u)). Proposition 29 Let K → L be an algebraic extension and let u ∈ L be algebraic over K. Let {u1 , . . . , uk } be the distinct roots of αx (fu,K ) ∈ L[x]. Then Gal(K(u)/K) = {τ1 , . . . , τk }, where each τi is defined by the condition τi (u) = ui . Proof. (exercise) Lemma 30 (Extension Lemma) Let α : K → M be a finite simple homomorphism and let u be a generator of this extension. Let β : K → L be any homomorphism. Let HomK (M, L) denote the set of homomorphisms γ : K(u) → L such that β = γ ◦ α. Then u : HomK (M, L) → L is a bijection onto the set of roots of βx (fu,K ) in L.

26

DRAFT : April 6, 2007

Proof. Begin by factoring β = β 00 ◦ β 0 where β 0 is an isomorphism and β 00 is an inclusion. Now, suppose γ ∈ HomK (M, L) and let v = u (γ), so v = γ(u). Then βx (fu,K )(v) = βx (fu,K )(β(v)) = β(fu,K (u)) = β(0) = 0, so v is a root of βx (fu,K ) in L. (under construction) /L O [

γ

K[x] K(u) o u O eKK KK KK ¯u KKK  α K[x]/(fu )

K

/ K 0 (v) 9 O s ss s s ss  sss ¯v / K 0 [x]/(fv ) K 0 [x]

β¯

∼ = β0

v

β 00

/ K0

It is not true that every algebraic extension is finite. To explore this issue further, we require another definition. Definition 31 Let α : K → L be a field extension and let X be a subset of L. Let K(X) denote the intersection (in L) of all subsets of L which contain X, α(K) and are fields. We say that L is finitely generated over K if L = K(X) for some finite subset S ⊆ L. Lemma 32 K(u, v) = K(u)(v). Proof. (exercise) Definition 31 leads to yet another characterization of elements of L which are algebraic over K. Proposition 33 Let K → L be a field homomorphism; let X be a finite subset of L. Then K → K(X) is finite (and thus algebraic) if and only if each u ∈ X is algebraic over K. Proof. Use Proposition 28, Exercise ?? and induction to see that K(u1 , . . . , ui ) = K(u1 , . . . , ui−1 )(ui ) for each 1 ≤ i ≤ n. Proposition 34 K → L is finite if and only if it is algebraic and finitely generated. Proof. Suppose K → L is finite. In Proposition 24 we have already seen that K → L is algebraic. Let n = dimK (L) and let (u1 , u2 , . . . , un ) beP a basis for L as a vector space over K. Then every element of L takes the form i ki · ui with ki ∈ K. Thus, L ⊆ K(u1 , u2 , . . . , un ), so L = K(u1 , u2 , . . . , un ), by Exercise ??. Thus, K → L is finitely generated.

DRAFT: April 6, 2007

27

Conversely, suppose K → L is algebraic and finitely generated. Then L = K(u1 , u2 , . . . , un ) where each ui is algebraic over K. By Exercise ?? (and induction) we know that K(u1 , . . . , ui ) = K(u1 , . . . , ui−1 )(ui ) for each 1 ≤ i ≤ n. Thus, K → L can be written as a composition of finite extensions. The result now follows from Proposition 22. Proposition 35 Let L be a field. The functor GALL takes finite extensions K → L to finite subgroups of Aut(L) and the functor FIXL takes finite subgroups of Aut(L) to finite extensions into L. Proof. We begin by showing that if K → L is finite then AutK (L) is finite. Let {u1 , . . . , un } be a basis for L over K. Then L = K(u1 , . . . , un ). Suppose σ ∈ AutK (L). Then σ is completely determined by the values {σ(u1 ), . . . , σ(un )}. For each 1 ≤ i ≤ n, let consider the minimal polynomial fui ,K for ui over K and observe that fui ,K (σ(ui )) = σ(fui ,K (ui )) = σ(0) = 0; thus, σ(ui ) is a root of fui ,K . Since there are finitely many roots of fui ,K for each i, there are only finitely many automorphisms σ ∈ AutK (L). Let H be a finite subgroup of Aut(L) and let K = LH ....(under construction)

2.5

Algebraic extensions and profinite extensions

Proposition 36 Suppose α = β ◦ γ. Then α is algebraic if and only if β and γ are algebraic. Proof. In Lemma 18 we showed that if α is algebraic then so are β and γ. Suppose, therefore, that β : M → L and γ : K → M are algebraic. Pick u ∈ L. ConsiderPfu,M ∈ M [x]× , which exists since β : M → L is algebraic. Write n fu,M = i=0 bi xi and consider the diagramme below. K(b0 , . . . , bn , u) 6 gNNN mmm NNN m mm m NNN m m m NN mmm K(b0 , . . . , bn ) K(u) hRRR o7 o RRR o o RRR ooo RRR RRR ooooo K Since γ : K → M is algebraic and bi ∈ M for each 1 ≤ i ≤ n, it follows from Proposition 33 that K → K(b0 , . . . , bn ) is finite. Since u is a root of βx (fu,M ) ∈ K(b0 , . . . , bn )[x], it follows that u is algebraic over K(b0 , . . . , bn ), so K(b0 , . . . , bn ) → b0 , . . . , bn , u) is finite. Thus, the two extensions on the lefthand side of the diagramme above are finite. It now follows from Proposition 22 that K → K(u) is finite, so u is algebraic over K (again, by Proposition 33).

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DRAFT : April 6, 2007

Definition 37 We say that a field homomorphism α : K → L is profinite if it is isomorphic (in the category of extensions of K) to a direct limit over a directed system of finite extensions of K. Theorem 38 Every algebraic extension is profinite. Proof. Let α : K → L be an algebraic extension. Let I denote the set of finite subsets of L, ordered by inclusion, and observe that I is directed. For each X ∈ I, let αX : K → K(X) denote the extension given in Definition 31. Since X is finite and each u ∈ X is algebraic, the extension αX is finite, by Proposition 33. If X, Y ∈ I and X ⊆ Y , let αX⊆Y : K(X) → K(Y ) denote the obvious map. The collection {αX⊆Y : K(X) → K(Y )}X,Y ∈I is a direct system (over I) in the category of extensions of K. Let αI : K → KI denote the direct limit of this system over I and recall that this comes equipped with maps αX : KX → KI of extensions of K. 3> LL N βX

∃!θ

6 KI mm{m{= N m m m α mm { mm {{{ Y mmm { α m { m m KX bEαX⊆Y / KO Y αI EE EE αY αX EEE K X

βY α

Now, for each X ∈ I there is a unique β X : K(X) → L (inclusion) such that α = β X ◦ αX . Accordingly, by the universal property of KI (see Proposition 11) there is a unique map θ : KI → L of extensions of K such that β X = θ ◦ αX for each X ∈ I. We claim that this is an isomorphism. Define θ 0 : L → KI u 7→ α{u} (u) = [u] Then, for each u ∈ L, θ ◦ θ0 (u) = θ(α{u} (u)) = θ ◦ α{u} (u) = β {u} (u) =u and

θ0 ◦ θ([u]) = θ0 ◦ θ(α{u} (u)) = θ0 ◦ θ ◦ α{u} (u) = θ0 ◦ β {u} (u) = θ0 (u) = [u]

DRAFT: April 6, 2007

29

so θ0 = θ−1 . This completes the proof of Theorem 38. Corollary 39 Every Galois extension is a profinite extension. Proof. Every Galois extension is an algebraic extension (see Definition 19) and every algebraic extension is profinite by Theorem 38. Proposition 40 Let K → K be a field homomorphism. The set of elements of L which are algebraic over K form a subfield of L. Proof. Suppose u and v are elements of L which are algebraic over K. The extension K → K(u, v) is finite by Proposition 33. Since K(u + v) is a subfield of K(u, v), K → K(u + v) is finite by Proposition 22. Thus, K → K(u + v) is algebraic, by Proposition 24. Thus, u + v is algebraic over K, by Proposition 28. Thus, the set of elements of L which are algebraic over K is closed under addition. A similar argument shows that this set is also closed under multiplication. LO K(u, v) jUU eLLL UUU jjt5 t: j j j LLL UUUUU jj ttt j UUUU j LLL j j t j t UUUU j j L tt UU jjjj K(u + v) K(u) jTT K(uv) i4 K(v) TTTT 8 eKK r iiii i r i TTTT KKK r i rr TTTT KK iiii TTTTKKK rririiiiii r r i TT riii K It only remains to show that the set of non-zero elements of L which are algebraic over Kis closed under inversion x 7→ x1 . To see this, suppose u ∈ L× is algebraic over K. Define fu−1 ,K ∈ K[x]× by fu−1 ,K (x) = xdegK K(u) fu,K (x−1 ). (A priori, fu−1 ,K ∈ K(x), but a moments thought will show you that fu−1 ,K is indeed a polynomial.) Since fu−1 ,K is non-zero and αx (fu−1 ,K )(u−1 ) = 0, it follows that u−1 is algebraic over K.

2.6

Finite subgroups are Galois

At last, our first theorem on Galois subgroups: all finite subgroups of Aut(L) are Galois subgroups! The proof will require several lemmas, which are rather delightful in the own right. Lemma 41 (Dedekind) Let G be a group and let K be a field. The set Homgroups (G, K × ) is linearly independent in the K-vector space of functions Homsets (G, K).

30

DRAFT : April 6, 2007

Proof. Suppose the lemma is false. Thus, there is a minimal finite set {χ1 , . . . , χn } of (distinct) characters of G such that n X

ai χi = 0,

(2.1)

i=1

where not all ai are 0. Since χ1 6= χ2 there is some h ∈ G such that χ1 (h) 6= χ2 (h). Evaluate the equation above at arbitrary g and multiply by χ1 (h), then evaluate the equation above at hg, and subtract: a1 χ1 (h)χ1 (g) + a1 χ1 (h)χ1 (g) +

n X i=2 n X

ai χ1 (h)χi (g) = 0 ai χi (h)χi (g) = 0

i=2 n X

ai (χ1 (h) − χi (h))χi (g) = 0

i=2

Since this clearly contradicts the minimality of the set {χ1 , . . . , χn }, this completes the proof of the lemma. Lemma 42 If K → L is a finite extension then AutK (L) ≤ dimK (L). Proof. We have already seen that if K → L is finite then AutK (L) is finite (see Proposition 35). Write AutK (L) = {σ1 , . . . , σm } and consider the matrix 

σ1 (u1 )  σ2 (u1 )  A= .  ..

σ1 (u2 ) σ2 (u2 ) .. .

··· ··· .. .

σm (u1 ) σm (u2 ) · · ·

 σ1 (un ) σ2 (un )   ..  .  σm (un )

The lemma claims that m ≤ n, so, for a contradiction, suppose n < m. Then the m rank of A is at most n, so the rows are Pmlinearly dependent in L . Consequently, there are a1 , . . . , am ∈ L such that i=1 ai σi (uj ) = 0 for all 1 ≤ j ≤ n and the ai arePnot all 0. Since each σi is determined by its values at the uj , it follows m that i=1 ai σi = 0. Now, observe that L× is a group and that each σ ∈ AutK (L) restricts to a group homomorphism σ|L× : L× → L× , which is a character. Since these characters are all distinct, it follows that set {σ1 |L× , . . . , σm |L× } is linearly independent, by Dedekind’s Lemma. But, from the paragraph above we have Pm × a σ | i i L = 0. This contradiction proves the lemma. i=1 Theorem 43 (Dedekind-Artin) Let L be a field. If H is a finite subgroup of Aut(L) then H is Galois, and dimLH (L) = H .

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31

Proof. Recall that a subgroup H of Aut(L) is Galois if H = GALL FIXL (H) and H = GALL (K) for some Galois extension K → L. Let K = LH . Then K → L is finite and Aut K (L) is finite, by Propo AutK (L) ≥ H . Now, sition ??. Since H ⊂ Aut (L), it follows that K dimK (L) ≥ Aut K (L) by Lemma 42, so dimK (L) ≥ H . For a contradiction, suppose H < dimK (L). Let n = H and write H = {σ1 , . . . , σn }. Since dimK (L) > n there exists a set {u1 , . . . un+1 } of elements from L which are linearly independent over K. Consider the matrix   σ1 (u1 ) σ1 (u2 ) · · · σ1 (un+1 )  σ2 (u1 ) σ2 (u2 ) · · · σ2 (un+1 )    A= .  .. .. ..   .. . . . σn (u1 ) σn (u2 ) · · · σn (un+1 ) Then the columns of A are linearly dependent in Ln . Let k be the cardinality of the smallest set of linearly dependent columns of A. Without loss of generality, we may assume the first k columns of A are linearly dependent. Thus, there are ai ∈ L such that k X ai σj (ui ) = 0. ∀1 ≤ j ≤ n, i=1

Without loss of generality we mayPassume a1 = 1. Now, if eachPcoefficient ai k k were in K, then we would have σj ( i=1 ai ui ) = 0, in which case i=1 ai ui = 0. Of course, this is impossible since {u1 , . . . un+1 } are linearly independent over K. Accordingly, for some 1 ≤ i ≤ k, ai is not in K. Now, pick σ ∈ H. Applying σ to the displayed equation above gives ∀1 ≤ j 0 ≤ n,

k X

σ(ai )σj 0 (ui ) = 0.

i=1

(Here we use the fact that there is a permutation j 7→ j 0 of n such that σ ◦ σj = σj 0 for each 1 ≤ j ≤ n.) Subtracting these equations gives ∀1 ≤ j ≤ n,

k X

(ai − σ(ai ))σj (ui ) = 0.

i=2

(Here we use the fact that a1 = 1 and σ(1) = 1.) Minimality of k implies ai = σ(ai ) for each 1 ≤ i ≤ k. Since σ ∈ H was arbitrary, we have ai ∈ LH = K for each i, which contradicts the conclusion of the paragraph above. This contradication completes the proof of the proposition. Theorem 44 A finite extension K → L is Galois if and only if AutK (L) = dimK (L). Proof. Suppose K → L is a finite Galois extension. Then Gal(L/K) is finite 35). Since K = FIXL GALL K, it follows from Theorem 43 that (by Proposition Gal(L/K) = dimK (L).

