Elimination of minimal primes Seminormal rings July 9, 2006

Printable version of these slides http://hlombardi.free.fr/publis/LectureDoc3.pdf

French detailed version http://hlombardi.free.fr/publis/NotesDeCoursSeminormal.pdf

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Plan • What is a seminormal ring ? • Projective modules, rank, Picard group. • Freeness criterium, computational meaning of the Traverso-Swan Theorem • Classical proof in the domain case – Gcd domains – Integrally closed domains – Seminormal domains • Making the proof constructive – Crucial lemma. Up side down – Proof of the crucial lemma. • The reduced case 2

References [Tra] Traverso C. Seminormality and the Picard group. Ann. Scuola Norm. Sup. Pisa, 24 (1970), 585–595. [Swan] Swan R. G. On Seminormality. (1980), 210–229.

Journal of Algebra, 67

[Coq] Coquand T. On seminormality. 2006. To appear in Journal of Algebra. http://www.cs.chalmers.se/∼coquand/min.pdf [Ric] Richman F. Non trivial uses of trivial rings. Proc. Amer. Math. Soc., 103 (1988), 1012–1014. 3

An exemplary constructive proof The Traverso-Swan Theorem says that a reduced ring A is seminormal if and only if the natural homomorphism

Pic A → Pic A[X] is an isomorphism ([Tra],[Swan]). We explain here the elementary constructive proof of this result. By Thierry Coquand in [Coq]. The method is: first give a classical proof as elementary as possible, next remove “purely ideal” objects (whose construction uses Choice Axiom) and Third Excluded Middle Principle (TEM). Here the purely ideal object is a minimal prime, and TEM is used in a proof by contradiction. 4

The meaning of the theorem is a concrete meaning: if the ring is seminormal and if you have a matrix P (X) representing an element of Pic A[X], you want to find a similitude between P (X) and P (0). Now the question is the following: does the proof by contradiction using a minimal prime is a miracle obtained by classical mathematics? If not, does the classical proof hides a computation?

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A general deciphering method is used in order to transform the classical argument in a constructive one. Replace purely ideal objects (here, a generic minimal prime) by finite approximations of these objects, and verify that the classical argument can be seen as a computation concerning these finite approximations. The constructive proof of Thierry Coquand is an illustration of the fact that a kind of Hilbert’s programme for abstract algebra works in practice.

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A citation of Henri Poincar´ e Quant ` a moi je proposerais de s’en tenir aux r` egles suivantes: 1. Ne jamais envisager que des objets susceptibles d’ˆ etre d´ efinis en un nombre fini de mots; 2. Ne jamais perdre de vue que toute proposition sur l’infini doit ˆ etre la traduction, l’´ enonc´ e abr´ eg´ e de propositions sur le fini; ´ viter les classifications et les d´ 3. E efinitions non pr´ edicatives.

Henri Poincar´ e, in La logique de l’infini (Revue de M´ etaphysique et de Morale 1909). R´ e´ edit´ e in Derni` eres pens´ ees, Flammarion.

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What is a seminormal ring? A discrete domain A is seminormal if when b2 = c3 6= 0, then the element a = b/c of the fraction field is in fact in A. Remark that a3 = b and a2 = c. In particular a normal domain is seminormal. An arbitrary commutative ring A is seminormal if when b2 = c3 there exists a ∈ A such that a3 = b and a2 = c. This implies that A is reduced: if b2 = 0 then b2 = 03, whence an a ∈ A with a3 = b and a2 = 0, so b = 0. In a ring if x2 = y 2 and x3 = y 3 then: (x − y)3 = (x3 − y 3) − 3(x2y − y 2x) = 3(x3 − y 3) = 0. So: In a reduced ring x2 = y 2 and x3 = y 3 imply x = y. Consequently if an a such that a3 = b and a2 = c exists, it is always unique. 8

The Picard group of the ring A A finitely generated projective module is a module M isomorphic to a direct summand in a free module of finite rank: M ⊕ M 0 ' Am. Equivalently, it is a module isomorphic to the image of a projection matrix. An A-linear map ψ : M → N between finitely generated projective modules with M ⊕ M 0 ' Am and N ⊕ N 0 ' An can be represented by e ψe : Am → An defined by ψ(x ⊕ x0) = ψ(x). In other words a finitely generated projective module over A is a square idempotent matrix P with coefficients in A, and a morphism from P to Q is a matrix H such that QH = H = HP . 9

