# Invariant Theory and David Hilbert

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 $S = k[x_1,\dots,x_n]$ and a group $G$ acting on $S$ as a $k$-automorphism, what are the elements of $S$ that are invariant under the action of $G$? Following familiar notation we denote this set $S^G$ and note that it certainly forms a subalgebra of $S$.

In the late 19th century it was found that $S^G$ could be described fully by a finite set of generators for several suggestive special cases of $G$. It soon became clear that the fundamental problem of invariant theory was to find necessary and sufficient conditions for $S^G$ 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 $R$ is Noetherian if every ideal of $R$ is finitely generated.

We shall also recall without proof Hilbert’s Basis Theorem, and draw an easy corollary.

Theorem A.2 If $R$ Noetherian then $R[x]$ Noetherian.

Corollary A.3 If $S$ is a finitely generated algebra over $R$, with $R$ Noetherian, then $S$ Noetherian.

Proof We’ll first show that any homomorphic image of $R$ is Noetherian. Let $I$ be an ideal in the image under than homomorphism $f$. Then $f^{-1}(I)$ an ideal in $R$. Indeed if $k\in f^{-1}(I)$ and $r\in R$ then $f(rk)=f(r)f(k)\in I$ so $rk \in f^{-1}(I)$. Hence $f^{-1}(I)$ finitely generated, so certainly $I$ finitely generated, by the images of the generators of $f^{-1}(I)$.

Now we’ll prove the corollary. Since $S$ is a finitely generated algebra over $R$, $S$ is a homomorphic image of $R[x_1,\dots,x_n]$ for some $n$, by the obvious homomorphism that takes each $x_i$ to a generator of $S$. By Theorem A.2 and induction we know that $R[x_1,\dots,x_n]$ is Noetherian. But then by the above, $S$ is Noetherian. $\blacksquare$

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 $R$-module $M$ is Noetherian if every submodule $N$ is finitely generated, that is, if every element of $N$ can be written as a polynomial in some generators $\{f_1,\dots,f_n\}\subset N$ with coefficients in $R$.

Theorem A.5 If $R$ Noetherian and $M$ a finitely generated $R$-module then $M$ Noetherian.

Proof Suppose $M$ generated by $f_1,\dots,f_t$, and let $N$ be a submodule. We show $N$ finitely generated by induction on $t$.

If $t=1$ then clearly the map $h:R\rightarrow M$ defined by $1\mapsto f_1$ is surjective. Then the preimage of $N$ is an ideal, just as in A.3, so is finitely generated. Hence $N$ is finitely generated by the images of the generators of $h^{-1}(N)$.  (*)

Now suppose $t>1$. Consider the quotient map $h:M \to M/Rf_1$. Let $\tilde{N}$ be the image of $N$ under this map. Then by the induction hypothesis $\tilde{N}$ is finitely generated as it is a submodule of $M/Rf_1$. Let $g_1,\dots,g_s$ be elements of $N$ whose images generate $\tilde{N}$. Since $Rf_1$ is a submodule of $M$ generated by a single element, we have by (*) that it’s submodule $Rf_1\cap N$ is finitely generated, by $h_1,\dots,h_r$ say.

We claim that $\{g_1,\dots,g_s,h_1,\dots,h_r\}$ generate $N$. Indeed given $n \in N$ the image of $n \in N$ is a linear combination of the images of the $g_i$. Hence subtracting the relevant linear combination of the $g_i$ from $n$ produces an element of $N \cap Rf_1$ which is precisely a linear combination of the $h_i$ by construction. This completes the induction. $\blacksquare$

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 $k$ to be a field of characteristic $0$ and $G$ to be a finite group, or one of the linear groups $\textrm{ GL}_n(k),\ \textrm{SL}_n(k)$. As in our notation above, we take $S=k[x_1,\dots,x_n]$.

Suppose also we are given a group homomorphism $\phi:G \to \textrm{GL}_r(k)$, which of course can naturally be seen as the group of invertible linear transformations of the vector space $V$ over $k$ with basis $x_1,\dots,x_r$. This is in fact the definition of a representation of $G$ on the vector space $V$. As is common practice in representation theory, we view $G$ as acting on $V$ via $(g,v)\mapsto \phi(g)v$.

If $G$ is $\textrm{SL}_n(k)$ or $\textrm{GL}_n(k)$ we shall further suppose that our representation of $G$ is rational. That is, the matrices $g \in G$ act on $V$ as matrices whose entries are rational functions in the entries of $g$. (If you’re new to representation theory like me, you might want to read that sentence twice)!

We now extend the action of $g\in G$ from $V$ to the whole of $S$ by defining $(g,f)\mapsto f(g^{-1}(x_1),\dots,g^{-1}(x_r),x_{r+1},\dots,x_n)$. Thus we may view $G$ as an automorphism group of $S$. The invariants under $G$ are those polynomials left unchanged by the action of every $g \in G$, and these form a subring of $S$ which we’ll denote $S^G$.

Enough set up. To proceed to more interesting territory we’ll need to make another definition.

Definition A.6 A polynomial is called homogeneous, 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 $S^G$ was critically important to the theory of invariants. We may write $S^G$ as a direct sum of the vector spaces $R_i$ of homogeneous forms of degree $i$ that are invariant under $G$. We say that $S^G$ may be graded by degree and use this to motivate our next definition.