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Conversely, suppose K → L is a finite extension and AutK (L) = dimK (L). Let M = FIXL GALL (K). Since AutK (L) is finite it follows that GALL (K) = GALL FIXL (K), by Theorem 43. Let M = FIXL (K). Then AutK (L) = AutM (L). Since AutK (L) = dimK (L) by hypothesis and since AutM (L) = dimM (L) by Theorem 43, it follows that dimK (L) = dimM (L), in which case M = K. But now, K = FIXL GALL (K) so K → L is Galois.

2.7

Splitting extensions

Definition 45 Let K be a field and f a non-zero element of K[x]. We say f splits over α : K → M if αx (f ) factors into linear polynomials over M [x]. If S ⊆ K[x]× and each f ∈ S splits over α, we say S splits over α. Proposition 46 Suppose f ∈ K[x]× splits over α : K → M . Let K → N denote the extension of K obtained by adjoining the roots of αx (f ) ∈ M [x] to K and denote this extension by αN (so αN is exactly α with image restricted to N ). Suppose f also splits over β : K → L. Then there is a unique θ : N → L such that β = θ ◦ αN . = MO 8/ L ppp I p p pp α1 pppθ1 p p p

N GO α αN

θ

NO 1

α1

β

K Proof. Write αx (f ) = c(x − u1 ) · · · (x − udeg(f ) ) where u1 , . . . , udeg(f ) ∈ M . Thus, f splits over α : K → K(u1 , . . . , udeg(f ) ). (Since the coefficient of xdeg(f ) is c and f ∈ K[x], it follows that c = α(a) for a unique a ∈ K[x].) In other words, N = K(u1 , . . . , udeg(f ) ). Now, suppose f also splits over β : K → L. Suppose deg(f ) = 1. Then N = K(u1 ) = α(K). Thus αN is an isomor−1 phism. Let θ = β ◦ αN . Clearly θ ◦ α = β and θ is unique with this property. Suppose Proposition 46 is true for all f ∈ K[x] with deg(f ) < n. Observe that u1 ∈ M is algebraic over K and fu1 ,K |f . Since f splits in N and L, the same is true of fu1 ,K . Without loss of generality, we assume that the roots of f1 := fu1 ,K in N are {u1 , u2 , . . . , uk }. We also assume (without loss of generality) that u1 is not contained in α(K), in which case the degree of f1 , which is k, is strictly less than n. Set N1 = K(u1 , u2 , . . . , uk ) and let α1 : K → N1 be the obvious extension. By the inductive hypothesis, there is a unique θ1 : N1 → L such that β = θ1 ◦ α1 . Now, let f 1 = αx (f )/f1 and observe that f 1 ∈ N1 [x]. Let α1 : N1 → N be the obvious map. Clearly f 1 splits over N and (θ1 )x (f 1 ) splits

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33

over L. Since the degree of f 1 is strictly less than n it follows from the inductive hypothesis again that there is a unique θ : N → L such that θ1 = θ ◦ α1 . Precomposing with α1 yields θ1 ◦ α1 = θ ◦ α1 ◦ α1 . Since θ1 ◦ α1 = β and since α1 ◦ α1 = αN it follows that β = θ ◦ α. It is fairly clear from this construction that θ is unique with this property, but just to be clear, suppose θ0 : N → L has the property β = θ0 ◦ α. With N1 and α1 as above, define θ10 := θ0 ◦ α1 . Applying the induction hypothesis to the bottom triangle (in the diagramme above), we see that θ10 = θ1 ; then, applying the induction hypothesis to the top triangle we see that θ0 = θ. Definition 47 An extension α : K → N is a splitting extension for S ⊆ K[x]× if: S splits over α : K → N ; and if S splits over β : K → L then there is a unique θ : N → L such that β = θ ◦ α. When S is a singleton {f }, we say α : K → N is a splitting extension for f . If S is finite then, without changing the splitting extension, we can replace S with the singleton {f } where f is the product of all the polynomials in S. In other words, if S is finite we can assume, without loss of generality, that it is a singleton. Lemma 48 If α : K → N is a splitting extension for f ∈ K[x]× . Then α is a finite extension, and thus algebraic. Proof. Now, let α : K → L be any splitting extension for f ∈ K[x]× . Then L = K(u1 , . . . , un ), where each ui is a root of f . Since each ui is algebraic, it follows from Proposition 33 that K → N is finite (and thus algebraic). Lemma 49 Suppose S ⊆ T ⊆ K[x]× . Let α : K → N be a splitting extension for S and let β : K → M be a splitting extension for T . Then there is a unique θ : N → M such that β = θ ◦ α. Proof. Klar. Proposition 50 Splitting fields are unique, up to isomorphism; more precisely, if α : K → N and β : K → M are splitting extensions for S ⊆ K[x]× , then there is a unique θ : N → M such that β = θ ◦ α. Proof. Since α is a splitting extension for S and S splits over β, it follows from Definition 47 that there is a unique θ : N → M such that β = θ ◦ α; likewise, since β is a splitting extension for S and S splits over α, there is a unique θ0 : M → N such that α = θ0 ◦ β. Since, θ0 ◦ θ : N → N and idN both make the triangle below commute, it follows from the universal property appearing in Definition 47 that θ0 ◦ θ = idN . By similar arguments, θ ◦ θ0 = idM , so θ is an isomorphism. θ 0 ◦θ /N N `A }> AA } } AA }}α α AA } } K

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Of course, all this begs the question: given f ∈ K[x]× , does a splitting field for f even exist? Needless to say, the answer is ‘yes’. We start with the case of finite extensions.

Lemma 51 If f ∈ K[x]× then there is a splitting field extension K → N for f , and for any splitting extention K → N of f , dimK (N ) ≤ deg(f )!. Proof. Suppose f ∈ K[x]× is irreducible. Then the ideal (f ) generated by f is maximal. Let N = K[x]/(f ) and let α : K → N be the obvious field homomorphism. Then α : K → N is finite and dimK (N ) = deg(f ), so the lemma is proved in this case. Suppose f ∈ K[x]× is not irreducible. Since K[x] is a UFD and a PID, there is some irreducible f1 ∈ K[x] with divides f in K[x]. Let N1 = K[x]/(f1 ) and let α1 : K → N1 be the obvious homomorphism; observe that degK (N1 ) ≤ deg(f ). Let f 1 = f /f1 and let f2 be an irreducible factor for (α1 )x (f 1 ) ∈ N1 [x]. Let N2 = N1 [x]/(f2 ) and let α2 : N1 → N2 be the obvious field homomorphism; observe that degN1 (N2 ) ≤ deg(f ) − 1. This process necessarily terminates with some field, say N , and we define α : K → N by α := · · · α2 ◦ α1 and observe that dimK (N ) ≤ deg(f )!, as promised. [It remains to show that α has the universal property appearing in Definition 47.]

Proposition 52 If S ⊆ K[x]× then there is a splitting field extension α : K → N for S.

Proof. Let I denote the set of all finite products of elements of S and observe that I is partially ordered by f ≤ g ⇐⇒ f |g. In fact, I is a directed set, since f, g ∈ I implies f ≤ f g and g ≤ f g. Now, for every f ∈ I, let αf : K → Nf be a splitting extension for f ∈ K[x]× . (The existence is such an extension was established in Lemma 51.) If f ≤ g in I then g = f h for some h ∈ S so by Lemma 46 there is a unique θ : Nf → Ng such that αg = θ ◦ αf . We denote this map of extensions of K by αf ≤g . We claim that (αf ≤g )f,g∈I is a direct system in the category of extensions of K. Since αf = idNf ◦ αf and αf ≤f is unique with this property (see Definition 47), it follows that αf ≤f = idNf , which is certainly a map of extensions of K. idNf

Nf O αf

K

% / Nf = {{ {{ { { αf {{ αf ≤f

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35

Now, suppose f ≤ g ≤ k and consider the diagramme below. αf ≤k

Nf O αf

K

αf ≤g

α

/ Ng g≤h /6 Nk = mm | αg || mmm m m | m || mmm αk |m|mmm

Again, from the uniqueness part of Definition 47 it follows that αf ≤k = αg≤k ◦ αf ≤g . Note too that these are all extensions of K. This completes the demonstration that (αf ≤g )f,g∈I is a direct system of extensions of K. Now, set N := − lim N ; more precisely, let α : K → N be the direct limit −→ f f ∈I

of the direct system (αf ≤g )f,g∈I of extensions of K (refer to Section 1.6 for the definition of a direct limit). Observe that N comes equipped with maps αf : Nf → N for each f ∈ I. Moreover, for each f ∈ I, αf ◦ αf : K → N is independent of f . This defines α : K → N . We claim that α : K → N is a splitting field field for S. To see this, we must verify the universal property appearing in Definition 47. [details omitted]

2.8

Algebraic closure

One splitting field is particularly important: Definition 53 A splitting extension for S = K[x]× is called an algebraic clo¯ sure for K and often denoted K → K. ¯ is defined up to isomorphism only, since for It is important to observe that K × each f ∈ K[x] we chose a splitting extension for f . However, Proposition 52 ¯ does exist, for each field K. This alone has shown that an algebraic closure K important consequences, as we shall see shortly. The following lemma justifies the use of the word algebraic in this context. Proposition 54 Every algebraic closure is algebraic. ¯ be an algebraic closure for K. Pick x ∈ K. ¯ Proof. Let K be a field and let K Then there is some f ∈ K[x]× , some splitting extension αf : K → Kf and some u ∈ Kf such that x = αf (u) (the equivalence class of u, see the proof of Proposition 52). Now, αf : K → Kf is finite, and hence algebraic, by Lemma 48. Let fu,K ∈ K[x]× be the minimal polynomial for u ∈ Kf . Then αx (fu,K )(x) = (αf )x (αf )x (fu,K )(αf (u)) = αf ((αf )x (fu,K )(u)) = αf (0) = 0.

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This shows that x is algebraic over K. Since x was arbitrary, it follows that the ¯ is algebraic. extension α : K → K Corollary 55 Every splitting extension is an algebraic extension. ¯ be an Proof. Let α : K → N be a splitting extension and let β : K → K algebraic closure. From Definition 53 and Lemma 49 it follows that there is ¯ such that β = θ ◦ α. Proposition 54 shows that β is a (unique) θ : N → K algebraic and Lemma 18 shows that α is algebraic. Proposition 56 Let α : K → L be an algebraic homomorphism and let β : ¯ be an algebraic closure of K. Then there is a unique field homomorK →K ¯ such that β = θ ◦ α. phism θ : L → K Proof. Recall from the proof of Theorem 38 that L = −lim K(X), where I −→ X∈I

denotes the set of finite subsets of L; more precisely, recall the definition of the direct system {αX⊆Y : K(X) → K(Y )}X,Y ∈I of extensions αX : K → K(X) of K and that α : K → L may be identified with the direct limit of this direct system in the category of extensions of K. Recall also that L comes equipped with maps αX : K(X) → L. From Proposition 46 it follows that, for each X ∈ I ¯ such that β = θX ◦ αX ; moreover, it follows from the there is a θX : K(X) → K universal property of θX promised in Proposition 46 that if X ≤ Y in I then the following diagramme commutes. ¯ K E O Y33i  33  33  θ  33 θ  θX  Y < L bEE 333  y  y E  yy EE 33 EE 3  yyy X Y  E3 y α α y / K(Y ) K(X) αX≤Y bEE y< EEαX αY yy EE y y EE y yy K

β

¯ such that β = θ ◦ α, as claimed. Consequently, there is a unique θ : L → K We are now able re-visit a question that arose in Chapter 1 concerning the existence of push-outs in the category of fields. Proposition 57 Suppose α : K → N and β : K → M are algebraic extensions. Then a push-out (α0 , β 0 ) of (α, β) exists. Moreover, for every push-out, both α0 : M → L and β 0 : N → L are algebraic extensions.

