Revisiting Zariski Main Theorem from a constructive point of view

# Revisiting Zariski Main Theorem from a constructive point of view

Alonso M. E., Coquand T., Lombardi H.
January 2016
###### Abstract

This paper deals with the Peskine version of Zariski Main Theorem published in 1965 and discusses some applications. It is written in the style of Bishop’s constructive mathematics. Being constructive, each proof in this paper can be interpreted as an algorithm for constructing explicitly the conclusion from the hypothesis. The main non-constructive argument in the proof of Peskine is the use of minimal prime ideals. Essentially we substitute this point by two dynamical arguments; one about gcd’s, using subresultants, and another using our notion of strong transcendence. In particular we obtain algorithmic versions for the Multivariate Hensel Lemma and the structure theorem of quasi-finite algebras.

Note.

This paper appeared in Journal of Algebra 406, (2014), 46–68

Here, we have fixed two typos.

At the end of the proof of Proposition 4.8, we write

and

In the proof of Lemma 4.9. In line 7 of the proof we write:

We have

Keywords. Zariski Main Theorem, Multivariate Hensel Lemma, Quasi finite algebras, Constructive Mathematics

## 1 Introduction

The paper is written in the style of Bishop’s constructive mathematics, i.e. mathematics with intuitionistic logic (see [4, 5, 14, 16]).

A partial realization of Hilbert’s program has recently proved successful in commutative algebra, see e.g., [1, 6, 7, 8, 9, 11, 14, 19] and  with references therein, and this paper is a new piece of realization of this program.

We were mainly interested in an algorithm for the Multivariate Hensel Lemma (MHL for short). Let us see what is the aim of the computation on a simple example.

We consider the local ring , . We take the equations

 −a+x+bxy+2bx2=0,      −b+y+ax2+axy+by2=0

and we want to compute a solution of the system in the henselization of . In other words, we have to find a Hensel equation (i.e.  monic, and ) such that, when adding the Hensel zero of to we are able to compute and .

Surprinsingly there is no direct proof of the result. Moreover elementary elimination techniques do not work on the above example. So we have to rely on the proof of MHL via the so called Zariski Main Theorem (ZMT for short), as for example in . Note that there are many versions of ZMT (e.g. [13, 20]) and we are interested in the ZMT à la Peskine as in .

We will give a solution of the above example in section 4.4.

This paper deals with the Peskine proof of ZMT published in 1965  and discusses some applications. Peskine statement is purely algebraic avoiding any hypothesis of noetherianity. The argument we give for Theorem 1.3 follows rather closely Peskine’s proof. The main non-constructive argument in the proof of Peskine is the use of minimal prime ideals. Note that the existence of minimal prime ideals in commutative rings is known to be equivalent to Choice Axiom. Essentially we substitute this point by two dynamical arguments; one about gcd’s, using subresultants, section 2.3, proof of Proposition 2.22, and another using our notion of strong transcendence, section 2.2 (in classical mathematics: to be transcendent over all residual fields).

In sections 4 and 5, we give a constructive treatment of two classical applications of ZMT: the Multivariate Hensel Lemma, and structure theorem of quasi finite algebras.

Being constructive, each proof in this paper can be interpreted as an algorithm for constructing explicitly the conclusion from the hypothesis.

###### Theorem 1.1

(ZMT à la Peskine, particular case)
Let be a ring, a detachable maximal ideal of and . If is an extension of such that is a finite -algebra then there exists such that , , …, are integral over .

In  an equivalent formulation (Proposition 13.4) of Peskine version of the Zariski Main Theorem can be written as the following lemma.

Proposition. Let be a residually discrete local ring and . If is an extension of such that , is integrally closed in and has a nontrivial zero-dimensional component as a -algebra, then .

The last hypothesis can be given in a concrete way: there exists an idempotent of such that is a nontrivial finite -algebra. This means that the residual variety has at least one isolated point.

The following corollary of Theorem 1.1 is a weakened form of the previous proposition.

###### Corollary 1.2

Let be a residually discrete local ring and . If is an extension of such that , is integrally closed in and is a finite -algebra then .