Projective modules of rank 1 A finitely generated projective module given by an idempotent matrix Vk+1 k P is of rank k if det(I + T P ) = (1 + T ) . This implies P = 0. A projective module of constant rank 1 is given by a matrix P V such that 2 P = 0 (all 2 by 2 minors are zero) and Tr(P ) = 1 (this implies P 2 = P ). Isomorphism classes of rank 1 projective modules form an abelian group with the law inherited from ⊗, this is the PICARD GROUP Pic A. The “inverse” is given by the transposed matrix tP , which defines the dual module. 10

Freeness criterium Lemma 1. Let P ∈ An×n a projection matrix of rank 1. T.F.A.E. 1. P has a free image 2. There exists a column vector f and a row vector g such that gf = 1 and f g = P . Moreover in this case f and g are unique, up to the multiplication by a unit, under the sole condition f g = P . 

3. Matrices

   

0 0 ... 0

0

··· P

0





   

  In+1,1 :=  

and

1 0 ... 0

0

···

0

    

0n,n

are similar. 11

Pic Ev0

Pic j

Pic A −→ Pic A[X1, . . . , X`] −→ Pic A Ev0 ◦ j = Id. So we have the following equivalence: Pic j is surjective (i.e., any projective module of constant rank 1 over A[X1, . . . , X`] = A[X] is isomorphic to a module obtained by a scalar extension of a projective module of constant rank 1 over A) if and only if

Pic Ev0 is injective (i.e. any projective module of constant rank 1 over A[X] which becomes a free module over A through Ev0, is free over A[X]). In this case one writes Pic A = Pic A[X]. 12

Theorem 2. (Traverso-Swan) For a reduced ring and ` ≥ 1, Pic A = Pic A[X1, . . . , X`] if and only if A is seminormal.

Computational meaning Theorem 3. (Traverso-Swan) For a reduced ring and ` ≥ 1 t.f.a.e. 1. A is seminormal. 2. For any projection matrix of rank 1, P = (mij ) ∈ A[X]n×n such that   1 01,n−1   P (0, . . . , 0) =   0n−1,1

0n−1,n−1

   = In,1 , 

there exist f1, . . . , fn, g1, . . . , gn ∈ A[X] such that figj = mij for all i, j. 13

The condition is necessary I.e., if A is reduced and Pic A = Pic A[X] then A is seminormal. Schanuel Example. Let b, c ∈ A reduced with b2 = c3. Let B = A[a] = A + aA a reduced ring containing A with a3 = b, a2 = c. One considers f1 = 1 + aX, f2 = cX 2 = g2 and g1 = (1 − aX)(1 + cX 2). We have f1g1 + f2g2 = 1. So the matrix M (X) = (figj ) is idempotent of rank 1. Its coefficients are in A[X] and M (0) = I2,1. Its image is free over B[X]. If it is free over A[X] there exist fi0’s and gj0 ’s in A[X] with fi0gj0 = figj . By unicity fi0 = ufi with u invertible in A[X], which implies invertible in A. With i = 1 one gets a ∈ A. 

NB: One can take B = A[T ]

.D

E 2 3 T − c, T − b . red

If an a is already in A, we get by unicity B = A. 14

If A is a gcd domain, Pic A = {1} = Pic A[X] Recall that if A is a gcd domain, then so is the polynomial ring A[X]. So we have to show that Pic A = {1} . Proof Let P = (mi,j ) be an idempotent matrix of rank 1. W.l.o.g. m1,1 is regular. Let f the gcd of the elements in the first row. We have m1,j = f gj with the gcd of the gj ’s equal to 1. Since f is regular and m1,1mi,j = m1,j mi,1 one gets g1mi,j = mi,1gj . So g1 divides all the mi,1gj , this implies it divides their gcd mi,1. One writes mi,1 = g1fi. Since g1f1 = m1,1 = f g1 we get f1 = f . Finally the equality m1,1mi,j = m1,j mi,1 gives f1g1mi,j = f1gj g1fi and mi,j = figj . 15