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37

α ¯ γ ¯

NO

β0

/L O

β¯

α0

α

K

/K ¯ > N

β

/M

¯ and β¯ : M → K ¯ as objects of the Proof. Use Proposition 56 to view α ¯:N →K ¯ let γ¯ : L → K ¯ be the co-product of α category sub(K); ¯ and β¯ in that category ¯ comes (which exists by Proposition 10). Recall from Section 1.5 that γ¯ : L → K equipped with maps α0 : N → L and β 0 : M → L. The universal property of co-products guarantees that (α0 , β 0 ) is a push-out of (α, β). Finally, it follows directly from Proposition 36 that α0 andβ 0 are algebraic.

2.9

Normal extensions

Definition 58 An algebraic extension K → N is normal if fu,K splits over N for each u ∈ N . Proposition 59 An extension α : K → N is normal if and only if it is a splitting field for some S ⊆ K[x]× . Likewise, α : K → N is normal if and only if, for every irreducible f ∈ K[x], either αx (f ) ∈ N [x] has no roots or αx (f ) splits completely in L[x]. Proof. (under construction) Proposition 60 If α : K → N is a normal extension and if α = β ◦ γ, then β is a normal extension. Proof. (mid-term exercise) By contrast, if α is normal and α = β ◦ γ, it does not follow that γ is normal. (mid-term exercise) Theorem 61 (Isomorphism Extension Theorem) Suppose α : K → N is normal and τ : K → M is an isomorphism. Let (α0 , σ) be a push-out of (α, τ ). Then α0 : M → L is normal and σ : N → L is an isomorphism. Proof. Since push-outs are unique up to isomorphism (exercise), to prove the theorem it is enough to construct a push-out with the properties claimed. Since normal extensions are splitting extensions by Proposition 59, α is a splitting extension for some S ⊆ K[x]× . Let T = β(S) and let α0 : M → L be a splitting extension for T ⊆ M [x]× . Since S splits over α0 ◦ τ there is a unique σ : N → L such that α0 ◦ τ = σ ◦ α and since T splits over α ◦ τ −1 there is a unique σ 0 : L → N such that α ◦ τ −1 = σ 0 ◦ α0 . By the universal property of splitting extensions, σ is an isomorphism (argue as in the proof of Proposition 50) and (α0 , σ) is a push-out of (α, τ ).

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Theorem 62 If α : K → N is normal and u ∈ N , then for every root v of fu,K in N there is some σ ∈ AutK (N ) such that σ(u) = v. Proof. Use Proposition 59 to work with splitting extensions. Then, prove the theorem for finite extensions using induction and Lemma 30. Then, prove the theorem for general splitting extensions using the direct limit construction appearing in Propositions 52 and 50. Details omitted.

2.10

Restriction

Suppose α : K → L and β : N → L are normal extensions and consider the morphism from α to β in sub(L) given by the commuting triangle (on the lefthand side) below. In this section we show the following: if γ : K → N is a normal extension then AutN (L) is a normal subgroup of AutK (L), and the quotient of AutK (L) by AutN (L) is AutK (N ). Lemma 63 Suppose β : N → L is normal and σ : L → L is an isomorphism. Then there is a unique isomorphism σ|N : N → N such that (β, σ|N ) is a pull-back of (β, σ). Proof. (exercise) = LO

σ

β α

NO

β

σ|N

γ

K

/L O a /N O

α

γ

idK

/K

Theorem 64 (Restriction Theorem) Suppose α : K → L and γ : K → N are normal and α = β ◦ γ. Then 1

/ AutN (L)

/ AutK (L) σ7→σ|N/ AutK (N )

/1

is a short exact sequence. Proof. Use Lemma 63 and Theorem 61. Corollary 65 Let E : F be a Galois extension and suppose K is a Galois extension of F contained in E. Then [K : F ] = [GalE : F : GalE : K].

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39

Proof. If K : F is finite then [K : F ] equals the cardinality of GalK : F , by [?, Th’m 10.4]. By Proposition ?? and the Isomorphism Theorem for Groups we have GalK : F ∼ = GalE : F /GalE : K, so [K : F ] = [GalE : F : GalE : K] as desired. On the other hand, if K : F is not finite then [K : F ] is not finite. If the index [GalE : F : GalE : K] were finite, then GalK : F would be finite, by Proposition ??. But in that case, K : K GalK:F would be finite by [?, Th’m 11.3], so K : F would be finite, by Lemma ??, which is the desired contradiction. This concludes the proof of Corollary 65. ¯ be an algebraic closure Proposition 66 Suppose K → L is algebraic and let L ¯ → AutK (L) is of L. Then K → L is normal if and only if restriction AutK (L) defined. Proof. (exercise)

2.11

Separable extensions

Definition 67 An irreducible polynomial f ∈ K[x] is separable if it has no repeated roots in any splitting field for f over K. An arbitrary polynomial is separable if its irreducible factors are separable; otherwise, it is inseparable. An extension K → L is separable if it is algebraic and fu,K ∈ K[x] is separable for each u ∈ L. Remark 68 Note that the second part of this definition makes use of the UFD property of K[x]. Lemma 69 If f ∈ K[x]× and gcd(f, f 0 ) = 1 then f is separable. If f ∈ K[x]× is irreducible and separable then gcd(f, f 0 ) = 1. Proof. (exercise) Exercise 70 Show that if the characteristic of K is 0, then every f ∈ K[x] is separable. Exercise 71 Find an example of an inseparable polynomial. Exercise 72 If K → N is normal and separable and u ∈ N , then the splitting field M for fu,K in N is simple and the roots of fu,K in N form a basis for M over K. Proposition 73 A normal extension is separable if and only if it is a splitting field for separable polynomials. Moreover, any extension which is normal and separable is a direct limit of finite, normal and separable extensions. Proof. Klar

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Proposition 74 If α is separable and α = β ◦ γ, then β and γ are separable. Proof. (mid-term exercise)

Add:

Separable closure K → K sep

2.12

Galois extensions

In this Section we prove one of the main results of the course: an extension is Galois if and only if it is normal and separable. Lemma 75 Any extension which is finite, normal and separable is also Galois. Proof. Suppose α : K → N is finite, normal and separable; we will prove that K → N is Galois, by induction on dimK (N ). First, suppose dimK (N ) = 1. Then α : K → N is an isomorphism, which is Galois by Lemma 20. Next, suppose dimK (N ) > 1 and suppose the lemma is true for all extensions of degree less than dimK (N ). Since K → N is normal and separable and finite, it is the splitting field for a separable polynomial f in K[x], by Proposition 73. Let u be a root of αx (f ) which is not in α(K). By Lemma 72, the splitting field M for fu,K in N is a simple extension and the roots {u1 , . . . , un } of fu,K in N form a basis for M over K. Let β : M → N and γ : K → M be the obvious maps. By Propositions 22, 60 and 74, β : M → N is finite, normal and separable; thus, β is Galois by the induction hypothesis. Since γ : K → M is the splitting field for fu,K and since fu,K is separable (because K → N is separable), it follows that γ : K → M is also finite, normal and separable, and thus Galois by the inductive hypothesis. For each 1 ≤ i ≤ k, let τi be the automorphism in AutK (M ) such that τi (u) = ui , which exists by Proposition 29. By Theorems 61 and 62, for each 1 ≤ i ≤ k there is some σi ∈ AutK (N ) such that σi (u) = ui > NO

σi

β

β α

MO γ

K

/N O

/M = | γ || | || || τi

Now, consider the subgroup AutM (N ) → AutK (N ) and the cosets  σi AutM (N ) 1 ≤ i ≤ k ⊆ AutK (N )/AutM (N ). where σi is given by an application of Lemma 5. If σi AutM (N ) were equal to σj AutM (N ) then σi−1 ◦ σj ∈ AutM (N ), in which case σi−1 ◦ σj (u) = σi−1 (uj ) =

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41

u, whence σi (u) = uj and this is only possible if i = j; consequently, the cosets {σ1 AutM (N ), . . . , σk AutM (N )} are all distinct, so the coset space above contains at least k elements. Therefore, AutK (N ) definition of index = [AutK (N ) : AutM (N )] AutM (N ) work above ≥ k AutM (N ) ≥ k dimM (N ) M → N is Galois = dimK (M ) dimM (N ) Lemma 72 and Proposition 28 = dimK (N ) Proposition 22 Now, AutK (N ) ≥ dim K (N ). Since AutK (N ) ≤ dimK (N ) by Lemma 42, it follows that AutK (N ) = dimK (N ). Thus, AutK (N ) is Galois, by Theorem 44. Lemma 76 Any extension which is normal and separable is also Galois. Proof. Suppose α : K → L is normal and separable. Then, up to isomorphism, α is a direct limit of finite, normal separable extensions, each of which is Galois by Lemma 75. Suppose u ∈ LAutK (L) . Then σ(u) = u for all σ ∈ AutK (L). Since K → L is normal and separable, there is some finite, normal and separable K → N such that u ∈ N , by Proposition 73. Then σ|N (u) = u for all σ ∈ AutK (L). Since restriction to a normal extension is surjective by Theorem 64, it follows that τ (u) = u for each τ ∈ AutK (N ); in other words, u ∈ N AutK (N ) . Since K → N is finite, normal and separable it is Galois by Lemma 75, so N AutK (N ) = K. Thus, u ∈ K. Since this argument applies to every u ∈ L, it follows that LAutK (L) = K, whence K → L is Galois. Lemma 77 Any extension which is Galois is also normal and separable. Proof. Suppose α : K → N is Galois. We will show that fu,K is a separable polynomial which splits over N , for every u ∈ K, whence K → N is separable and normal. Pick u ∈ N and onsider the orbit  AutK (N )u = σ(u) σ ∈ AutK (N ) of u under the action of AutK (N ) on N . 1 This is a finite set: if σ ∈ AutK (N ) then αx (fu,K )(σ(u)) = σ(αx (fu,K (u)) = σ(0) = 0, so σ(u) is a root of fu,K in N , of which there are a finite number. Note that we have made implicit use of the fact that u is algebraic over K, since α : K → N is algebraic. Now, consider the polynomial fu ∈ N [x] defined by Y gu = (x − v). v∈AutK (N )u 1 [Add

a comment on group actions, and explain that the trick below used to construct a separable polynomial below is general and important.]

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Notice that gu is the product of the distinct linear factors of αx (fu,K ). Since any σ ∈ AutK (N ) permutes the set AutK (N )u, it follows that σ(gu ) = gu , so the coefficients of gu are in the field N AutK (N ) . Since K → N is Galois, it follows that N AutK (N ) = αN , so gu ∈ α(K)[x]. Let fu be the unique polynomial in K[x] such that αx (fu ) = gu . Now, αx (fu )(u) = 0, so fu,K divides fu ; since the degree of fu is at most that of fu,K , and since both polynomials are monic, it follows that fu = fu,K . Thus, fu,K is separable. Since u was arbitrary, fu,K is separable for every u ∈ N × , so K → N is a separable extension, by Definition 67. The fact that fu = fu,K also shows that fu,K splits over N , for each u ∈ N × ; since K → N is Galois, and hence algebraic, it follows that K → N is normal, by Proposition 59. This shows that any extension which is Galois is also normal and separable. Theorem 78 (The ‘Galois is Normal and Separable’ Theorem) An extension is Galois if and only if it is normal and separable. Proof. Simply assemble Lemmas 75, 76 and 77. Corollary 79 The separable closure K → K sep is a galois extension.

Combine the lemma and proposition, then move both:

Lemma 80 Let K → L be an extension and let H be a subgroup of Aut(L). Then −1 ∀σ ∈ AutK (L), σ(LH ) = LσHσ . −1

Proof. Kix σ ∈ AutK (L). Suppose u ∈ LσHσ . Then στ σ −1 (u) = u for all τ ∈ H. Let v = σ −1 (u). Then τ (v) = v for all τ ∈ H, so v ∈ LH . Since −1 σ(v) = u, it follows that u ∈ σ(LH ), whence LσHσ ⊆ σ(LH ). Conversely, suppose u ∈ σ(LH ). Then σ −1 (u) ∈ LH . Let v = σ −1 (u). Then τ σ −1 (u) = −1 σ −1 (u) for all τ ∈ H so στ σ −1 (u) = u for all τ ∈ H. Thus, u ∈ LσHσ , so −1 σ(LH ) ⊆ LσHσ . This concludes the proof of Lemma 80. Theorem 81 Suppose K → L and N → L are galois. If AutN (L) is normal in AutK (L) then K → N is normal. Proof. Suppose AutN (L) is a normal subgroup of AutK (L). Using Lemma 80 we see that σ(LAutN (L) ) = LAutN (L) for all σ ∈ AutK (L). Thus, K → LAutN (L) is normal. Since K → L is Galois it follows immediately that N → L is Galois, so LAutN (L) = N , by definition. Thus, K → N is normal.

Chapter 3

The Fundamental Theorem Until now we have studied the group AutK (L) of automorphism associated to a homomorphism K → L of fields; this is not the Galois group of K → L. Rather, the Galois group of the field homomorphism K → L is a topological group, obtained by equipping AutK (L) with the Krull (or profinite) topology, which is defined below; this topological group will be denoted Gal(L/K).

