Commutative Algebra/Hilbert's Nullstellensatz

Zariski's lemma

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Definition 24.1 (Finitely generated algebra):

Let   be a ring. An  -algebra   is called finitely generated, iff there are elements   such that   is already all of  ; that is  .

  being a finitely generated  -algebra thus means that we may write any element of   as a polynomial   for a certain   (where polynomials are evaluated as explained in chapter 21).

Lemma 24.2 (Artin–Tate):

Let   be ring extensions such that   is a Noetherian ring, and   is finitely generated as an  -module and also finitely generated as an  -algebra. Then   is finitely generated as an  -algebra.

Proof:

Since   is finitely generated as an  -module, there exist   such that   as an  -module. Further, since   is finitely generated as  -algebra, we find   such that   equals  . Now by the generating property of the  , we may determine suitable coefficients   (where   ranges in   and   in  ) such that

 ,  .

Furthermore, there exist suitable   ( ) such that

 .

We define  ; this notation shall mean:   is the algebra generated by all the elements  . Since the algebra operations of   are the ones induced by its ring operations,  , being a subalgebra, is a subring of  . Furthermore,   and  . Since   is a Noetherian ring,   is also Noetherian by theorem 16.?.

We claim that   is even finitely generated as an  -module. Indeed, if any element   is given, we may write it as a polynomial in the  . Using  , multiplying everything out, and then using   repeatedly, we can write this polynomial as a linear combination of the   with coefficients all in  . This proves that indeed,   is finitely generated as an  -module. Hence,   is Noetherian as an  -module.

Therefore,  , being a submodule of   as  -module, is finitely generated as an  -module. We claim that   is finitely generated as an  -algebra. To this end, assume we are given a set of generators   of   as an  -module. Any element   can be written

 ,  .

Each of the   is a polynomial in the generators of   (that is, the elements  ) with coefficients in  . Inserting this, we see that   is a polynomial in the elements   with coefficients in  . But this implies the claim. 

Theorem 24.3 (Zariski's lemma):

Let   be a field extension of a field  . Assume that for some   in  ,   is a field. Then every   is algebraic over  .

Proof 1 (Azarang 2015):


Before giving the proof of the lemma, we recall the following two well-known facts.

Fact 1. If a field   is integral over a subdomain  , then   is a field.

Fact 2. If   is any principal ideal domain (or just a UFD) with infinitely many (non-associate) prime elements, then its field of fractions is not a finitely generated  -algebra.

Proof of the Lemma: We use induction on   for arbitrary fields   and  . For   the assertion is clear. Let us assume that   and the lemma is true for less than  . Now to show it for  , one may assume that one of  , say  , is not algebraic over   and since   is a field, by induction hypothesis, we infer   are all algebraic over  . This implies that there are polynomials   such that all  's are integral over the domain  . Since   is integral over  , by Fact 1,   is a field. Consequently,  , which contradicts Fact 2.



Proof 2 (Artin–Tate):

If all of the generators of   over   are algebraic over  , the last paragraph of the preceding proof shows that   is a finite field extension of  . Hence, we only have to consider the case where at least one of the generators of   over   is transcendental over  .

Indeed, assume that  . By reordering, we may assume that   are transcendental over   ( ) and   are algebraic over  . We have  , and furthermore   since   is a field extension of   containing all the elements  . Hence,  .

Since all the   are algebraic over  , they are also algebraic over  . Assume that there exists a polynomial   such that  . Then   is algebraic over  ; for, the part of the monomials not being a power of   may be seen as coefficients within that field. Hence, we may lower   by one and still obtain that   are algebraic over  . Repetition of this process eventually terminates, or otherwise   would be algebraic over  , and   would be a finite tower of algebraic extensions ( ,   and so on) and thus a finite field extension.

Therefore, we may assume that   are algebraically independent over  . In this case, the map

 

is an isomorphism (it is a homomorphism, surjective and injective), and hence,   is a unique factorisation domain (since   is).