###### Proof.

By Theorem 1.1 we find such that and . We have then and hence is invertible in . Hence are in and . ∎

Remark. The hypothesis that is integrally closed in is necessary, even if we weaken the conclusion to “ is finite over ”. Let be a DVR with , the ring is finitely generated over , and , but is not finite over . If is the integral closure of in , we cannot apply Corollary 1.2 with replacing because is not a maximal ideal of (in fact ).

In fact we shall prove a slightly more general version of Theorem 1.1, without assuming to be a detachable maximal ideal.

###### Theorem 1.3

(ZMT à la Peskine, variant)
Let be a ring with an ideal and be an extension of such that is a finite -algebra, then there exists such that , , …, are integral over .

Remark. In fact, the hypothesis that the morphism is injective is not necessary: it is always possible to replace and by their images in , and the conclusion remains the same.

###### Corollary 1.4

Let be a ring with an ideal and be an extension of such that is a finite -algebra, then there exists a finite extension of inside and such that .

###### Proof.

Take . ∎

We shall also give a proof of the following “global form” of Zariski Main Theorem.

Theorem 5.41 (ZMT à la Raynaud, )
Let be rings such that the inclusion morphism is zero dimensional (in other words, is quasi-finite over ). Let be the integral closure of in . Then there exist elements in , comaximal in , such that all .
In particular for each , . Moreover letting which is finite over , we get also for each .

We give now the plan of the paper.

In section 2 we give some preliminary results and the proof of a Peskine “crucial lemma”.

In section 3 we give the constructive proof for Theorem 1.3.

In section 4 we give a constructive proof for the Multivariate Hensel Lemma (Theorem 4.32). A usual variant is the following corollary.

Corollary 4.33 Let be a Henselian local ring. Assume that a polynomial system in has residually a simple zero at . Then the system has a (unique) solution in with coordinates in .

Section 5 is devoted to structure theorem of quasi-finite algebras: we give a proof of Theorem 5.41, moreover Proposition 5.40 explains the constructive content of the hypothesis in Theorem 5.41.

Acknowledgements. First and third authors are partially supported by Spanish GR MTM2011-22435. Third author thanks the Computer Science and Engineering Department at University of Gothenburg for several invitations. This article has been discussed in the course of a researching stay of the first author at the Department of Mathematics of the University of Franche-Comté. She thanks the Department for its kind invitation.

## 2 Peskine crucial lemma

In this section we give a constructive proof of a crucial lemma in the proof of Peskine. This is Proposition 2.22 in the following.

### 2.1 Basic tools for computing integral elements

Let be rings and let be an ideal of . We say that is integral over if and only if it satisfies a relation with in . The integral closure of in is the ideal of elements of that are integral over .

###### Lemma 2.5

(Lying Over, concrete form)

1. If is integral over then the integral closure of in is .

As a consequence .

2. If is integral over and then .

###### Proof.

1. See  Lemma 5.14.

2. Use item 1 with and . ∎

Algorithm: Let be in : , with and . Let be generators of as an -module. The multiplication by in is expressed on by a matrix with coefficients in . The characteristic polynomial of is with ’s , and .

###### Definition 2.6

We denote (or ) the ideal of generated by the coefficients of ( is called the -content ideal of in ).

###### Lemma 2.7

(Kronecker)

Let where is the subring generated by .

1. (simple form) If divides in then are integral over (more precisely they are integral over the ideal generated by in ).

2. (general form) If in , a coefficient of and a coefficient of then is integral over the ideal generated by in .

3. (Gauss-Joyal) If in then .

###### Proof.

1. Considering the splitting algebra of over , we can assume . We have then integral over and hence also since they are (symmetric) polynomials in .

2. This is deduced from 1 by homogeneization arguments.

3. This is an immediate consequence of 2.

###### Lemma 2.8

If and satisfies an equation with then is integral over .

###### Lemma 2.9

(see ) Let and satisfies an equation with . We take

 un=an,un−1=unx+an−1,……,u0=u1x+a0=0

We get the following results.

1. and are integral over and as ideals of .

2. Let be an ideal of s.t.  and is integral over then there exists s.t.  and are integral over .

###### Proof.

1. Lemma 2.8 shows that is integral over . It follows that is integral over . We have then

 un−1xn−1+an−2xn−2+⋯+a0=0

so that, again by Lemma 2.8, is integral over and so over . In this way, we get that , , , , …, , are all integral over .