3.1

Galois groups are profinite

In Chapter 2 we saw that every Galois extension is profinite (see Definition 37) and the proof of Theorem 78 may have suggested to you that Galois groups are profinite groups (see Definition 15). This is indeed true, as we now show. Theorem 82 Every Galois group is a profinite group; more precisely, for any Galois extension K → L, AutK (lim L )∼ lim AutK (Lf ), =← −−→ f −− f ∈I

f ∈I

where I denotes the directed poset of separable polynomials in K[x]× which split over L and where Lf denotes the splitting field for f in L, for each f ∈ I. The direct limit is taken in the category of extensions of K and the inverse limit is taken in the category of subgroups of Aut(L). Proof. Suppose α : K → L is Galois. Let I denote the directed poset of separable polynomials in K[x]× which split over L; thus, α is a splitting extension for I. For each f ∈ I, let αf : K → Lf , αf : Lf → L and αf ≤g : Lf → Lg be as in the proof of Proposition 52. For each f ∈ I, let %f : AutK (L) → AutK (Lf ) denote the restriction from K → L to K → Lf map (see Section 2.10); likewise, for each f ≤ g in I let %f ≤g : AutK (Lg ) → AutK (Lf ) denote the restriction from K → Lg to K → Lf . Then %f ≤g ◦ %g = %f for each f ≤ g in I. It now follows from the universal property of the inverse limit (see Definition 12) 43

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that there is a unique map θ : AutK (L) → ← lim AutK (Lf ) making the following −− f ∈I

diagramme commute, AutK (L) ∃!θ

x

%f

lim AutK (Lf ) ←−− f ∈I h h h pp ϕf hhhhh hhh pϕpgp h h p %g h p h h p h x thhhh xpp AutK (Lf ) o %f ≤g AutK (Lg ) q where ϕf : ← lim AutK (Lf ) → AutK (Lf ) is defined by ϕ((σf )f ∈I ) = σf . In fact, −− f ∈I

θ admits a very simple description: AutK (L) → ← lim AutK (Lf ) −− f ∈I

σ 7→ (σ|Lf )f ∈I . To see that θ is an isomorphism, we exhibit its inverse θ0 : ← lim AutK (Lf ) → −− f ∈I

AutK (L), which is defined as follows. Recall that L ∼ lim L in the category of =− −→ f f ∈I

extensions of K. Thus, for each u ∈ L there is some f ∈ I such that u ∈ Lf ; in this case define θ0 ((σf )f ∈I )(u) = σf (u). It remains to verify that this is 0 well-defined and that θ ◦ θ0 = id← lim AutK (L) and θ ◦ θ = idAutK (L) . − − f ∈I

Corollary 83 For every field K, AutK (K sep ) ∼ lim AutK (Kf ), =← −− f ∈I

with notation as above.

3.2

The Krull topology

Definition 84 Let K → L be Galois. The Krull topology for AutK (L) is the coarsest topology for AutK (L) such that if f ∈ K[x]× is separable and splits over K → L, then the restriction map %f : AutK (L) → AutK (Lf ) is continuous, where AutK (Lf ) is given the discrete topology. Lemma 85 If K → L is galois then the Krull topology for AutK (L) is characterized by the following statement: a neighbourhood base at σ ∈ AutK (L) is  σAutLf (L) f ∈ I , where I denotes the set of all separable polynomials in K[x]× which split over K → L (so Lf ranges over all finite Galois extensions of K contained in L).

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45

Proof. Suppose K → L is Galois and σ ∈ AutK (L). Definition 84 says that the Krull topology for AutK (L) is generated by sets of the form %−1 f (τ ), where f ranges over all separable polynomials in K[x]× which split over L and where τ ∈ AutK (Lf ). By the Restriction Theorem (Theorem 64), there is some σ ∈ AutK (L) such that %−1 f (τ ) = σAutLf (L). Thus, a neighbourhood base at σ ∈ AutK (L) is given by sets of the form σAutLf (L), with f as above. Proposition 86 If K → L is galois then AutK (L), equipped with the Krull topology, is a topological group. Q Proof. Let m : AutK (L) AutK (L) → AutK (L) be multiplication and let i : AutK (L) → AutK (L) be inversion. We must prove that these functions are continuous. Since multiplication m is a group homomorphism, in order to show that m is continuous it is sufficient to show that m is continuous at the identity 1L ∈ AutK (L). In light that m−1 (AutN (L)) Q of Lemma 85 is it sufficient to show−1 is open in AutK (L) AutK (L). Suppose (σ1 , σ2 ) ∈ m (AutN (L)). Then σ1 ◦ σ2 ∈ AutN (L); in other words, %N (σ1 ◦ σ2 ) = idN , where %N denotes restriction in the sense of section 2.10, from which we recall that %N is a group homomorphism. Now, consider n o Y U := (σ1 ◦ σ10 , σ2 ◦ σ20 ) ∈ AutK (L) AutK (L) σ 0 , τ 0 ∈ AutN (L) . This is an open neighbourhood of (σ1 , σ2 ) in AutK (L)

Q

AutK (L). Moreover,

%N ◦ m(σ1 ◦ σ10 , σ2 ◦ σ20 ) = %N (σ1 ◦ σ10 ◦ σ2 ◦ σ20 ) = %N (σ1 ) ◦ ϕN (σ10 ) ◦ ϕN (σ2 ) ◦ ϕN (σ20 ) = %N (σ1 ) ◦ idN ◦ ϕN (σ2 ) ◦ idN = %N (σ1 ◦ σ2 ) = idN . Thus, (σ1 , σ2 ) ∈ U ⊆ m−1 (AutN (L)). This completes the proof that m is continuous. In order to show that inversion i : Gal(L/K) → Gal(L/K) is continuous, suppose σ ∈ Gal(L/N ) and let N be a normal extension of K contained in L. If σ1 ∈ i−1 (σGal(L/N )) then σ1−1 ∈ σGal(L/N ) so σ1−1 = σ ◦ τ for some τ ∈ Gal(L/N ); equivalently, σ1 = τ −1 ◦σ −1 , so σ1 ∈ Gal(L/N )σ −1 . Since K → N is normal, Gal(L/N )σ −1 = σ −1 Gal(L/N ), so i−1 (σGal(L/N )) = σ −1 Gal(L/N ). Since this is open in Gal(L/K), it follows that inversion i is continuous.

3.3

Galois (topological) groups

Definition 87 If K → L is Galois then Gal(L/K) denotes the topological group AutK (L) equipped with the Krull topology.

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Notice that we have just passed from the category of groups to the category of topological groups! We will have more to say about this in Section 3.6; in the meantime we re-visit Theorems 64 and 82. Theorem 88 If K → L is Galois and N is an intermediate field with K → N normal, then 1

/ Gal(L/N )

/ Gal(L/K)

%N

/ Gal(N/K)

σ7→σ|N

/1

is a short exact sequence in the category of topological groups. Proof. We have already seen in Theorem 64 that the sequence above is exact in the category of groups. To prove Theorem 88, therefore, we must show that inclusion ϕN : Gal(L/N ) ,→ Gal(L/K) and restriction %N : Gal(L/K) → Gal(N/K) are continuous. Since these are group homomorphisms, it is sufficient to show that they are continuous at idGal(L/N ) and idGal(L/K) respectively. Begin with ϕN . From Proposition 86 (and the fact that ϕN is a group homomorphism) it follows that, to show that ϕN is continuous at idGal(L/N ) , is it sufficient to show that ϕ−1 N (Gal(L/Lf ) is open in Gal(L/N ), for each f in the set of non-zero separable polynomials in K[x] which split over L. LO N Lf aDD {= DD { { DD { { DD { {{

N aD DD DD DD D

K

y< yy y yy yy

Lf

A moment’s reflection reveals that ϕ−1 N (Gal(L/Lf ) = Gal(L/N Lf ), and that K → N Lf is finite galois. (In fact, N → N Lf is the splitting field for f ∈ N [x]× in L.) Thus, ϕ−1 N (Gal(L/Lf ) is open in Gal(L/N ), whence ϕN is continuous. Next we show that %N is continuous. Again, from Proposition 86 (and the fact that ϕN is a group homomorphism) it follows that, to show that %N is continuous at idGal(L/K) , is it sufficient to show that %−1 N (Gal(N/Ng ) is open in Gal(L/K), for each g in the set of non-zero separable polynomials in K[x] which split over N . Since %−1 N (Gal(N/Ng )) = Gal(L/Ng ) and since K → Ng is finite galois, it follows that %−1 N (Gal(N/Ng )) is open in Gal(L/K), whence %N is continuous.

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47

Theorem 89 For any Galois extension K → L, Gal(lim L /K) ∼ lim Gal(Lf /K), =← −−→ f −− f ∈I

f ∈I

where I denotes the directed poset of separable polynomials in K[x]× which split over L and where Lf denotes the splitting field for f in L, for each f ∈ I. The direct limit is taken in the category of extensions of K and the inverse limit may be taken in the category of topological groups, or subgroups of Aut(L), or topological subgroups of Gal(L) (see Section 3.6).

3.4

Closed subgroups

Proposition 90 If K → L is Galois and N in an intermediate field, then Gal(L/N ) is a closed subgroup of Gal(L/K). Proof. It is clear that Gal(L/N ) is a subgroup of Gal(L/K). To show that Gal(L/N ) is closed in Gal(L/K) we shall show that the complement of Gal(L/N ) in Gal(L/K) is open. Let σ be an element of Gal(L/K) which is not contained in Gal(L/N ). Since σ 6∈ Gal(L/N ) here is some u ∈ N such that σ(u) 6= u. Since K → L is profinite there is some finite Galois extension M of K contained in L such that u ∈ M . Since M/K is finite and normal, σ(u) ∈ M by Theorem 64. Then σGal(L/M ) is an open neighbourhood of σ in Gal(L/K). We claim that σGal(L/M ) ∩ Gal(L/N ) = ∅. To see this, suppose τ is contained in the intersection of σGal(L/M ) and Gal(L/N ). Then τ ∈ Gal(L/N ), so τ (u) = u, since u ∈ N . On the other hand, τ = ση for some η ∈ Gal(L/M ), so τ (u) = ση(u) = σ(u) 6= u since u ∈ M . This is the desired contradiction, showing that σGal(L/M ) ∩ Gal(L/N ) is indeed empty. We have now shown that every element of Gal(L/K) which is not contained in Gal(L/N ) is contained in an open subset of Gal(L/K) which does not intersect Gal(L/N ). This means that the complement of Gal(L/N ) in Gal(L/K) is open, from which it follows immediately that Gal(L/N ) is closed in Gal(L/K). This concludes the proof of Proposition 90. Proposition 91 If K → L is Galois and H is a closed subgroup of Gal(L/K), then H is galois. Proof. First, note that H ⊆ Gal(L/LH ). Pick σ ∈ Gal(L/LH ). We will show that every open neighbourhood of σ in Gal(L/LH ) intersects H, and therefore that σ is in the closure of H. Since H is closed, it will follow that σ ∈ H, and therefore that Gal(L/LH ) ⊆ H. Thus, H = Gal(L/LH ). To simplify notation slightly, let N = LH . Let U be an open neighbourhood of σ in Gal(L/N ). Since galois extensions are profinite, there is some finite

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 Galois extension M/N such that σGal(L/M ) ⊆ U . Define H 0 = τ |M τ ∈ H 0 and note that H 0 is a subgroup of Gal(M/N ). Then M H ⊇ M Gal(M/N ) and 0 M Gal(M/N ) = N . Let N 0 = M H and suppose u ∈ N 0 . Then τ 0 (u) = u for all τ 0 ∈ H 0 . But τ 0 ∈ H 0 implies τ 0 = τ |M for some τ ∈ H. Since N 0 /M we have u ∈ M , so τ 0 (u) = τ (u). Therefore, τ (u) = u for all τ ∈ H which implies u ∈ LH = N . We have now shown that u ∈ N 0 implies u ∈ N , so N 0 ⊆ N . Since the reverse inclusion is obvious, it follows that N 0 = N , or in other words, 0 0 0 M H = N . Thus, Gal(M/M H ) = Gal(M/N ), so H 0 = Gal(M/M H ), whence 0 H = Gal(M/N ). We have just shown that the map ρ:H τ

→ Gal(M/N ) 7 → τ |M

is surjective. Now, note that L/N is Galois (since L/K is Galois) and consider the map % : Gal(L/N ) → Gal(M/N ) σ 0 7→ σ 0 |M . If η ∈ Gal(L/M ) then σ 0 = ση ∈ Gal(L/N ) and %(σ 0 ) = σ 0 |M = σ|M . Since this is an element of Gal(M/N ) and ρ is surjective, there is some τ ∈ H such that ρ(τ ) = %(ση); thus there is some τ ∈ H such that τ |M = σ|M . This shows that H ∩ σGal(L/M ) is not empty, Q and concludes the proof of the Proposition. Observe that Gal(L/K) ⊂ f ∈I Gal(Lf , K), where I is the directed poset of separable f ∈ K[x]× which split over L, and that, Q by definition, Gal(L/K) is given the subspace topology, where the product f ∈I Gal(Lf , K) is taken in the category of topological groups. Theorem 92 Let K → L be Galois. Then the topological group Gal(L/K) is compact and admits a neighbourhood base of compact open sets. Proof. Let K → L be Galois and let I be as above. Since each Gal(Lf /K) is finite, and thus compact, it follows from Tychonoff’s Theorem that the product Q Gal(L f , K) is also compact. Observe that the complement of Gal(L/K) fQ ∈I in f ∈I Gal(Lf , K) is ( ∪f ∈I ∪f ≤g

(σh ) ∈

) Y

Gal(Lh /K) %f ≤g (σg ) 6= σf

.

h∈I

This Q is a union of open sets, and this open. Therefore, Gal(L/K) is closed in f ∈I Gal(Lf , K). Since any closed subset of a compact topological space is compact, it follows immediately that Gal(L, K) is compact. Recall from Lemma ?? that, for each σ ∈ Gal(L/K), a base at σ is given by  σGal(L/Lf ) f ∈ I .

