# If $M$ is locally free of finite rank, is the symmetric part of its tensor power?

Let $R$ be a commutative ring with unity, and let $M$ be locally free of rank $n$. (Specifically, suppose there exist $\{r_i: i\in I\}\subset R$ such that $\sum_i r_iR=R$ and $M_{r_i}$ is free of rank $n$ for each $i\in I$.)

Consider the subalgebra $(M^{\otimes d})^{S_d}\subset M^{\otimes d}$ consisting of elements fixed by the action of $S_d$ permuting the tensor factors. Is $(M^{\otimes d})^{S_d}$ also locally free of finite rank? Certainly if $M$ is free then so is $(M^{\otimes d})^{S_d}$, but in the general case I am not sure whether $$R_{r_i} \otimes (M^{\otimes d})^{S_d} \cong (M_{r_i}^{\otimes_{R_i}d})^{S_d}.$$

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Your isomorphism holds in general. Namely if $N$ is a $R$-module with the action of a finite group $G$ and if $f\in R$, then $$R_f\otimes N^G=(N_f)^G$$ as submodules of $N_f$. Indeed $R_f\otimes N^G\subset (N_f)^G$ is clear. Conversely, suppose $v=x/f^n\in (N_f)^G$. For any $\sigma\in G$, we have $\sigma(x)/f^n=x/f^n$. So $f^m(\sigma(x)-x)=0$ for some $m\in \mathbb N$ (depending on $\sigma$). As $G$ is finite, there is an $m_0$ big enough such that $f^{m_0}(\sigma(x)-x)=0$ for all $\sigma\in G$. Then $f^{m_0}x\in N^G$ and $$v=f^{m_0}x/f^{n+m_0}\in R_f\otimes N^G.$$

Of course, if $R$ is a domain $N$ has no $R$-torsion ($fx=0$ with $f\in R$ and $x\in N$ implies $f=0$ or $x=0$), then the proof is much simpler and we don't need the finiteness of $G$.

Generalization Let $N, G$ be as above. Let $R\to R'$ be a flat extension. Then $G$ acts naturally on $R'\otimes N$ (trivially on $R'$) and we have
$$R'\otimes_R N^G = (R'\otimes_R N)^G$$ as submodules of $R'\otimes N$.

Proof: For any $\sigma\in G$, denote by $\sigma_N : N\to N$ the image of $\sigma$ in $\mathrm{Aut}_R(N)$. We have $$N^G=\cap_{\sigma\in G} \ker(\sigma_N - 1).$$ As $R'$ is flat and $G$ is finite, $$R'\otimes \cap_{\sigma\in G} \ker(\sigma_N - 1)=\cap_{\sigma\in G} (R'\otimes \ker(\sigma_N-1))=\cap_{\sigma\in G} \ker(\sigma_{R'\otimes N} - 1)=(R'\otimes N)^G.$$ Hence $R'\otimes N^G=(R'\otimes N)^G$.

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Thank you so much! This seems to be a case of the question becoming much simpler when the extraneous details are removed. – Owen Biesel Nov 1 '12 at 22:35
@QiL Unless you have other solution, I can't see how this one works for $R$ domain and $G$ infinite. – user26857 Nov 1 '12 at 23:41
@navigetor23: I meant when $N$ has no torsion over $R$. Thanks ! – user18119 Nov 1 '12 at 23:44
@OwenBiesel: you are right, sometimes general facts are easier to proof than special facts. This also apply to my proof above that I will generalize :) – user18119 Nov 1 '12 at 23:51