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Consider a three-dimensional irreducible representation $V$ of $S_4$, corresponding to symmetries of cube. Let $p$ be canonical projection $p: V \rightarrow V/S_4$.

My question: is $p$ flat?

I want to add two comments. Possibly you do not need all this to answer my question.

Firstly, I already asked a question of this type. Ring of polynomials as a module over symmetric polynomials

Secondly, I can prove that $k[V]$ is not free module over $k[V]^{S_4}$. The idea of proof is follows. I calculated generating function $$Z_s (t) = \sum_n \dim( S^n(V)^{S_4} ) t^n = \frac{1-t^3+t^6}{(1-t)^3 (1+t)(1+t+t^2)(1+t+t^2+t^3)}$$

Remind you that $$Z(t) = \sum_n \dim( S^n(V) ) t^n = \frac{1}{(1-t)^3 }$$

If this module was free it would be generated by homogeneous generators. If there are $k_i$ generators of degree $i$ then

$$Z(t) = Z_s (t) (\sum _i k_i t^i)$$

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quinque
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1 Answers1

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It follows from Chevalley–Shephard–Todd that if $G \subset GL(V)$ is a finite group whose order is prime to the characteristic, then $V \to V/G$ is flat if and only if $G$ is generated by pseudoreflections if and only if $V/G$ is regular if and only if $k[V]^G$ is a polynomial ring if and only if $k[V]$ is free over $k[V]^G$.

So in this case, since you proved $k[V]$ is not free over $k[V]^{S_4}$, then $V \to V/S_4$ is not flat.

EDIT: If $V \to V/G$ is flat, then since $V$ is regular, $V/G$ is regular by flat descent (see http://en.wikipedia.org/wiki/Flat_morphism and EGA IV2, Corollaire 6.5.2) so the equivalent statements of CST all hold.

On the other hand, if any of the equivalences in CST hold, then $k[V]$ is free over $k[V]^G$ and so is flat.

Dori Bejleri
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