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49

 Since Gal(L/Lf ) = %−1 f (idGal(Lf /K) ) and since idGal(Lf /K) is closed and open in Gal(Lf /K), it follows that σGal(L/Lf ) is an open and a closed neighbourhood of σ in Gal(L/K). Since Gal(L/K) is compact (as we have just seen), it follows that σGal(L/Lf ) is a compact open neighbourhood of σ in Gal(L/K). A topological space is totally disconnected if every point is its own connected component. Every galois group Gal(L/K) is totally disconnected [exercise]. In fact a topological group is a profinite topological group if and only if it is compact and totally disconnected topological group; this provides an intrinsic characterization of profinite groups [exercise].

3.5

The Fundamental Theorem

Consider the category int(L/K) of fields intermediate between K and L. In this category, an object is a field M equipped with maps M → L and K → M , and a map from K → N → L to K → M → L is a map M → N such that the following diagramme commutes. > L `A || AAA | AA | AA || || /N M `B > BB } } BB } } BB B }}} K We will often refer to objects and maps in int(L/K) by M and M → N respectively, making implicit use of the forgetful functor from int(L/K) to the category of fields. Now suppose K → L is Galois. Since int(L/K) is a subcategory of sub(L), we may consider the restriction of the Galois functor GALL : M → AutM (L) from sub(L) to int(L/K). However, as we have seen in this chapter, when K → L is Galois and M is intermediate, the group AutM (L) carries a topology and we use the symbol Gal(L/M ) to denote the topological group formed by equipping AutM (L) with the Krull topology. Moreover, as we have also seen in this chapter, the topological group Gal(L/M ) is a closed subgroup of Gal(L/K). This leads us to define the topological Galois functor and its adjoint. Definition 93 Suppose K → L is Galois. The topological Galois functor GALL K : int(L/K) → topsubGal(L/K) is the functor taking the intermediate field K → M → L to the topological subgroup Gal(L/M ) → Gal(L/K). The functor taking a topological subgroup H → Gal(L/K) to the intermediate field K → LH → L will be denoted FIXL K. L Exercise 94 Show that (GALL K , FIXK ) is an adjoint pair of contravariant functors. Show that if K → L is finite Galois, then GALL K is an equivalence. In the

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next chapter we will see some examples that illustrate that GALL K need not be an equivalence, when K → L is infinite Galois. Theorem 95 Suppose K → L is Galois. The topological Galois functor GALL K : M → Gal(L/M ) is an equivalence between int(L/K) and the category of closed subgroups of Gal(L/K). A map M → N in int(L/K) is normal if and only if Gal(L/N ) → Gal(L/M ) is normal.

3.6

The Galois Equivalence

We are now is a position to solve a problem expressed in Chapter 1: how to characterize the category of galois subgroups of Aut(L). Let L be a field and let J op denote the set of all galois subfields of L. Equip J op with the partial order defined as follows: α : N → L is greater than β : M → L if there is some γ : N → M such that α = β ◦ γ. Observe that J op is directed, since the pull-back of two normal subgroups is a normal subgroup: ; L cH ww O HHH HH w HH ww H ww N c ;M / www

N ∩M For each α ≥ β in J op , let ϕβ≤α denote the obvious map Gal(L/M ) → Gal(L/N ); we may view this as a map in the category of topological groups and also in the category of subgroups of Aut(L). Then  ϕβ≤α : Gal(L/M ) → Gal(L/N ) β ≤ α in J op is a direct system in the the category of topological groups and also in the category of subgroups of Aut(L). Define Gal(L) := −−lim Gal(L/N ); −−→ N ∈J op

this is a topological group and also a subgroup of Aut(L). Now, the galois functor GALL induces a new functor, which we denote by the same symbol, from the category of subfields of L to the category of topological subgroups of Gal(L): GALL : sub(L) → topsubGal(L) K 7→ Gal(L/K). Likewise, let FIXL denote the functor from the category of topological subgroups of Gal(L) to the category of subgroups of L obtained by restricting the functor

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51

denoted by the same symbol in Chapter 1: FIXL : topsubGal(L) → sub(L) H 7→ LH . Theorem 96 (The Galois Equivalence) The adjoint pair (GALL , FIXL ) of functors defined above defines an equivalence of categories between Galois subfields of L and closed subgroups of Gal(L); moroever, a map of galois subfields of L is normal if and only if the corresponding map in topsubGal(L) is normal.

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Chapter 4

Some important Galois groups 4.1

The Pruffer ring

ˆ is defined by Definition 97 The Pr¨ ufer ring Z ˆ = lim Z/nZ, Z ← −−− − n∈N×

where N× is given the partial order such that n≤m if n|m; here, Z/mZ → Z/nZ is the obvious epimorphism. The inverse limit is taken in the category of topological rings. ˆ consists of N× with the The inverse system appearing in the definition of Z partial order indicated above together with ring homomorphisms ∀ n|m

rn|m : Z/mZ → Z/nZ k + mZ 7→ k + nZ.

ˆ also comes equipped with ring homomorphisms The Pruffer ring Z ∀n

ˆ → Z/nZ rn : Z a 7→ an ,

ˆ is equipped with the weakest topology making it a topological ring such and Z that each rn is continuous, where Z/nZ is given the discrete topology. ˆ First, recall that Recall some basic facts concerning the topology for Z. ˆ ˆ ˆ Z comes with maps rn : Z → Z/nZ taking a = (an )n∈N× to an , and that Z is equipped with the coarsest topology making these maps continuous, where ˆ is Z/nZ is given the discrete topology. Thus, a neighbourhood base at a ∈ Z given by sets of the form \ US,a = rn−1 (an ), n∈S

53

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where S is a finite set of non-zero natural numbers. In fact, since N× is directed ˆ we can when equipped with the partial order appearing in the definition of Z, improve this result. ˆ is given by sets of the form Lemma 98 A neighbourhood base at a ∈ Z o n −1 ˆ bm = am , rm (am ) = b ∈ Z where m ranges over the set of positive integers. ˆ since rm is Proof. Note that the set appearing in the lemma is open in Z, continuous and {am } is an open subset of Z/mZ. To prove this lemma we must show that any open neighbourhood of a contains a set of the form above. To that end, let S be an arbitrary finite set of positive integers. Set m be the least common multiple of the n as n ranges over S; that is, m = lcm{n n ∈ S}. If b ∈ U{m},a then bm = am . For each n ∈ S we have n|m, so bn = rn|m (bm ) = ˆ therefore, b ∈ r−1 (an ) for each n ∈ S, so rn|m (am ) = an , by the definition of Z; n b ∈ US,a . This concludes the proof of the lemma. ˆ Proposition 99 Z is dense in Z. Proof. For each integer k let φ(k)n denote coset k + nZ ∈ Z/nZ. If n|m then the image of k + mZ under rn|m : Z/mZ → Z/nZ is clearly k + nZ, so φ(k) is ˆ This defines an element of Z. ˆ φ:Z → Z k 7→ (k + nZ)n∈N× . It is clear that φ is a homomorphism of rings. Moreover, using the Fundamental Theorem of Arithmetic we see that if k 6= k 0 then there is some n ∈ N× such that k + nZ 6= k 0 + nZ, so φ is a monomorphism. ˆ it is now sufficient to show that for each a To show that φ(Z) is dense in Z ˆ and for each positive integer m there is some k ∈ Z such that φ(k)m = am . in Z But this is very simple: let k be a integer representative for the coset am ∈ Z/mZ and consider φ(k) ∈ Z. Then φ(k)m = am . This concludes the proof that Z is ˆ dense in Z. Definition 100 For any prime p, the ring Zp of p-adic integers is defined by Zp = ← lim Z/pn Z, −− n∈N

where N× is given the usual order and where Z/pm Z → Z/pn Z is the obvious epimorphism when n ≤ m. The inverse limit is taken in the category of topological rings.

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55

To begin, let us fix some notation. Recall that the inverse system appearing in the definition of Zp consists of N× with the usual partial order together with ring homomorphisms ∀n≤m

s(p)n≤m : Z/pm Z → Z/pn Z k + pm Z 7→ k + pn Z.

Thus, s(p)n≤m = rpn |pm . Recall also that Zp itself comes equipped with ring homomorphisms ∀n

s(p)n : Zp → Z/pn Z a 7→ an ,

and that Zp is equipped with the weakest topology making it a topological ring such that each s(p)n is continuous, where Z/pn Z is given the discrete topology. Proposition 101 There is an isomorphism of topological groups Y ˆ∼ Zp , Z = p

where the product on the right-hand side is taken over all primes p. ˆ For each prime p and n ∈ N× , define a(p)n = Proof. Let a be an element of Z. × apn . Note that if n ≤ m in N then the image of a(p)m under the obvious map sn≤m : Z/pm Z → Z/pn Z is sn≤m (a(p)m ) = rpn |pm (apm ) = apn = a(p)n . This shows that a(p) is a p-adic number and defines ˆ → Zp ϕp : Z a 7→ a(p). Now, let ϕ be the unique function ˆ → ϕ:Z

Y

Zp

p

such that πp ◦ ϕ = ϕp for each prime p where Y πp : Zp → Zp p

is projection (please excuse the notation). We now and multipliQ show that ϕ is a group homomorphism. Since addition Q cation in Zp are defined component-wise, and since ϕ = ϕp , to show that ϕ p

p

is a group homomorphism it is enough to show that ϕp is a ring homomorphism. But this is clear: since sn ◦ ϕp (a + b)

= = = =

sn (ϕp (a + b)) rpn (a + b) rpn (a) + rpn (b) sn ◦ ϕp (a) + sn ◦ ϕp (b),

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it follows that ϕp (a + b) = ϕp (a) + ϕp (b). (In fact, it is equally clear that ϕ is a ring homomorphism.) We now show that ϕ is a injective. Suppose a ∈ ker ϕ. Then ϕp (a) = 0 for each prime p, so a(p)n = 0 for each prime p and for each positive integer n. Now, let m be any positive integer and factor m into a product of prime mt 1 m2 powers m = pm 1 p2 · · · pt . Then am = 0 by the Chinese Remainder Theorem since am ∈ Z/mZ is uniquely determined by the finite family of congruences i m am ≡ apmi ( mod pm i ) for each i = 1, . . . , t, and api i = a(pi )mi = 0 for each i i = 1, . . . , t. This shows that ker ϕ = {0}, so ϕ is injective. Q We now show that ϕ is surjective. Pick b ∈ Zp and write b(p) for πp (b) p

as above. Let m be any positive integer and factor m into a product of prime mt 1 m2 powers m = pm 1 p2 · · · pt . Using the Chinese Remainder Theorem again, let am be the unique element of Z/mZ such that rn|m (am ) = b(pi )mi , for each ˆ To see this, suppose 1 ≤ i ≤ t. We claim that (am )m∈N× is an element of Z. m1 m2 mt n|m and write m = p1 p2 · · · pt and above and n = pn1 1 pn2 2 · · · pnt t . It is sufficient to note that the diagramme rn|m

/ Z/nZ

Z/mZ

∼ =

∼ =

Q  mi Z/pi Z

1≤i≤t

/

Q

 Z/pni i Z

1≤i≤t ni 6=0

commutes, where the vertical arrows are given by the Chinese Remainder TheQ orem and the lower arrow is s(pi )ni |mi . In other words, rn|m (am ) = an . 1≤i≤j

Since ϕ(a) = b, this completes the proof that ϕ is surjective. We now show that ϕ is an isomorphism of topological groups. It was shown in lecture thatQa profinite group is a topological group. We will take as granted the fact that Zp is also a topological group when equipped with the product p

topology. Thus, to show that ϕ is an isomorphism of topological groups it is sufficient to show that ϕ is continuous, having already shown that ϕ is an isomorphism of groups. First, let ` be a prime number and fix n ∈ N× and k ∈ Z. −1 n Consider the basic open Q set V = s(`)−1 n (k + ` Z) in Z` . Let U = π` (V ). Then U is a basic open set in Zp . In fact, note that p

( U=

b∈

Y p

) n Zp b(`)n = k + ` Z .