2. Let be the image of in . So with integral over . Item 1. shows that . Lying Over item 2 gives with ’s . Let , then and are clearly integral over . ∎

###### Lemma 2.10
1. If is integral over and is a monic polynomial in such that is in then there exists in such that is integral over .

2. If is integral over and is a polynomial in such that is in then there exists in and such that is integral over .

###### Proof.

1. We write in . We do the Euclidian division of by and get . We can then write . This shows that we have in and hence that is integral over . Since is integral over we get that is integral over and hence over .

2. We have an equation for of the form . Let be the greatest exponent of in this expression. By multiplying by we get an equality of the form

 aℓtn+q1(ax)tn−1+⋯+qn(ax)=0

and hence, by Lemma 2.8, is integral over . Sowe we have such that is integral over for all .
We write and by multiplying by a suitable power of we get an with and monic. We can then apply item 1.

###### Corollary 2.11

If is integral over and is integrally closed in and then there exists such that .

Next lemma is a kind of glueing of integral extensions.

###### Lemma 2.12

Let and . If are integral over and integral over then for big enough and the elements are integral over .

###### Proof.

We write and . Let be the highest power of that appears in these expressions. We have that and are integral over and so over , and we take . ∎

### 2.2 Strong transcendence

Let be a -algebra and . We say that is strongly transcendent over in if for all and such that , we have (each time it is needed, stands for the image of in ).

Note that the definition strongly depends on and . Moreover from an equality in we deduce only that in .

###### Lemma 2.13
1. If is a -algebra, strongly transcendent over in and a monoid of , then is strongly transcendent over in .

2. If is a -algebra and is strongly transcendent over in and , then is strongly transcendent over in .

###### Lemma 2.14

If , a reduced -algebra, strongly transcendent over in and are integral over then .

###### Proof.

We have and for some in . So is a polynomial with constant coefficient and leading coefficient . Since , is transcendent over in , and is reduced, it follows that we have in . We get in

 (ux)Q1(ux)=0 with Q1(T)=(Q(T)−cℓ)/T.

Now we consider the reduced ring . In this ring is strongly transcendent over . We have in

 Q1(ux)=0 with Q1(0)=cℓ−1 and P(u)=0.

So in . Similarly we deduce in . So in and finally in . ∎

###### Lemma 2.15

If is a reduced -algebra and is strongly transcendent over in and and is integral over then is strongly transcendent over in .

###### Proof.

Assume an equality with and ’s in . Passing to we get with ’s integral over . So is integral over , and thus over too. So  and are integral over . By Lemma 2.14 in , so in . We finish by induction on . ∎

### 2.3 Crucial lemma

###### Context 2.16

We fix now the following context, which comes from Corollary 2.11: integral over of degree and integrally closed in . We define .

###### Lemma 2.17

(Context 2.16)
If we have if and only if .

###### Proof.

This is clear since all elements of can be written . ∎

###### Lemma 2.18

(Context 2.16)
If and and , then there exists such that .

###### Proof.

We have by Lemma 2.17

 (a0+⋯+akxk)u,(a0+⋯+akxk)ut,…,(a0+⋯+akxk)utn−1∈R[x].

All elements are integral over and is integrally closed in . Hence by Corollary 2.11 we find such that . ∎

We consider now the radical of in .

###### Corollary 2.19

(Context 2.16)
If and and , then

###### Proof.

We have such that . By Lemma 2.18 we have such that and hence . It follows that and so and we get successively . ∎

Summing up previous results in Context 2.16 and using the notion of strong transcendence.

###### Proposition 2.20

Assume with integral over and is integrally closed in . We take . If we take and , then is a reduced ring with a subring such that is integral over and is strongly transcendent over in .

###### Proof.

Clear. The last assertion comes from Corollary 2.19. ∎

###### Proposition 2.21

Assume that is a reduced ring with a subring such that is integral over and is strongly transcendent over in . Let be an ideal of such that . Then .
Equivalently, if is the localization of at the monoid , then is a trivial ring.

The proof is given after the crucial lemma.

