Commutative Algebra/Hilbert's Nullstellensatz

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Zariski's lemma[edit]

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 occuring 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[edit]

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[edit]

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[edit]

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[edit]

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.