DRAFT: April 6, 2007

57

Then ϕ−1 (U )

n ˆ a∈Z n ˆ = a∈Z n ˆ = a∈Z n ˆ = a∈Z =

o ϕ(a) ∈ V o ϕ` (a) ∈ U o ϕ` (a)n = k + `n Z o a`n = k + `n Z

n = r`−1 n (k + ` Z).

ˆ Moreover, since any basic open set in Q Zp takes the This is clearly open in Z. p

form \


n π`−1 s(`)−1 n (k + ` Z) ,

(`,n,k)∈S

where S is a finite set of triples as above, it follows that ϕ is continuous. ˆ We conclude this Section with someQexamples of closed subgroups of Z. −1 For each `, the projection map π` : p Zp → Z` is continuous, so π` (0) is a ˆ Observe that we may identify π −1 (0) with Q 0 Z`0 and closed subgroup of Z. ` 6=` ` ˆ `. with Z/Z Likewise, for each prime `, define Y π ` : Z` → Zp . p

as follows. For each `-adic integer c, let π ` (c) be the element of

Q

Zp such that

p

πp (π ` (c)) = 0 for all p 6= ` and π` (π ` (c)) = c; thus, π ` (c) = (0, 0, . . . , 0, c, 0, . . . , 0, 0), with c is the `th spot. It is clear from this description that π ` is a monomorphism. We now show that the image of π ` is closed. First, recall from lecture that any profinite group is Hausdorff. Thus, singleton sets are closed sets in Zp . In particular, {0} is closed in Zp , for each prime p, so πp−1 ({0}) is closed. Since \ π ` (Z` ) = π`−1 0 ({0}) `0 6=`

and since anQarbitrary intersection of closed sets is closed, it follows that π ` (Z` ) is closed in Z`0 . Now, define ϕ` by the diagramme `0

Q π` / Z`0 Z` B 0 ` BB BB ϕ BB ` BB ϕ  ˆ Z.

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DRAFT : April 6, 2007

Since ϕ is an isomorphism of topological groups and since all the groups in question are abelian, this concludes the proof that we may view Z` as a closed ˆ subgroup of Z. ˆ ∗ = lim (Z/nZ)∗ and Z∗ = lim (Z/pn Z)∗ . Show Exercise 102 Show that Z p ← −−− − ← −−− − ×

×

n∈N n∈N Q Q ˆ∗ ∼ ˆ ∗. that Z = p Z∗p and that Z∗` and p6=` Z∗p are closed subgroups of Z

4.2

¯ p /Fp ) Some subgroups of Gal(F

¯ p be a fixed algebraic Definition 103 Let Fp denote the field Z/pZ and let F ¯ ¯ closure of Fp . Let Frobp : Fp → Fp be the function defined by Frobp (u) = up ; this is called the Frobenius automorphism. In order to justify the nomenclature above, observe that Frobp (u + v) = (u + v)p = up + p

p−1   X p k=1 p

k

uk v p−k + v p

=u +v = Frobp (u) + Frobp (v),
since kp is an integer divisible by p for each 1 ≤ k ≤ p − 1. Since it was already clear that Frobp (uv) = Frobp (u)Frobp (v) and Frobp (1) = 1, it follows that ¯p → F ¯ p is a field homomorphism. To see that Frobp is an isomorphism, Frobp : F ¯ p there is some x ∈ F ¯ p such that xp = v, since F ¯ p is observe that, for each v ∈ F ¯ p /Fp ). algebraically closed. Thus, Frobp ∈ Gal(F ¯ p is Galois. Let Fpn be the extension of Fp fixed by the Theorem 104 Fp → F ¯ subgroup of Gal(Fp : Fp ) generated by Frobnp . Then Fpn is the unique subfield of ¯ p with cardinality pn . Moreover, Fp → Fpn is Galois and F Gal(Fpn /Fp ) ∼ = Z/nZ. Consequently,

¯ p /Fp ) ∼ ˆ Gal(F = Z.

¯ p fixed by the group generated by Frobn . Proof. Let Fpn be the subfield of F p ¯ p fixed by Frobn ; thus, Then Fpn is the subfield of elements of F p ¯ p upn = u}. Fpn = {u ∈ F n

Let fpn = xp − x. Then fpn is an element of Fp [x] which splits over Fpn . Since ¯ p : Fp is normal, Fpn is contained in the splitting field for fpn over Fp . In F fact, from the equation above it is clear that Fpn is precisely the set of roots of

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59

fpn . Since the formal derivative of fpn is −1, it follows from Lemma 69 that fpn is separable, so there are exactly pn roots of fpn . It follows that Fpn has pn elements and that Fpn is the splitting field for fpn over Fp , so Fpn : Fp is Galois. ¯ p with Finally, to see that Fpn is the unique extension of Fp contained in F n cardinality p , suppose K is an extension with the same property. Then K × is n a group of order pn − 1, when up −1 = 1 for all u ∈ K. It follows immediately that K = Fp . We now show that Gal(Fpn /Fp ) ∼ = Z/nZ. Since Fpn has pn elements and Fp has p elements, it follows immediately that [Fpn : Fp ] = n. By [insert reference], it follows that Gal(Fpn /Fp ) is a group of order n. To see that Gal(Fpn /Fp ) ∼ = Z/nZ is it sufficient to find an element of Gal(Fpn /Fp ) with order n. Let Frobp,n denote the image of Frobp under the restriction map ¯ p /Fp ) → Gal(Fpn /Fp ). Gal(F n

Then Frobnp,n (u) = Frobnp (u) = up = u for all u ∈ Fpn so Frobnp,n = idGal(Fpn /Fp ) . m Let m be the order of Frobp,n ; then m divides n. Then up = u for all u ∈ Fpn , in which case Fpn ⊆ Fpm . Since m divides n, this is only possible if n = m, so Z/nZ → Gal(Fpn /Fp ) k

7→ Frobkp,n

is an isomorphism. ¯ p is the splitting field for fpn = Since every finite normal extension of Fp in F pn × ¯ x − x, for some n ∈ N , it nfollows Fp d→ pFnp is a splitting field for the set n x − x = −1 are relatively prime {xp − x n ∈ N× }. Since xp − x and dx ¯ p [x], it follows from [insert reference] that {xpn − x n ∈ N× } is a set of in F ¯ p is Galois, by [insert reference]. It now separable polynomials. Thus, Fp → F follows from Theorem 89 that ¯ p /Fp ) ∼ Gal(F Gal(Fpn /Fp ) = ←lim −−− − n∈N×

∼ ˆ Z/nZ =: Z. = ←lim −−− − n∈N×

We will write

¯ p /Fp ) ∼ ˆ ϑ : Gal(F =Z

for this isomorphism. Q Exercise 105 Let ` be a prime number. Then Z` and p6=` Zp are closed norˆ and thus of Gal(F ¯ p /Fp ). Describe the corresponding Galois mal subgroups of Z, ¯ extensions of Fp contained in Fp . ¯ p /Fp ) Definition 106 The Weil group for Fp is the subgroup WFp ⊂ Gal(F generated by Frobp . ¯ p /Fp ) and WF ∼ Proposition 107 WFp is a dense proper subgroup of Gal(F Z. p =

60

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¯ p : Fp would be finite, which Proof. Observe that WFp ∼ = Z, since otherwise F it is not. Thus we have the following commutative diagramme of continuous maps. / Gal(F ¯ p /Fp ) WFp ∼ =

 Z

ϑ

φ

 /Z ˆ

ˆ (see Section 4.1) it follows that WF is a dense Now, since Z is dense in Z p ¯ subgroup of Gal(Fp /Fp ). To see that this subgroup is proper observe that ¯ p /Fp ) is uncountable (use Cantor’s diagonalization argument!) while Z is Gal(F countable. ˆ In this In Section 4.1 we saw that Z` is a closed normal subgroup of Z. ∼ ˆ ¯ section we have seen that Gal(Fp /Fp ) = Z (isomorphism of topological groups). ¯ p /Fp ). Consider the field Thus, we may view Z` as a closed subgroup of Gal(F Z` Z` ¯ ¯ Fp . By Proposition 91, Fp → Fp is Galois, and ¯ p /F ¯ Z` ) ∼ Gal(F p = Z` . ¯ Z` /Fp ) is the cokernel of the restriction map From Theorem 88 we see that Gal(F p ¯ p /Fp ) → Gal(F ¯ p /F ¯ Z` ). Gal(F p In summary we have the following isomorphic short exact sequences: 1

¯ p /F ¯ Z` ) / Gal(F p

/ Gal(F ¯ p /Fp )

¯ Z` /Fp ) / Gal(F p

/1

0

 / Z`

 /Z ˆ

 / Z/Z ˆ `

/0

0

 Z/pn Z / ←lim −−− −

 Z/mZ / ←lim −−−−

 lim Z/mZ ←−−−−−−

n∈N×

m∈N×

/

m∈N× (m,`)=1

/0

¯ Z` we use the direct limit description of Theorem 88. In order to describe F p ¯ Z` ? From the diagramme above we see (`, n) = 1. Which Fpm are contained in F p Thus ¯ Z` = ∪ Fpm . F p

m∈N× (`,m)=1

ˆ ` = Q 0 Z`0 as a closed subgroup of Gal(F ¯ p /Fp ). Likewise, we may view Z/Z ` 6=` ˆ ` ˆ ` ¯ pZ/Z ¯ pZ/Z is Galois, and Consider the field F . By Proposition 91, Fp → F ˆ

¯ p /F ¯ Z/Z` ) ∼ ˆ `. Gal(F = Z/Z p

DRAFT: April 6, 2007

61 ˆ Z/Z

¯ p ` /Fp ) is the cokernel of the restriction From Theorem 88 we see that Gal(F map ˆ ` ¯ p /Fp ) → Gal(F ¯ p /F ¯ Z/Z Gal(F ). p

In summary we have the following isomorphic short exact sequences: 1

ˆ ` / Gal(F ¯ p /F ¯ Z/Z ) p

/ Gal(F ¯ p /Fp )

ˆ ` / Gal(F ¯ Z/Z /Fp ) p

/1

0

 / Z/Z ˆ `

 /Z ˆ

 / Z`

/0

 lim Z/`n Z ←−−−−−

 Z/mZ / ←lim −−−−

 n Z/` Z / ←lim −−− −

/0

0

/

n∈N× (n,`)=1

m∈N×

n∈N×

Thus ˆ ` ¯ Z/Z = F p

∪ Fp`n .

n∈N×

The following diagramme summarizes the situation. ¯ = FO p dIII { { II Z `0 6=` II ` {{ II {{ { I { { ∪ Fpn ∪ Fp`m ˆ Z gcd(n,`)=1 m aCC : u CC uu CC uQ u C u Z` CC Z`0 uu uu `0 6=` Fp Q

4.3

Z`0

¯ Some subgroups of Gal(Q/Q)

We have already seen (from the mid-term assignment) that Q(µn ) is splitting field for xn − 1 ∈ Q[x] and that Gal(Q(µn )/Q) ∼ = (Z/nZ)∗ . Proposition 108 Let Q → Q(µ) be a splitting field for {xn − 1 ∈ Q[x] n ∈ N× }. Then Q → Q(µ) is a Galois extension and ˆ ∗. Gal(Q(µ)/Q) ∼ =Z Proof. Since xn − 1 and nxn−1 are relatively prime in Q[x], for each n ∈ N× , it follows from [insert reference] that Q → Q(µ) is a splitting field for a set of

62

DRAFT : April 6, 2007

separable polynomials, and thus Galois by [insert reference]. Now, N× , partially ordered by division, is a directed poset, and {Q(µn ) → Q(e2πi/m ) n|m} is a directed system, so Q(µ) =

∪ Q(µn ).

n∈N×

Now it follows from Proposition 89 that Gal(Q(µ)/Q) ∼ Gal(Q(µn )/Q). = ←lim −−− − n∈N×

Since Gal(Q(µn )/Q) ∼ = (Z/nZ)∗ , follows from Exercise 102 that ˆ ∗, Gal(Q(µ)/Q) ∼ (Z/nZ)∗ = Z = ←lim −−− − n∈N×

as desired. We can now use some ideas from Section 4.2 to construct some interesting fields intermediate between Q and Q(µ). ˆ ∗ . Thus, Q → Exercise 102 shows that Z∗` is a closed normal subgroup of Z Z∗ ` Q(µ) is Galois and we have the following isomorphic short exact sequences: 1

/ Gal(Q(µ)/Q(µ)Z∗` )

/ Gal(Q(µ)/Q)

1

 / Z∗`

 /Z ˆ∗

1

 n (Z/` Z)∗ / ←lim −−− −

 ∗ (Z/mZ) / ←lim −−−−

n∈N×

/ Gal(Q(µ)Z∗` /Q)

/

m∈N×

/

Q `0 6=`

Z∗`0

 ∗ lim (Z/mZ) ←−−−−−− m∈N× (m,`)=1

Thus ∗

Q(µ)Z` =

∪

n∈N× (n,`)=1

Q(µn )

∗

In other words, the extension Q → Q(µ)Z` is a splitting field for {xn − 1 ∈ Q[x] n ∈ N× , gcd(n, `) = 1}. Thus, ∗

Q(µ)Z` = Q(µ` ), ¯ which are prime to `. where µ` denotes the set of roots of unity in Q

/1

/1

/ 1.

