Hochster–Roberts theorem

Remarks

A group is linearly reductive when every rational representation is a direct sum of irreducible ones. In characteristic $0$ this holds exactly for the reductive groups; in characteristic $p > 0$ it holds for tori and for finite groups of order prime to $p$, for example. Proof idea: linear reductivity gives the Reynolds operator, a $k[V]^G$-linear projection $k[V] \to k[V]^G$, so $k[V]^G$ is a direct summand of the regular ring $k[V]$. Hochster and Roberts showed that such summands are Cohen–Macaulay here, first in characteristic $p$ with the Frobenius map, and then in characteristic $0$ by reduction to characteristic $p$.

Example: let $G = \set{1, g}$ act on $k[x, y]$ by $g \cdot x = -x$, $g \cdot y = -y$, where the characteristic of $k$ is not $2$. Then $k[x, y]^G = k[x^2, xy, y^2]$ is a free module over the polynomial ring $k[x^2, y^2]$ with basis $1, xy$, which is the Cohen–Macaulay property in graded form.

Linear reductivity cannot be dropped: for a finite group whose order is divisible by the characteristic of $k$, the ring of invariants of a linear action need not be Cohen–Macaulay.

Going deeper

For a student

The Cohen–Macaulay property makes the Hilbert series of $k[V]^G$ a rational function with a numerator with nonnegative coefficients, and it controls the free resolution of the invariant ring. The proof is an early major use of reduction to positive characteristic, the method that later gave tight closure. The result fits with the finite generation of rings of invariants of reductive groups, the positive side of Hilbert's fourteenth problem. Boutot later proved the stronger fact that, in characteristic $0$, these invariant rings have rational singularities.