Health warning: this post is part of a more advanced series on commutative algebra. It may be a little tricky for the layman to understand!
David Hilbert was perhaps the greatest mathematicians of the late 19th century. Much of his work laid the foundations for our modern study of commutative algebra. In doing so, he was sometimes said to have killed the study of invariants by solving the central problem in the field. In this post I’ll give a sketch of how he did so.
Motivated by Galois Theory we ask the following question. Given a polynomial ring and a group acting on as a -automorphism, what are the elements of that are invariant under the action of ? Following familiar notation we denote this set and note that it certainly forms a subalgebra of .
In the late 19th century it was found that could be described fully by a finite set of generators for several suggestive special cases of . It soon became clear that the fundamental problem of invariant theory was to find necessary and sufficient conditions for to be finitely generated. Hilbert’s contribution was an incredibly general sufficient condition, as we shall soon see.
To begin with we shall recall the alternative definition of a Noetherian ring. It is a standard proof that this definition is equivalent to that which invokes the ascending chain condition on ideals. As an aside, also recall that the ascending chain condition can be restated by saying that every nonempty collection of ideals has a maximal element.
Definition A.1 A ring is Noetherian if every ideal of is finitely generated.
We shall also recall without proof Hilbert’s Basis Theorem, and draw an easy corollary.
Theorem A.2 If Noetherian then Noetherian.
Corollary A.3 If is a finitely generated algebra over , with Noetherian, then Noetherian.
Proof We’ll first show that any homomorphic image of is Noetherian. Let be an ideal in the image under than homomorphism . Then an ideal in . Indeed if and then so . Hence finitely generated, so certainly finitely generated, by the images of the generators of .
Now we’ll prove the corollary. Since is a finitely generated algebra over , is a homomorphic image of for some , by the obvious homomorphism that takes each to a generator of . By Theorem A.2 and induction we know that is Noetherian. But then by the above, is Noetherian.
Since we’re on the subject of Noetherian things, it’s probably worthwhile introducing the concept of a Noetherian module. The subsequent theorem is analogous to A.3 for general modules. This question ensures that the theorem has content.
Definition A.4 An -module is Noetherian if every submodule is finitely generated, that is, if every element of can be written as a polynomial in some generators with coefficients in .
Theorem A.5 If Noetherian and a finitely generated -module then Noetherian.
Proof Suppose generated by , and let be a submodule. We show finitely generated by induction on .
If then clearly the map defined by is surjective. Then the preimage of is an ideal, just as in A.3, so is finitely generated. Hence is finitely generated by the images of the generators of . (*)
Now suppose . Consider the quotient map . Let be the image of under this map. Then by the induction hypothesis is finitely generated as it is a submodule of . Let be elements of whose images generate . Since is a submodule of generated by a single element, we have by (*) that it’s submodule is finitely generated, by say.
We claim that generate . Indeed given the image of is a linear combination of the images of the . Hence subtracting the relevant linear combination of the from produces an element of which is precisely a linear combination of the by construction. This completes the induction.
We’re now ready to talk about the concrete problem that Hilbert solved using these ideas, namely the existence of finite bases for invariants. We’ll take to be a field of characteristic and to be a finite group, or one of the linear groups . As in our notation above, we take .
Suppose also we are given a group homomorphism , which of course can naturally be seen as the group of invertible linear transformations of the vector space over with basis . This is in fact the definition of a representation of on the vector space . As is common practice in representation theory, we view as acting on via .
If is or we shall further suppose that our representation of is rational. That is, the matrices act on as matrices whose entries are rational functions in the entries of . (If you’re new to representation theory like me, you might want to read that sentence twice)!
We now extend the action of from to the whole of by defining . Thus we may view as an automorphism group of . The invariants under are those polynomials left unchanged by the action of every , and these form a subring of which we’ll denote .
Enough set up. To proceed to more interesting territory we’ll need to make another definition.
Definition A.6 A polynomial is called homogeneous, a homogeneous form, or merely a form, if each of its monomials with nonzero coefficient has the same total degree.
Hilbert noticed that the following totally obvious fact about was critically important to the theory of invariants. We may write as a direct sum of the vector spaces of homogeneous forms of degree that are invariant under . We say that may be graded by degree and use this to motivate our next definition.