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63

ˆ ∗ /Z∗ = Q 0 Z∗0 is a closed normal Likewise, Exercise 102 shows that Z ` ` 6=` ` ˆ ∗ . Thus, Q → Q(µ)Zˆ ∗ /Z∗` is Galois and we have the following subgroup of Z isomorphic short exact sequences: 1

/ Gal(Q(µ)/Q(µ)Zˆ ∗ /Z∗` )

/ Gal(Q(µ)/Q)

/ Gal(Q(µ)Zˆ ∗ /Z∗` /Q)

/1



 /Z ˆ∗

 / Z∗`

/1

 (Z/mZ)∗ / ←lim −−−−

 (Z/mZ)∗ / ←lim −−−−

/ 1.

/

1

1

/

Q

`0 6=`

Z∗`0

 n lim (Z/` Z)∗ ←−−−−−− ×

n∈N (m,`)=1

m∈N×

m∈N×

Thus ˆ ∗ /Z∗

Q(µ)Z

`

n

∪ × Q(e2πi/` ).

=

n∈N

ˆ ∗ /Z∗

In other words, the extension Q → Q(µ)Z

`

is a splitting field for

n {x` − 1 ∈ Q[x] n ∈ N}. Thus, ˆ ∗ /Z∗

Q(µ)Z

`

= Q(µ`∞ ),

¯ where µ`∞ denotes the set of `n th roots of unity in Q. The following diagramme summarizes the situation. ¯ Q O

Q(µ) v: O cGGG Z∗ v v GG ` vv GG vv G v v ∗ ˆ Q(µ`∞ ) Z Q(µl ) dII w; II w II ww II ww Q Z∗`0 Z∗ w II ` ww `0 6=` Q Q

`0 6=`

4.4

Z∗ `0

p-adic numbers

Throughout this section, p is a fixed prime. Recall the definition of the cring Zp from Section 4.1. Observe that Zp is a cring and an integral domain, because ... (under construction).

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Definition 109 The field of fractions of Zp is denoted Qp . When equipped with the coarsest topology making Zp → Qp a continuous homomorphism of topological fields, Qp is called the field of p-adic numbers. We have already seen (in Section 4.1) that Zp is a topological ring with neighbourhood base at 0Zp given by sets of the form pn Zp . From Definition 109 it follows that this is also a neighbourhood base at 0Qp . The inclusion Z → Zp studied in Section 4.1 now extends to a field homomorphism Q → Qp . In this Section we show that Qp is a complete normed field, that Q is a dense subfield ¯ is dense in any algebraic closure Q ¯ p of Qp . In order to do of Qp , and that Q this, we must say a few more words about the cring Zp . Proposition 110 Zp is a local ring; specifically, pZp is the unique maximal ideal in Zp , and Zp /pZp ∼ = Z/pZ. Also, ∩n∈N pn Zp = {0}. Finally, if a ∈ Zp and a 6= 0 then there is a unique unit u ∈ Z∗p and n ∈ N such that a = upn . Proof. It is clear that pZp is an ideal and that Zp /pZp ∼ = Z/pZ, so pZp is certainly maximal. To see that pZp is the only maximal ideal in Zp , we show that the compliment of pZp in Zp is the set of units in Zp . Suppose a ∈ Zp is a unit. Then there is some b ∈ Zp such that ab = 1. Write a = (ai )i∈N× and b = (bi )i∈N× . Then ai bi = 1Z/pi Z for each i ∈ N× . In particular, a1 6= 0, so a6∈pZp . Conversely, suppose a6∈pZp . Then a1 6= 0 where a = (ai )i∈N× . Observe that a1 ∈ Z/pZ, which is a field. Thus, a1 is a unit in Z/pZ. Since a1 is a unit in Z/pZ and since a1 is the image of a2 under the surjective cring homorphism Z/p2 Z → Z/pZ, it follows that a2 is a unit in Z/p2 Z. Since a2 is a unit in Z/p2 Z and since a2 is the image of a3 under the surjective cring homorphism Z/p3 Z → Z/p2 Z, it follows that a3 is a unit in Z/p3 Z. Continuing in this manner (i.e., by induction) it follows that ai is a unit in Z/pi Z for each i ∈ N× . It follows that Z∗p = {a ∈ Zp a1 6= 0}. Now, suppose I is a maximal ideal of Zp and that I 6= pZp . Then there must be some a ∈ I with a6∈pZp (since otherwise, I ⊂ pZp , in which case it is not maximal). But a6∈pZp implies a ∈ Z∗p in which case I = Zp , which also contradicts the definition of a maximal ideal. The second claim of the Proposition is also clear. Observe that pm Zp is precisely the set of a ∈ Zp such that a = (ai )i∈N× with ai = 0 for 1 ≤ i ≤ m. Thus, n < m implies pm Zp ⊂ pn Zp , and if a ∈ ∩n∈N pn Zp then ai for all i ∈ N× , so a = 0. The final claim of the Proposition follows from the work already done. Every local ring carries a canonical topology, called the adic topology, which is the unique topology such that the maximal ideal is open, and the ring is a

DRAFT: April 6, 2007

65

topological ring. In the adic topology for Zp , a neighbourhood base at 0 is given by {pn Zp n ∈ N}, which is also a neighbourhood base for O in the profinite topology. In other words, the profinite topology is precisely the adic topology for Z. This leads directly to the following definition. Definition 111 The p-adic norm p : Qp → R≥0 is defined by 0 p = 0 and a = p−n , where n is the largest integer such that pn |a. p Proposition 112 Qp is a complete, normed field and Q is dense in Qp . Proof. (under construction) The field Qp of p-adic numbers may also be constructed as the completion on Q with respect to the p-adic norm; thus, Qp may be identified, as a topological field, with the quotient of the ring of sequences N → Q which are Cauchy with respect to the p-adic norm, by the ideal of null sequences with respect to the p-adic norm [exercise]. This analytic description is very important, and stresses the parallel with the field R, which is, by definition, the quotient of the ring of sequences N → Q which are Cauchy with respect to the usual norm, by the ideal of null sequences with respect to the usual norm. Since the usual norm is often denoted ∞ in number theory, the field R is often denoted Q∞ in number theory, to emphasize its genesis and the fact that, in many regards, it stands on equal footing with the p-adic fields Qp . Exercise 113 (exam question) Let K → L be a finite galois extension and consider the function NL/K : L → L defined by NL/K (u) :=

Y

σ(u).

σ∈Gal(L/K)

This function enjoys the following properties, for each u and v in L: (i) NL/K (u) ∈ K (ii) NL/K (uv) = NL/K (u)NL/K (v) n/d

(iii) NL/K (u) = (−1)d a0 , where fu,K = xd + ad−1 xd−1 + · · · + a1 x + a0 and n = dimK (L), (iv) NL/K (u) = det(Tu ), where Tu : L → L is function Tu (v) = uv, as in the mid-term test. ¯ p according to Proposition 114 The p-adic norm p extends from Qp to Q the definition u := NK /Q (u) 1/[Ku :Qp ] , u p p p ¯ p. where Ku is the splitting field for fu,Qp in Q Proof. (exam question)

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Remark 115 Likewise, the usual norm extends from R to C according to the definition u := NK /R (u) 1/[Ku :R] , ∞

∞

u

where Ku is the splitting field for fu,R in C. ¯ p /Qp ) and u ∈ Q ¯ p , then σ(u) = u . Proposition 116 If σ ∈ Gal(Q p p Proof. (exam question) ¯ p /Qp ) is continuous with respect to . Corollary 117 σ ∈ Gal(Q p ¯ of Q; let us now fix and algebraic We have fixed an algebraic closure Q → Q ¯ ¯ closure Qp → Qp of Qp . Since Qp → Qp is a splitting field for Qp [x]× , it is also ¯ →Q ¯p a splitting field for Q[x]× . Thus, by Definition 47 there is a unique θ : Q making the following diagramme commute. ?

¯p Q

∃!θ

¯ Q _@@ @@ @@ @@

`@@ @@ @@ @@ > }} } } }} }}

Qp

Q ¯ is a subfield of Q ¯ p , we may equip it with the topology inherited from Since Q ¯ p . Then, it follows from Corollary 117 that every σ ∈ Gal(Q/Q) ¯ Q is continuous ¯ given by the extension of the p-adic norm to with respect to the topology for Q ¯ p. Q ¯ is dense in Q ¯p Proposition 118 Q Proof. (under construction) For your edification, we present some Galois groups which are related to the cyclotomic extensions already considered above. Sadly, we do not have time to provide the proofs. ¯p Q O Qp (µ) dII ∗ t9 II Zp t t II tt II t t I tt Qp (µp∞ ) Qp (µp ) : eJJ u JJ u JJ uu u J u JJ Z∗ uu ˆ p J uu Z Qp ˆ Z

DRAFT: April 6, 2007

4.5

67

¯ Decomposition subgroups of Gal(Q/Q)

¯ p /Qp ). In Section 4.4 we saw that Q ¯ is a subfield of Q ¯ p , so Suppose σ ∈ Gal(Q ¯ ¯ we may consider the restriction of σ from Qp to Q. Note that that since σ fixes ¯ then u and σ(u) are both roots of fu,Q , so σ(u) ∈ Q. ¯ Qp then σ fixes Q. If u ∈ Q ¯ p /Qp ) → Gal(Q/Q) ¯ Definition 119 Let decp : Gal(Q be the group homomorphism defined by decp (σ) = σ|Q¯ . Remark 120 There is room for some confusion here. The notation σ|N was also used in the context of normal restriction (see Section 2.10). It is not the ¯ →Q ¯ p is normal (it is not even algebraic!). Nevertheless, the same case that Q ideas apply to the present situation. ¯ p /Qp ) → Gal(Q/Q) ¯ Proposition 121 The group homomorphism decp : Gal(Q is a continuous and injective. Proof. (under construction) ¯ p /Qp ) as a subgroup of Gal(Q/Q); ¯ Definition 122 We may thus regard Gal(Q ¯ p /Qp ) is called the decomposition subgroup of Gal(Q/Q) ¯ when we do so, Gal(Q at p. Note that we use the same arguments to define a continuous group monomor¯ phism dec∞ : Gal(C/R) → Gal(Q/Q). We may thus view Gal(C/R) as a sub¯ group of Gal(Q/Q); when we do so, the non-trivial element of Gal(C/R) is called ¯ complex conjugation on Q.

4.6

¯ Inertia subgroups of Gal(Q/Q)

In this section, p is a fixed prime. ¯ p := {u ∈ Q ¯ p u ≤ 1} and mp := {u ∈ Q ¯ p u < 1}. Definition 123 Z p p ¯ p is a local ring, and the unique maximal ideal in Z ¯ p is mp . Lemma 124 Z Moreover, ¯ p /mp ∼ ¯p. Z =F Proof. (exam question) ¯ p /Qp ) → Gal(F ¯ p /Fp ) be the group homomorDefinition 125 Let redp : Gal(Q phism defined by redp (σ)(u) = σ(u) ˙ + mp , ¯p ⊂ Q ¯ p for u ∈ F ¯ p by way of the isomorphism where u˙ is a representative in Z ∼ ¯ ¯ Zp /mp = Fp .

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DRAFT : April 6, 2007

Proposition 126 redp is a continuous, surjective group homomorphism. Proof. (under construction) ¯ Definition 127 When viewed as a subgroup of Gal(Q/Q), the kernel of redp ¯ is called the inertia subgroup of Gal(Q/Q) at p and denoted Ip ; thus, Ip = decp (ker redp ). ¯ Gal(Q/Q) O decp

Ip

/ Gal(Q ¯ p /Qp )

redp

/ Gal(F ¯ p /Fp )

Chapter 5

Galois Representations 5.1

Ramification

¯ Definition 128 A representation ρ of Gal(Q/Q) is said to be unramified at p if the inertia group Ip is contained in the kernel of ρ. If ρ is unramified at p then the restriction ρp of ρ to the decomposition ¯ p /Qp ) factors through the reduction map redp . Thus, if ρp is ungroup Gal(Q ramified at p, then ρp is completely determined by a continuous representation ¯ p /Fp ), also denoted ρp . In Section 4.2 we saw that the subgroup of of Gal(F ¯ p /Fp ) generated by Frobp is dense in Gal(F ¯ p /Fp ), from which it follows Gal(F that ρp is completely determined by ρp (Frobp ), when ρ is unramified at p. ¯ Definition 129 An element σ of Gal(Q/Q) is a lift of Frobenius at p if ¯ p /Qp ) and its image under it is contained in the decomposition subgroup Gal(Q ¯ ¯ the reduction map Gal(Qp /Qp ) → Gal(Fp /Fp ) is the Frobenius automorphism Frobp . It is common to denote a lift of Frobenius at p by Frobp also; the context will remove any ambiguity. ¯ Definition 130 A representation ρ of Gal(Q/Q) is a Galois representation if: the representation space V for ρ is a finite-dimensional vector space over a topological field; ρ is continuous with respect to the topology for GL(V ) inherited from the topological field; and ρ is unramified at all but finitely many primes p. If the topological field is C then ρ is a Complex representation, if the topological field is a finite extension of Q` then ρ is a `-adic representation, and if the topological field is a finite field then ρ is a finite representation.

