“There is changeability in process bu invariance in outcome.”
―Paul Watzlawick
Remark9.22.
If \(R\) is a graded ring, and \(G\) is a group acting on \(R\) by degree-preserving automorphisms, then \(R^G\) is a graded subring of \(R\text{,}\) meaning \(R^G\) is graded with respect to the same grading monoid. In particular, if \(G\) acts \(k\)-linearly on a polynomial ring over \(k\text{,}\) the invariant ring is \(\N\)-graded.
Using this perspective, we can now give a different proof of the finite generation of invariant rings that works under different hypotheses. The proof we will discuss now is essentially Hilbert’s proof. To do that, we need another notion that is very useful in commutative algebra.
Definition9.23.Direct Summand.
Let \(\varphi\!:R \to S\) be a ring homomorphism. We say that \(R\) is a direct summand of \(S\) if the map \(\varphi\)splits as a map of \(R\)-modules, meaning there is an \(R\)-module homomorphism
\begin{equation*}
\begin{CD}
R@>\varphi>> S\\\\
R@<<\rho< S
\end{CD}
\end{equation*}
such that \(\rho \varphi\) is the identity on \(R\text{.}\)
Remark9.24.
First, observe that the condition on \(\rho\) implies that \(\varphi\) must be injective, so we can assume that \(R \subseteq S\text{,}\) perhaps after renaming some elements. The condition on \(\rho\) is that \(\rho|_R\) is the identity and \(\rho(rs)= r\rho(s)\) for all \(r \in R\) and \(s\in S\text{.}\) We call the map \(\rho\) the splitting of the inclusion \(R \subseteq S\text{.}\) Note that given any \(R\)-linear map \(\rho\!: S \to R\text{,}\) if \(\rho(1)=1\) then \(\rho\) is a splitting: indeed, \(\rho(r)=\rho(r \cdot 1) = r \rho(1)=r\) for all \(r \in R\text{.}\)
Remark9.25.
The subring \(R\) of \(S\) is a direct summand of \(S\) if and only if there exists an \(R\)-submodule of \(S\) such that \(S = R \oplus M\text{.}\) In the language above, \(M = \ker \rho\text{.}\) Conversely, given a direct sum decomposition \(S = R \oplus M\text{,}\) the quotient map onto the first component is a splitting.
Definition9.26.Expansion and Contraction.
Let \(\varphi\! : R \to S\) be a ring homomorphism. Given an ideal \(I\) in \(S\text{,}\) we write \(I \cap R\) for the contraction of \(R\) back into \(R\text{,}\) meaning the preimage of \(I\) via \(\varphi\text{.}\) In particular, if \(R \subseteq S\) is a ring extension, then \(I \cap R\) denotes the preimage of \(I\) via the inclusion map \(R \subseteq S\text{.}\) Given a ring map \(R \to S\text{,}\) and an ideal \(I\) in \(R\text{,}\) the expansion of \(I\) in \(S\) is the ideal of \(S\) generated by the image of \(I\) via the given ring map; we naturally denote this by \(IS\text{.}\)
Lemma9.27.Direct Summand of Ideals Contract.
Let \(R\) be a direct summand of \(S\text{.}\) Then, for any ideal \(I \subseteq R\text{,}\) we have \(IS \cap R=I\text{.}\)
Proof.
Let \(\pi\) be the corresponding splitting. Clearly, \(I \subseteq IS \cap R\text{.}\) Conversely, if \(r \in IS \cap R\text{,}\) we can write \(r = s_1 f_1 + \cdots + s_t f_t\) for some \(f_i \in I\text{,}\)\(s_i \in S\text{.}\) Applying \(\pi\text{,}\) we have
be a chain of ideals in \(R\text{.}\) The chain of ideals in \(S\)
\begin{equation*}
I_1 S \subseteq I_2 S \subseteq I_3 S \subseteq \cdots
\end{equation*}
stabilizes, so there exist \(J\text{,}\)\(N\) such that \(I_n R = J\) for \(n \geqslant N\text{.}\) Contracting to \(R\text{,}\) we get that \(I_n = I_n S \cap R = J \cap R\) for \(n\geqslant N\text{,}\) so the original chain also stabilizes.
Notice in general a subring of a noetherian ring does not have to be noetherian.
Example9.29.Non-Noetherian Subring of Noetherian Ring.
If \(k\) is a field, \(S = k[x,y]\) is noetherian by Hilbert’s Basis Theorem , but we claim that the subring
\begin{equation*}
R = k[xy,xy^2,xy^3,\ldots]
\end{equation*}
is an ascending chain of ideals of \(R\) that does not stabilize. Notice that if we considered the same chain of ideals in \(S\text{,}\) then it does stabilize, and in fact it is the constant chain \((xy)\text{.}\)
Proposition9.30.\(K\)-Linear Action and Direct Summand.
Let \(k\) be a field, and \(R\) be a polynomial ring over \(k\text{.}\) Let \(G\) be a finite group acting \(k\)-linearly on \(R\text{.}\) Assume that the characteristic of \(k\) does not divide \(|G|\text{.}\) Then \(R^G\) is a direct summand of \(R\text{.}\)
Proof.
We consider the map \(\rho\!: R \to R^G\) given by
\begin{equation*}
\rho(r)=\frac{1}{|G|} \sum_{g\in G} g\cdot r.
\end{equation*}
First, note that the image of this map lies in \(R^G\text{,}\) since acting by \(g\) just permutes the elements in the sum, so the sum itself remains the same. We claim that this map \(\rho\) is a splitting for the inclusion \(R^G \subseteq R\text{.}\) To see that, let \(s\in R^G\) and \(r\in R\text{.}\) We have
The condition that the characteristic of \(k\) does not divide the order of \(G\) is trivially satisfied if \(k\) has characteristic zero.
Theorem9.32.Hilbert’s Finiteness Theorem for Invariants.
Let \(k\) be a field, and \(R\) be a polynomial ring over \(k\text{.}\) Let \(G\) be a group acting \(k\)-linearly on \(R\text{.}\) Assume that \(G\) is finite and that the characteristic of \(k\) does not divide \(|G|\text{,}\) or more generally, that \(R^G\) is a direct summand of \(R\text{.}\) Then \(R^G\) is a finitely generated \(k\)-algebra.
Proof.
Since \(G\) acts linearly on \(R\text{,}\)\(R^G\) is an \(\N\)-graded subring of \(R\) with \(R_0=k\text{.}\) Since \(R^G\) is a direct summand of \(R\text{,}\)\(R^G\) is noetherian by . By our characterization of noetherian graded rings in , \(R^G\) is finitely generated over \(R_0=k\text{.}\)
Remark9.33.
One important thing about this proof is that it applies to many infinite groups. In particular, for any linearly reductive group, including \(\mathrm{GL}_n(\C)\text{,}\)\(\mathrm{SL}_n(\C)\text{,}\) and \((\C^{\times})^n\text{,}\) we can construct a splitting map \(\rho\text{.}\)