Definition A.7 A graded ring is a ring together with a direct sum decomposition as abelian groups , such that .
This allows us to generalise our notion of homogeneous also.
Definition A.8 A homogeneous element of a graded ring is an element of one of the groups . A homogeneous ideal of is an ideal generated by homogeneous elements.
Be warned that clearly homogeneous ideals may contain many inhomogeneous elements! It’s worth mentioning that there was no special reason for taking as our indexing set for the . We can generalise this easily to , and such graded rings are often called -graded rings. We won’t need this today, however.
Note that if we have a unique expression for of the form with . (I have yet to convince myself why this terminates generally, any thoughts? I’ve also asked here.) We call the a homogeneous component of .
The next definition is motivated by algebraic geometry, specifically the study of projective varieties. When we arrive at these in the main blog (probably towards the end of this month) it shall make a good deal more sense!
Definition A.9 The ideal in a graded ring generated by all forms of degree greater than is called the irrelevant ideal and notated .
Now we return to our earlier example. We may grade the polynomial ring by degree. In other words we write with containing all the forms (homogeneous polynomials) of degree .
To see how graded rings are subtly useful, we’ll draw a surprisingly powerful lemma.
Lemma A.10 Let be a homogeneous ideal of a graded ring , with generated by . Let be a homogeneous element. Then we may write with homogeneous of degree .
Proof We can certainly write with . Take to be the homogeneous components of of degree . Then all other terms in the sum must cancel, for is homogeneous by assumption.
Now we return to our attempt to emulate Hilbert. We saw earlier that he spotted that grading by degree may be useful. His second observation was this. maps of -modules s.t. (1) preserves degrees and (2) fixes every element of . It is easy to see that this abstract concept corresponds intuitively to the condition that be a summand of the graded ring .
This is trivial to see in the case that is a finite group. Indeed let . Note that we have implicitly used that has characteristic zero to ensure that the multiplicative inverse to exists. In the case that is a linear group acting rationally, then the technique is to replace the sum by an integral. The particulars of this are well beyond the scope of this post however!
We finally prove the following general theorem. We immediately get Hilbert’s result on the finite generation of classes of invariants by taking .
Theorem A.11 Take a field and a polynomial ring graded by degree. Let be a -subalgebra of . Suppose is a summand of , in the sense described above. Then is finitely generated as a -algebra.
Proof Let be the ideal of generated by all homogeneous elements of degree . By the Basis Theorem is Noetherian, and an ideal of , so finitely generated. By splitting each generator into its homogeneous components we may assume that is generated by some homogeneous elements which we may wlog assume lie in . We’ll prove that these elements precisely generate as a -algebra.
Now let be the -subalgebra of generated by and take a general homogeneous polynomial. Suppose we have shown . Let be a general element of . Then certainly a sum of homogeneous components. But a summand of , so applying the given map we have that the homogeneous components are wlog in . Thus also, and we are done.
It only remains to prove which we’ll do by induction on the degree of . If then . Suppose so . Since the generate as a homogeneous ideal of we may write with homogeneous of degree by Lemma A.10. But again we may use the map obtained from our observation that a summand of . Indeed then for . But preserves degrees so 0f lower degree than . Thus by the induction hypothesis and hence as required.
It’s worth noting that such an indirect proof caused quite a furore when it was originally published in the late 19th century. However the passage of time has provided us with a broader view of commutative algebra, and techniques such as this are much more acceptable to modern tastes! Nevertheless I shall finish by making explicit two facts that help to explain the success of our argument. We’ll first remind ourselves of a useful definition of an algebra.
Definition A.12 An -algebra is a ring which has the compatible structure of a module over in such a way that ring multiplication is -bilinear.
It’s worth checking that this intuitive definition completely agrees with that we provided in the Background section, as is clearly outlined on the Wikipedia page. The following provide an extension and converse to Corollary A.3 (that finitely generated algebras over fields are Noetherian) in the special case that a graded ring.
Lemma A.13 a Noetherian graded ring iff Noetherian and a finitely generated algebra.
Lemma A.14 Let be a Noetherian graded ring, a summand of . Then Noetherian.
We’ll prove these both next time. Note that they certainly aren’t true in general when isn’t graded!