###### Proposition 2.22

(crucial lemma)
If and is integrally closed in and is integral over and ideal of such that then mod.  where .

###### Proof.

This follows from Propositions 2.20 and 2.21. ∎

Here begins the proof of Proposition 2.21.

Since is integral over we get a -monic polynomial in s.t. . As by Lying Over we get a polynomial

 Q(X,T)=XnTn+μ1(X)Xn−1Tn−1+⋯+μn(X) with ∈μi(X)∈IC[X]

s.t. . We need now to prove Lemma 2.23.

###### Lemma 2.23

Assume , that is transcendent over and that divides , with and . Then is a trivial ring.

###### Proof.

Since is transcendent over we have that divides . By taking we see that divides . If is big enough we can apply Lemma 2.7 and conclude that all coefficients of are integral over . Since it follows that is integral over , and so is a trivial ring. ∎

We consider the ring , we compute the subresultants of and in and we show that they are all in , i.e.  has to divide in .

The conclusion follows then from Lemma 2.23 with the image of in and .

We use results about subresultants given in Lemma 2.24 (for the general theory of subresultants, see [3, Chapter 4]) We consider one such subresultant assuming that all previous subresultants have been shown to be . We can assume to be invertible, replacing by . We let be the leading coefficient of and we show . We write . Since divides we have that are integral over by Lemma 2.7. By Lemma 2.10, are in with integral over . By Corollary 2.15 and Lemmas 2.23 and 2.13, we have in and hence in .

Here the proof of Proposition 2.21 is finished.

###### Lemma 2.24

Let be a reduced ring, a monic polynomial of degree , and a bound for the degree of . Let a nonnegative integer. The subresultant of and in degree , denoted is a well defined polynomial of degree : it does not depend on . We let . Let us denote the coefficient of in . Then we have:

1. belongs to the ideal of ().

2. Let , . If for and is invertible, then:

for .

divides and in .

###### Proof.

1. This is a classical result.

2. Since the results are well known when is a field, the lemma follows by using the formal Nullstellensatz. ∎

## 3 Proof of ZMT

It is more convenient for a proof “by induction on ” to use the following version 3.25.

###### Theorem 3.25

(ZMT à la Peskine, general form, variant)
Let be a ring with an ideal and be a finite extension of such that is a finite -algebra, then there exists such that are integral over .

Here, the precise hypothesis is , with finite over . Clearly Theorems 1.3 and 3.25 are equivalent.

#### Case n=1

###### Proposition 3.26

Let be a ring with an ideal and be a finite extension of such that is a finite -algebra, then there exists such that are integral over .

###### Proof.

Let a monic polynomial s.t. . By Lying Over . This provides such that and . Apply Lemma 2.9, item 2 with . ∎

#### The induction step

###### Proposition 3.27

Let be a ring with an ideal , an extension of with in such that is integral over and in such that is in . There exist such that meets and are integral over .

###### Proof.

By Lying Over and we can assume as well that .

We apply Proposition 2.22 with the integral closure of in , , . We get an with , . We have . Since , and by Lying Over with ’s in .

We write with , written as in and . We have and .

Let . In we have and by Gauss-Joyal. Remark that implies that .

If is a bound for the degree of , by Lemma 2.9 we get integral over s.t. avec integral over and .

Finally we get

###### Corollary 3.28

Let be a ring with an ideal , an extension of with in such that is integral over , a monic polynomial and in such that is in . There exist such that meets and are integral over .

###### Proof.

Let , then is integral over . Applying Proposition 3.27 with instead of , we get such that meets and are integral over . We say that this implies ’s are integral over . If the integral dependance of over is given by a polynomial of degree and is of degree , multiplying the equation by , one gets an integral dependance equation of over

Now we can prove Theorem 3.25.

###### Proof.

We give the proof for , and .
The induction from to follows the same lines as the induction from to .

First we apply Proposition 3.26 with instead of , replacing . We get with and integral over .

Let be a monic polynomial such that . By Lying Over is integral over . We take for big enough such that is integral over . By Lying Over again is in .

We apply Corollary 3.28 with , , , replacing by . We get such that meets and are integral over . Since , . As is finite over , by Lying Over item 2 we have that for some polynomials with coefficients in . Let