Now set  . Then  , and   is finitely generated as an  -algebra and finitely generated as a  -module (since it is a finite field extension of  ). Therefore, by lemma 24.2,   is finitely generated as an  -algebra. Let

 

be generators of   as  -algebra. Let   be all the primes occurring in the (unique) prime factorisations of  . Now   contains an infinite number of primes. This is seen as follows.

Assume   were the only primes of  . Since we have prime factorisation, the element   is divisible by at least one of  , say  . This means

 

for a certain  , which is absurd, since applying the inverse of the above isomorphism to  , we find that   is mapped to  , but the right hand side has strictly positive degree.

Hence, we may pick   prime. Then   can not be written as a polynomial in terms of the generators, but is nonetheless contained within  . This is a contradiction. 

Proof 3 (using Noether normalisation):

According to Noether's normalisation lemma for fields, we may pick   algebraically independent over   such that   is a finitely generated  -module. Let   be elements of   that generate   as an  -module. Then according to theorem 21.10 3.   1., the generators are all integral over  , and since the integral elements form a ring,   is integral over  . Hence,   is a field by theorem 21.11. But if  , then the   being algebraically independent means that the homomorphism

 

is in fact an isomorphism, whence   is not a field, contradiction. Thus,  , and hence   is finitely generated as an  -module. This implies that we have a finite field extension; all elements of   are finite  -linear combinations of certain generators. 

Hilbert's Nullstellensatz

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There are several closely related results bearing the name Hilbert's Nullstellensatz. We shall state and prove the ones commonly found in the literature. These are the "weak form", the "common roots form" and the "strong form". The result that Hilbert originally proved was the strong form.

Weak form

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The formulation and proof of the weak form of Hilbert's Nullstellensatz are naturally preceded by the following lemma.

Lemma 24.5:

Let   be any field. For any maximal ideal  , the field   is a finite field extension of the field  . In particular, if   is algebraically closed (and thus has no proper finite field extensions), then  .

Proof 1 (using Zariski's lemma):

  is a finitely generated  -algebra, where all the operations are induced by the ring structure of  ; this is because the set   constitutes a set of generators, since every element in   can be written as polynomials in those elements over  . Therefore, Zariski's lemma implies that   is a finite field extension of the field  . 

Proof 2 (using Jacobson rings):

We proceed by induction on  .

The case   follows by noting that   is a principal ideal domain (as an Euclidean domain) and hence, if   is a (maximal) ideal, then   for a suitable  . Now   is a field if   is maximal; we claim that it is a finite field extension of the field  . Indeed, as basis elements we may take  , where   is the degree of the generating polynomial of the maximal ideal  . Any element of   can thus be expressed as linear combination of these basis elements, since the relation

  (where  )

allows us to express monomials of degree   in terms of smaller ones.

Assume now the case   is proven. Let   be a maximal ideal. According to Jacobson's first criterion,   is a Jacobson ring (since   is, being a field). Now   and hence   is a maximal ideal of  . Thus, Goldman's second criterion asserts that   is a maximal ideal of  . Thus,   is a field, and, by the induction hypothesis, a finite field extension of  .

We define the ideal  . The following map is manifestly an isomorphism:

 

This map sends   to   (and, being an isomorphism, vice versa).

Furthermore, since  , the ideal   is maximal in  . Hence,   is maximal in   and thus   is a field. By the case   it is a finite field extension of the field  .


In general, any proper ideal of  , where   is a field, does not contain any constants (apart from zero), for else it would contain a unit and thus be equal to the whole of  . This applies, in particular, to all maximal ideals of  . Thus, elements of   of the form   are distinct for pairwise distinct  . By definition of addition and multiplication of residue class rings, this implies that we have an isomorphism of rings (and thus, of fields)

 .

Hence, in the case that   is algebraically closed, the above lemma implies   via that isomorphism.

Theorem 24.6 (Hilbert's Nullstellensatz, weak form):

Let   be an algebraically closed field. For any  , set

 ;

according to lemma 21.12,   is a maximal ideal.

The claim of the weak Hilbert's Nullstellensatz is this: Every maximal ideal   has the form   for a suitable  .

