1
$\begingroup$

For a finite group $G$ and complex representation V of degree $n$, I would like to know the precise definition of Primary invariants. Does any set of n algebraically independent homogeneous invariants qualify to be called primary ? Or does the set need to satisfy additional conditions ?

I would like to know specifically the following : For a set $f_1,f_2,\ldots,f_n$ of algebraically independent homogeneous invariants, is the full ring of invariants of G a free module over the subalgebra generated by $f_i$'s.

What I have done:For the group $S_5$ and its irreducible representation of degree 6 (This is not generated by reflections), I have found six homogeneous invariants . I also computed the degrees of them whose product is a multiple of the order of $S_5$.

$\endgroup$
  • $\begingroup$ This is cross-posted from stack exchange as there was no response even after 48 hours. math.stackexchange.com/questions/1777620/… $\endgroup$ – user 1 May 11 '16 at 6:49
  • 2
    $\begingroup$ You also need that the whole ring is finitely generated as a $K[f_1,\dots,f_n]$-module. I.e. that you have a "homogenous system of parameters". I recommend looking at a book on invariant theory, e.g. "Computational Invariant Theory" by Derksen and Kemper $\endgroup$ – Max Horn May 11 '16 at 7:18
  • $\begingroup$ @MaxHorn- I read the definitions in Derksen and Kemper and also in Sturmfels. The extra condition that you have mentioned does not seem to be explicitly mentioned there. Hence this question in this forum. So I am still in the dark. $\endgroup$ – Desikan May 11 '16 at 8:32
2
$\begingroup$

You should trust Max Horn. Definition 2.4.6 in our book has the finiteness condition explicitly. (This is the numbering in the first edition of the book.) For finite groups, this is equivalent to saying that the variety given by the primaries consists only of the origin (see Propos. 3.3.1).

A counter example arises if you take K[V] = K[x,y] with the trivial group action and choose $f_1=x$ and $f_2 = x y$. These are algebraically independent, but you need all the powers of $y$ to generate $K[V]$ as a module over $K[f_1,f_2]$.

If you have primary invariants in the correct sense, the invariant ring is a finitely generated free module over $K[f_1,\ldots,f_n]$ by Cohen-Macauly property. In the above counter example, the module is neither free nor finitely generated.

$\endgroup$
  • $\begingroup$ Thanks to both Max Horn and Gregor Kemper. Actually the answer to my question is clearly available in the book. Now I realised. Sorry for my earlier comment. $\endgroup$ – Desikan May 12 '16 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.