Let $G$ be a finite group with $n$ elements acting on $\mathbb{Q}[x_1,\cdots,x_n]$ through the regular representation. And let $\mathbb{Q}[x_1,\cdots,x_n]^G = \mathbb{Q}[g_1,\cdots,g_m]$ . Then the $g_i$ might be the roots of some polynomials $s_j(y_1,\cdots,y_m) \in \mathbb{Q}[y_1,\cdots,y_m]$ , $j=1,\cdots,r$ called "syzygies". What does the Hilbert Syzygy theorem state in this situation?
What does the Hilbert Syzygy theorem state?
0
$\begingroup$
finite-groups
invariant-theory
-
0The Syzgy theorem tells you that the complex which compute the syzygies, the syzygies of the syzygies and "higher relations" is identically $0$ after $n$ steps, where $n$ is the number of indeterminates. – 2017-01-10
-
0Thanks for your comment! Can you rephrase this in terms of polynomials? – 2017-01-10
-
0This is better explained in term of complexes. I can write an answer a bit later if no one is doing it. – 2017-01-10