5.2

`-adic cyclotomic characters

The best way to introduce `-adic representations is with an interesting example. 69

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DRAFT : April 6, 2007

Definition 131 Let ` be a prime. The `-adic cyclotomic character ¯ χ` : Gal(Q/Q) → Q× ` is the continuous homomorphism defined by the composition, ¯ Gal(Q/Q)

%

∼

/ Gal(Q(µ)/Q) =

/ Z∗`

/Z ˆ∗

/ Q× , `

ˆ ∗ → Z∗ where % is restriction, the isomorphism is from Section 4.3, the map Z ` ∗ ∗ is projection and Z` → Q` is inclusion. ¯ Lemma 132 The kernel of the `-adic cyclotomic character χ` is Gal(Q/Q(µ `∞ )) and the image of χ` is Z∗` . Proof. The following diagramme commutes ∼ =

/ Gal(Q(µ)/Q) ¯ Gal(Q/Q) PPP PPP PPP PPP  ( ∼ = Gal(Q(µ`∞ )/Q)

/Z ˆ∗  / Z∗` ,

where the three left-most arrows are restriction, the isomorphisms are from Section 4.3 and the right-most arrow in inclusion. ¯ Proposition 133 For every p 6= `, the `-adic cyclotomic character χ` : Gal(Q/Q) → Q× is unramified at p, and χ (Frob ) = p. Moreover, χ (conj) = −1. ` p ` ` Proof. (exam exercise)

5.3

Tate module of the algebraic group GL(1)

We now give a slightly different construction of the `-adic cyclotomic charac¯ ters of Gal(Q/Q) which will lead to a more general class of representations of ¯ Gal(Q/Q) in the next section. ¯ more Consider the algebraic curve X given by the equation xy = 1 over Q; ¯ precisely, let X = Spec(Q[x, y]/(xy − 1)). Then, as a set, ¯ × } ∪ {(0)}. X = {(x − a) a ∈ Q ¯ [insert reference to exercise] The Q-rational points ¯ = {(x − a) a ∈ Q ¯ ×} X(Q) ¯ ×. are naturally identified with Q ¯ defined by For each integer d, consider the cring homomorphism over Q ¯ y]/(xy − 1) → Q[x, ¯ y]/(xy − 1) φd : Q[x, x 7→ xd y 7→ y d .

DRAFT: April 6, 2007

71

Now consider

fd : X → X p 7→ φ−1 d (p).

−1 Thus, fd (x − a) = φ−1 d (x − a) and f (0) = (0). Since φd (x − a) is a (non-zero!) ¯ × . Thus, x − a divides prime ideal in A, it takes the form (x − b) for some b ∈ Q d d x − b, which forces a = b. In other words,

¯ ×, ∀a ∈ Q

fd (x − a) = (x − ad ).

¯ Restricting this map to Q-rational points gives ¯ → X(Q) ¯ fd : X(Q) a 7→ ad . Now, define  ¯ fp (a) = 1 . X[d] = a ∈ X(Q) ¯ If n|m and (x − a) is This is called the group of d-torsion points on X(Q). m/n m/n in X[m] then (x − a) = (x − a ) is in X[n], so fm/n restricts to fm/n : X[m] → X[n]. Now, {fm/n : X[m] → X[n] n|m} is an inverse system. Definition 134 Fix a prime ` and consider the case when d = `n for some n ∈ N. Then {fn≤m : X[`m ] → X[`n ] ∀n, m ∈ N, n ≤ m} is an inverse system (of groups) over a directed poset. The inverse limit T` X := ← lim X[`n ], −− n∈N

¯ is called the `-adic Tate module of the alegbraic group X over Q. Now, (x−a) ∈ X[d] implies ad = 1 which in turn implies a ∈ µd . Conversely, ... Thus, ¯ ∀d ∈ N, X[d] ∼ = µd (Q). Although there are many such isomorphisms, we may choose one for each d ∈ N is such a way that if n|m then X[n] o

X[m]

 µn o

 µm

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DRAFT : April 6, 2007

commutes. Thus, the Tate module T` XQ¯ is precisely Z` . ¯ Now, the Galois group Gal(Q/Q) acts on T` XQ¯ , and thus gives rise to a ¯ representation π` of Gal(Q/Q) on AutQ` (T` XQ¯ ⊗Z` Q` ), which is precisely the `-adic cyclotomic character χ` ! /`AutQ` (T` XQ¯ ⊗Z` Q` ) ¯ Gal(Q/Q) QQQ QQQ χ` QQQ ∼ = QQQ QQQ  ( × Q` π

5.4

Tate module of an elliptic curve

5.5

Complex Galois representations

5.6

Weil-Deligne group

5.7

Weil-Deligne representations

5.8

An introduction to the Langlands Programme

Chapter 6

Student Presentations 6.1

Damian Brotbek: L¨ uroth’s Theorem, 2007.03.28, 08:45-10:00 in MS452

6.2

Kelly Rose: Cubics and Quartics, 2007.03.30, 08:45-10:00 in MS452

6.3

Ryan Stratford: Ruler and Compass, 2007.03.30, 11:00-12:15 in MS528

6.4

Shuai Zhang: Solution by Radicals, 2007.04.04, 08:45-10:00 in MS452

6.5

Robert Parkinson: Galois Cohomology, 2007.04.11, 08:45-10:00 in MS452

6.6

Thomas Kuwahara: Inverse Galois Problem, 2007.04.13, 14:00-14:50 in ICT516

6.7

Matthew Musson: Kummer Theory, 2007.04.20, 14:00-14:50 in ICT516

6.8

Safa Ismail: Primitive Element Theorem, 2007.04.27, 08:45-10:00 in MS452 73

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DRAFT : April 6, 2007

Chapter 7

Tests and Exams 7.1

Take-home Test

Questions posted Thursday, 15 February 2007 at 16:00. Please submit your solutions to the following questions by Thursday, 1 March 2007 at 12:30. (Electronic submissions to the course website are preferred.) The rules of the game: do not discuss these questions with anyone. √ √ Question 135 Consider the extension Q → Q( 4 2, 3). Is this galois? Find the dimension and the Galois group of this extension. Question 136 Find the minimal polynomial fu,Q for u = e2πi/7 ∈ C over Q and find the splitting field F for this polynomial. Find the group AutQ (F ). Find the minimal polynomial for v = cos(2π/7) ∈ R over Q find the splitting field E for this polynomial. Find the group AutQ (E). n Question 137 Describe the galois group of the splitting √ field F in C for x −1 ∈ × Q[x], where n ∈ √ N . Find an integer n so that Q( 5) is a subfield of F . In this case, is Q( 5) = F ∩ R?

Question 138 Consider the polynomial f = ax4 + bx3 + cx2 + bx + a ∈ Q[x]. Find a condition on a, b, c ∈ Q which ensures that f is irreducible. Find a splitting field extension F for f ∈ Q[x]. Compute AutQ (F ) and dimQ (F ). Question 139 Suppose K → K(u) is a finite simple extension. Define Tu : K(u) → K(u) by Tu (a) = au. Then Tu is clearly K-linear. Show that the characteristic polynomial of Tu is the minimal polynomial for u over K. Question 140 If K is a field then Spec(K) = {(0)}, so the set of prime ideals of K is a singleton. Is the converse true? More precisely, if A is a non-zero cring and Spec(A) is a singleton, does it follow that A is a field? 75

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DRAFT : April 6, 2007

Question 141 Prove or disprove: if α and β are galois (resp. normal, separable, finite) and α = β ◦ γ, then γ is galois (resp. normal, separable, finite). Question 142 Prove or disprove: if α and γ are galois (resp. normal, separable, finite) and α = β ◦ γ, then β is galois (resp. normal, separable, finite). Question 143 Prove or disprove: if β and γ are galois (resp. normal, separable, finite) and α = β ◦ γ, then α is galois (resp. normal, separable, finite). Question 144 Let α : K → N and β : K → M be galois extensions. Show that a push-out (α0 : M → L, β 0 : N → L) exists Q in the category of fields. Show that both α0 and β 0 are galois. Consider α0 β 0 in the category of subfields of L; is this β 0 ◦ α : K → L?

7.2

Take-home Exam

Questions posted Thursday, 06 April 2007 at 17:00. Answer question 145, one of questions 146 and 147, question 148, one of questions 149 and 150, and question 151, for a total of five answers. Please submit your solutions by Thursday, 26 April 2007 at 12:30. (You are welcome to submit your solutions to the course website using the ‘electronic drop box’ feature.) Please do not discuss these questions with anyone, except perhaps with me for clarification. You are welcome to use any definitions and results proved in lecture or in the posted course notes. You are also welcome to consult the literature for ideas and results to help you answer these questions, but you may not use results from the literature unless you also include a proof in your solution, suitably adapted (definitions and notation) to this course.

Question 145 Let α : K → L be galois and let β : K → M be arbitrary. Show that a push-out (α0 , β 0 ) exists (in the category of fields). (Note: No over-field is given!) Is α0 galois? Question 146 Let K → L be Galois. Let M1 and M2 be extensions of K contained in L and let H1 and H2 be closed subgroups of Gal(L/K). (i) Prove: Gal(L/M1

a

M2 ) = Gal(L/M1 )

Y

Gal(L/M2 ).

(ii) Prove: LH1 (iii) Prove: if the subgroup H1

`

Q

H2

= LH1

a

LH2 .

H2 is closed in Gal(L/K), then Y ‘ LH1 H2 = LH1 LH2 .

DRAFT: April 6, 2007 (iv) Prove: if Gal(L/M1 )

77 `

Gal(L/M1

GalL : M2 is a closed subgroup of GalL : K then

Y

M2 ) = Gal(L/M1 )

a

Gal(L/M2 ).

` Here, M1 M2 denotes the coproduct ofQM1 and M2 in the category of fields intermediate between K and L and M1 M2 denotes the product of M1`and M2 in the category of fields intermediate between K and L; likewise, H1 H2 denotes Q the coproduct of H1 and H2 in the category of subgroups of Gal(L/K) and H1 H2 denotes the product of H1 and H2 in the category of subgroups of Gal(L/K). Question 147 Let K → L be Galois. Let M1 and M2 be intermediate fields. ` 1. Suppose K → M1 is Galois. Show that M2 → M1 M2 is Galois and that a Y Gal(M1 M2 /M2 ) ∼ M2 ). = Gal(M1 /M1 2. Suppose Q K → M1 and ` K → M2 are Galois. Show that K → M1 and M1 M2 → M1 M2 are Galois. Show that Gal(M1

a

Q

M2

M2 /K)

is the pull-back in the category of groups of the pair of restriction maps Y ρ1 : Gal(M1 /K) → Gal(M1 M2 /K) Y ρ2 : Gal(M2 /K) → Gal(M1 M2 /K). Question 148 Let K → L be a finite galois extension and consider the function NL/K : L → L defined by NL/K (u) =

Y

σ(u).

σ∈Gal(L/K)

Show that this function enjoys the following properties, where u and v are arbitrary elements of L: (i) NL/K (u) ∈ K (ii) NL/K (uv) = NL/K (u)NL/K (v) n/d

(iii) NL/K (u) = (−1)d a0 , where fu,K = xd + ad−1 xd−1 + · · · + a1 x + a0 and n = dimK (L). (iv) NL/K (u) = det(Tu ), where Tu : L → L is function Tu (v) = uv, as in the mid-term test.

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DRAFT : April 6, 2007

¯p → Question 149 Let p : Qp → R≥0 be the p-adic norm.1 Define p : Q R≥0 by u = NK /Q (u) 1/[Ku :Qp ] , u p p ¯ p → R≥0 is a ¯ p . Show that : Q where Ku is the splitting field for fu,Qp in Q p ¯ p /Qp ) then norm. Then, prove the following: if σ ∈ Gal(Q σ(u) = u , p p ¯ p . Conclude that σ is continuous with respect to the topology for for each u ∈ Q ¯ ¯ p. Qp determined by the extension of the p-adic norm to Q ¯ p denote the set of u ∈ Q ¯ p such that u ≤ 1 and let mp Question 150 Let Z p ¯ p such that u < 1. Show that denote the set of u ∈ Q p ¯p. ¯ p /mp ∼ Z =F Question 151 Show that the `-adic cyclotomic character χ` is unramified at every prime p 6= `, and that χ` (Frobp ) = p for every p 6= `. Also, show that χ` (conj) = −1, where conj is complex conjugation.

˛˛ ˛ ˛ 1 ˛ ˛ : A → R≥0 such that ˛a˛ = 0 if and only if a = 0, ˛ ˛ ˛ ˛ A norm ˛ ˛˛ ˛ on a ˛cring ˛A is˛ a˛ function ˛ab˛ = ˛a˛˛b˛ and ˛a + b˛ ≤ ˛a˛ + ˛b˛ for all a, b ∈ A.