If A is integrally closed, Pic A = Pic A[X] Proof Let P (X) = (mi,j (X))i,j=1,...,n an idempotent matrix of rank 1 with P (0) = In,1. Let K be the fraction field of A. Over K[X] the module Im P (X) is free. So there exist f = (f1(X), . . . , fn(X)) and g = (g1(X), . . . , gn(X)) in K[X]n such that mi,j = figj for all i, j. We can assume f1(0) = g1(0) = 1. Since f1gj = m1,j Kronecker’s theorem implies that the coefficients of gj ’s are integral over the ring generated by the coefficients of m1,j ’s. So the gj ’s are in A[X]. Samething for the fi’s. 16

If A is a seminormal domain, Pic A = Pic A[X] Proof Let P (X) = (mi,j (X))i,j=1,...,n an idempotent matrix of rank 1 with P (0) = In,1. There exist f1(X), . . . , fn(X), g1(X), . . . , gn(X) in K[X]n such that mi,j = figj pour tous i, j. Moreover f1(0) = g1(0) = 1. Let B be the subring of K generated by A and by the coefficients of the polynomials fi and gj . Kronecker’s theorem implies that B is a finite extension of A (i.e., B is a finitely generated A-module). We have to show that A = B. Let us denote a the conductor of A in B, i.e., the set {x ∈ B | xB ⊆ A}. It is an ideal of A and B. We have to show that a = h1i. 17

If A is a seminormal domain, then Pic A = Pic A[X]

Lemma 4. If A ⊆ B, A seminormal and B reduced, then the conductor a of A in B is a radical ideal of B.

Proof We have to show that if u ∈ B and u2 ∈ a then u ∈ a. Let c ∈ B, we have to show that uc ∈ A. We know that u2c2 ∈ A and u3c3 = u2(uc3) ∈ A since u2 ∈ a. Since (u3c3)2 = (u2c2)3 there exists a ∈ A such that a2 = (uc)2 and a3 = (uc)3. As B is reduced this implies a = uc, and uc ∈ A.

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If A is a seminormal domain, then Pic A = Pic A[X]

Lemma 5. Let A ⊆ B, B = A[c1, . . . , cq ] reduced finite over A. Let a the conductor of A in B. √ Assume that a is a radical ideal, i.e. a = a. Then a = {x ∈ A | xc1, . . . , xcq ∈ A}.

Proof Indeed if xci ∈ A then x`c`i ∈ A for all `. So for N great enough xN y ∈ A for all y ∈ B. So x is in the radical of a. (if d is a bound for the equations of integral dependance of the ci’s over A, one can take N = (d − 1)q). 19

If A is a seminormal domain, then Pic A = Pic A[X]

End of the proof, in classical mathematics. Assume a 6= h1i. Let C = A/a ⊆ B/a = C0. Let p be a minimal prime of C, P the corresponding ideal of A, and S = C \ p. Since p is a minimal prime and since C is reduced S −1C = L is a field, contained in the reduced ring S −1C0 = L0. If x is an object defined over A we note x what it becomes through the change of ring A → L0. The module M is defined by the matrix P whose coefficients are in L[X]. Since L is a field, Im P is free over L[X]. This implies by unicity (Lemma 1), and since f1(0) = g1(0) = 1, that the coefficients of fi and gj are in L[X]. This means there exists s ∈ A \ P such that the sfi’s and sgj ’s have their coefficients in A. By Lemma 5, this implies s ∈ a. This is absurd. 20

Making the proof constructive (domain case) Is there a computation hidden in the classical proof? Proof by contradiction? If the ring A/a is not trivial it contains a minimal prime, and so on. . . , and finally the ring is trivial. But up side down! the proof maybe proves directly that A/a is trivial. What about the minimal prime? Run the argument dynamically . . . 21

Making the proof constructive, 2.

What means “run the argument dynamically?” Since the proof is finite, it uses only a finite amount of information concerning the “purely ideal” minimal prime, and this minimal prime can be replaced by a finite number of finite approximations, concerning all the cases that can occur in the proof. Remind Poincar´ e’s citation about infinite ideal objects. What is a finite approximation for the localization in a minimal prime? it is the localization in ONE element!

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Making the proof constructive, 3.

The contraposed form of the proposition (A) Let C be a reduced ring and P a projective module of constant rank 1 over C[X]. If C is not trivial, there must exist a nontrivial localization S −1C for which P becomes free. is the following (B) Let C be a reduced ring and P a projective module of constant rank 1 over C[X]. If each localization S −1C of C for which P becomes free is trivial, then C itself is trivial. The classical proof used (A) with C = A/a for proving 1 ∈ a. But (B) is sufficient! 23

Making the proof constructive, 4.