Proof:

Let   be any maximal ideal of  . According to the preceding lemma, and since   is algebraically closed, we have   via an isomorphism that sends elements of the type   to  . Now this isomorphism must send any element of the type   to some element   of  . But further, the element   is sent to  . Since we have an isomorphism (in particular injectivity), we have  . Thus   for suitable  . Since the ideal   is maximal (lemma 21.12), we have equality:  . 

Common roots form

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Theorem 24.7 (Hilbert's Nullstellensatz, common root form):

Let   be an algebraically closed field and let  . If

 ,

then there exists   such that  .

Proof:

This follows from the weak form, since   is contained within some maximal ideal  , which by the weak form has the form   for suitable   and hence  ; in particular,  , that is,   is a common root of  . 

Strong form

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Theorem 24.8 (Hilbert's Nullstellensatz, strong form):

Let   be an algebraically closed field. If   is an arbitrary ideal, then

 ;

recall:   is the radical of  .

In particular, if   is a radical ideal (that is,  ), then

 .

Note that together with the rule

 

for any algebraic set   (that was established in chapter 22), this establishes a bijective correspondence between radical ideals of   and algebraic sets in  , given by the function

 

and inverse

 .

Proof 1 (using Jacobson rings):

Certainly, a field is a Jacobson ring. Furthermore, from Goldman's first criterion (theorem 14.4) we may infer that   is a Jacobson ring as well. Let now   be a polynomial vanishing at all of  , and let   be any maximal ideal of   that contains  . By the weak Nullstellensatz,   has the form   for a suitable  .

Now we have  , since any polynomial in   can be written as a  -linear combination of the generators  . Hence,   is not all of  ; due to the constant functions, only the empty set has this ideal of vanishing. This, in combination with the fact that   and the maximality of   implies  .

Furthermore,  , and hence  . Therefore,  .

Since   was arbitrary,   is thus contained in all maximal ideals containing   and hence, since   is Jacobson,  . However, the other direction   is easy to see (we will prove this in the first paragraph of the next proof; there is no need to repeat the same argument in two proofs). Thus,  . 

Proof 2 (Rabinowitsch trick):

First we note  : Indeed, if  , then   for all  . Hence also   for all   since a field does not have nilpotent elements except zero (in fact, not even zero divisors). This implies  .

  is the longer direction. Note that any field is Noetherian, and thus, by Hilbert's basis theorem, so is  . Hence,  , being an ideal of  , is finitely generated. Write

 .

Let  . Consider the polynomial ring  , which is augmented by an additional variable. In that ring, consider the polynomial  . The polynomials   have no common zero (where the polynomials   are seen as polynomials in the variables   by the way of  ), since if all the polynomials   are zero at   (where the variable   does not matter for the evaluation of  ), then so is  . Hence, in this case,  .

Now we may apply the common roots form of the Nullstellensatz for the case of   variables. The polynomials   have no common zero, and therefore, the common roots form Nullstellensatz implies that the ideal   must be all of  . In particular, we can find   such that

 .

Passing to the field of rational functions  , we may insert   for   (recall that we assumed  ) to obtain

 ,

where we left out the variables of   so that it still fits on the screen. Now  , whence

 .

Multiplying this equation by an appropriate power of  , call it  , sufficiently large such that we clear out all denominators, and noting that the last variable does not matter for  , yields that   equals an  -linear combination of   and is thus contained within  . Hence,  . 

Note how Yuri Rainich ("Rabinowitsch") may have found this trick. Perhaps he realized that the weak Nullstellensatz is a claim for arbitrary  , and for the proof of the strong Nullstellensatz, we can do one   at a time, using the infinitude of cases of the common roots form Nullstellensatz. That is, compared to a particular dimensional case in the strong Nullstellensatz, the infinitude of cases for the common roots form Nullstellensatz are not so weak at all, despite the common roots form being a consequence of the weak Nullstellensatz. This could have given Rainich evidence that using more cases, one obtains a stronger tool. And indeed, it worked out.

 
A diagram depicting the different paths to Hilbert's Nullstellensatz covered in this wikibook.