Definition A.7 A graded ring is a ring $R$ together with a direct sum decomposition as abelian groups $R = R_0 \oplus R_1 \oplus \dots$, such that $R_i R_j \subset R_{i+j}$.

This allows us to generalise our notion of homogeneous also.

Definition A.8 A homogeneous element of a graded ring $R$ is an element of one of the groups $R_i$. A homogeneous ideal of $R$ 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 $\mathbb{N}$ as our indexing set for the $R_i$. We can generalise this easily to $\mathbb{Z}$, and such graded rings are often called $\mathbb{Z}$-graded rings. We won’t need this today, however.

Note that if $f \in R$ we have a unique expression for $f$ of the form $f = f_0 + f_1 + \dots + f_n$ with $f_i \in R_i$. (I have yet to convince myself why this terminates generally, any thoughts? I’ve also asked here.) We call the $f_i$homogeneous component of $f$.

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 $R$ generated by all forms of degree greater than $0$ is called the irrelevant ideal and notated $R_+$.

Now we return to our earlier example. We may grade the polynomial ring $S=k[x_1,\dots,x_n]$ by degree. In other words we write $S=S_0\oplus S_1 \oplus \dots$ with $S_i$ containing all the forms (homogeneous polynomials) of degree $i$.

To see how graded rings are subtly useful, we’ll draw a surprisingly powerful lemma.

Lemma A.10 Let $I$ be a homogeneous ideal of a graded ring $R$, with $I$ generated by $f_1,\dots,f_r$. Let $f\in I$ be a homogeneous element. Then we may write $f = \sum f_i g_i$ with $g_i$ homogeneous of degree $\textrm{deg}(f)-\textrm{deg}(f_i)$.

Proof We can certainly write $f = \sum f_i G_i$ with $G_i \in R$. Take $g_i$ to be the homogeneous components of $G_i$ of degree $\textrm{deg}(f)-\textrm{deg}(f_i)$. Then all other terms in the sum must cancel, for $f$ is homogeneous by assumption. $\blacksquare$

Now we return to our attempt to emulate Hilbert. We saw earlier that he spotted that grading $S^G$ by degree may be useful. His second observation was this. $\exists$ maps $\phi:S\to S^G$ of $S^G$-modules s.t. (1) $\phi$ preserves degrees and (2) $\phi$ fixes every element of $S^G$. It is easy to see that this abstract concept corresponds intuitively to the condition that $S^G$ be a summand of the graded ring $S$.

This is trivial to see in the case that $G$ is a finite group. Indeed let $\phi (f) = \frac{1}{|G|}\sum_{g\in G} g(f)$. Note that we have implicitly used that $k$ has characteristic zero to ensure that the multiplicative inverse to $|G|$ exists. In the case that $G$ 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 $R=S^G$.

Theorem A.11 Take$k$ a field and $S=k[x_1,\dots,x_n]$ a polynomial ring graded by degree. Let $R$ be a $k$-subalgebra of $S$. Suppose $R$ is a summand of $S$, in the sense described above. Then $R$ is finitely generated as a $k$-algebra.

Proof Let $I\subset R$ be the ideal of $R$ generated by all homogeneous elements of degree $> 0$. By the Basis Theorem $S$ is Noetherian, and $IS$ an ideal of $S$, so finitely generated. By splitting each generator into its homogeneous components we may assume that $IS$ is generated by some homogeneous elements $f_1,\dots,f_s$ which we may wlog assume lie in $I$. We’ll prove that these elements precisely generate $R$ as a $k$-algebra.

Now let $R'$ be the $k$-subalgebra of $S$ generated by $f_1,\dots,f_s$ and take $f\in R$ a general homogeneous polynomial. Suppose we have shown $f\in R'$. Let $g$ be a general element of $R$. Then certainly $g\in S$ a sum of homogeneous components. But $R$ a summand of $S$, so applying the given map $\phi$ we have that the homogeneous components are wlog in $R$. Thus $g\in R'$ also, and we are done.

It only remains to prove $f \in R'$ which we’ll do by induction on the degree of $f$. If $\textrm{deg}(f)=0$ then $f\in K\subset R'$. Suppose $\textrm{deg}(f)>0$ so $f\in I$. Since the $f_i$ generate $IS$ as a homogeneous ideal of $S$ we may write $f = \sum f_i g_i$ with $g_i$ homogeneous of degree $\textrm{deg}(f)-\textrm{deg}(f_i)<\textrm{deg}(f)$ by Lemma A.10. But again we may use the map $\phi$ obtained from our observation that $R$ a summand of $S$. Indeed then $f=\sum \phi(g_i)f_i$ for $f,\ f_i \in R$. But $\phi$ preserves degrees so $\phi(g_i)$ 0f lower degree than $f$. Thus by the induction hypothesis $\phi(g_i) \in R'$ and hence $f\in R'$ as required. $\blacksquare$

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 $R$-algebra $S$ is a ring $S$ which has the compatible structure of a module over $R$ in such a way that ring multiplication is $R$-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 $R$ a graded ring.

Lemma A.13 $S=R_0\oplus R_1 \oplus \dots$ a Noetherian graded ring iff $R_0$ Noetherian and $S$ a finitely generated $R_0$ algebra.

Lemma A.14 Let $S$ be a Noetherian graded ring, $R$ a summand of $S$. Then $R$ Noetherian.

We’ll prove these both next time. Note that they certainly aren’t true in general when $S$ isn’t graded!