Here is THE crucial lemma which eliminates the minimal prime. It uses only localizations in ONE element (inversing finitely many elements ammounts to inverse their product).

Lemma 6. Let C be a reduced ring and P = (mi,j ) ∈ C[X]n×n an idempotent matrix of rank 1 such that P (0) = In,1. Assume the following is true: ∀a ∈ C, if

Im P

is

free

over

C[1/a][X], then a = 0.

Then C is trivial, i.e., 1 = 0 in C. 24

Using THE lemma, in order to make the proof constructive We show here that the end of the proof page 20 becomes constructive if we have THE lemma. We have the reduced ring C = A/a ⊆ B/a = C0. It is sufficient to prove that the ring C and the matrix P mod a satisfy the hypotheses of THE lemma. Let a ∈ A such that Im P is free over C[1/a][X]. Let C[1/a] = L ⊆ C0[1/a] = L0. If x is an objet which is defined over A let us denote x what it becomes through the change of ring A → L0. The module M is free over L[X] and this implies by unicity (Lemma 1), since L is reduced, that the fi’s and gj ’s are in L[X]. This means there exists N ∈ N such that the aN fi’s and aN gj ’s have their coefficients in A. Using Lemma 5, this implies a ∈ a, so a = 0 in C. 25

Proof of THE lemma A classical proof should be: Assume C is nontrivial and let p be a minimal prime. Since C is reduced, Cp is a field. So Im P is free over Cp[X]. This means mi,j = figj where the fi’s and gj ’ are in Cp[X]. So there exists a ∈ / p such that the fi’s and gj ’s are in C[1/a][X]. So a = 0: a contradiction. This proof can be reread in a dynamical way, replacing the minimal prime by finite approximations.

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Proof of THE lemma, 2.

If the ring C were a field, the fi’s and gj ’s could be computed by an algorithm which is deduced from the constructive proofs which were given for the field case. This algorithm uses the disjunction “a is zero or invertible”, for elements a which are computed by the algorithm from the coefficients of polynomials mi,j . Since C is only a reduced ring, without equality test to 0 nor invertibility test, the algorithm for the fields, when one runs it for C, has to be replaced by a tree in which one opens two branches each times a question “is a zero or invertible?” is asked by the algorithm. 27

Proof of THE lemma, 3.

So we get a tree, a HUGE tree, but finite tree. Assume we have put always the branch “a invertible” on the left, and the branch “a = 0 on the right”. We want to show that C = 0. Consider the leftmost branch. At the corresponding leave we have a1, . . . , an invertible and the module P is free over C[1/(a1 · · · an)][X]. Conclusion: in the ring C, we have a1 · · · an = 0. 28

Proof of THE lemma, 4.

Go up one step. In the ring C[1/(a1 · · · an−1)], we have an = 0. So we see the computation under the node: a1, . . . , an−1 invertible, an = 0. We consider the leftmost branch under this node. At the leave we have made invertible the elements a1, . . . , an−1, b1, . . . , bk (if k = 0, bk = an−1). The module P is free over C[1/(a1 · · · bk )][X]. Conclusion: in the ring C, we have a1 · · · bk = 0. And so on. At the end of this process all branches are dead and P is free over C[X] = C[1/1][X]. So 1 = 0. This is the end of the constructive proof in the domain case! 29

What about the case A reduced? In classical mathematics we could say: any reduced ring is a subring of a product of fields. E.g., the product of localizations at minimal primes. Then one makes a photocpy of the proof given when “A domain (subring of a field K)” for the case “A reduced, subring of K, product of fields”.

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What about the case A reduced? 2

Quasi inverses. In constructive mathematics, the ring K (product of fields) is not accessible. But one can construct the subring L ⊆ K generated by A and the quasi inverses of elements of A. In a commuative ring C, two elements a and b are said quasi inverses if one has: a2b = a,

b2a = b

On says that b is the quasi inverse of a (it is uniquely determined by a). 31

What about the case A reduced? 3

A ring in which every element has a quasi inverse is called Von Neuman regular, or absolutely flat, or zero-dimensional reduced. These quasi fields are very similar to fields. One can now make a photocpy of the proof given when “A domain (subring of a field K)” for the case “A reduced, subring of L, Von Neuman regular”.

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