# Spanning set of symmetric invariants of tensor powers

Let $$M$$ be a module over a commutative ring $$R$$ and let the symmetric group $$\Sigma_n$$ act on $$M^n$$ and $$M^{\otimes n}$$ by permuting factors. For $$v \in M^n$$, let $$\text{Stab}(v)$$ denote the stabilizer subgroup of $$v$$, and let $$\text{Sym}(v) = \sum_{[\sigma] \in \Sigma_n / \text{Stab}(v)} \sigma \bigotimes_i v_i \in \left( M^{\otimes n} \right)^{\Sigma_n}$$ where the exponent of $$\Sigma_n$$ denotes the invariants. Notice that $$\text{Stab}(v) \leq \mathop{\text{Stab}}\left( \bigotimes_i v_i\right)$$ but these subgroups are often unequal.

Is it the case that $$\left(M^{\otimes n} \right)^{\Sigma_n}$$ is spanned by the image of $$\text{Sym} \colon M^n \to M^{\otimes n}$$? What if $$M$$ is projective? If the answers are negative, I am interested in counterexamples. I have no reason to expect that the symmetric invariants are spanned by this set, but I have been unable to construct a counterexample.

Here is a counterexample. Let $$k$$ be a field of characteristic $$2$$, let $$R=k[x,y,z]$$, and let $$M$$ be the ideal $$(x,y,z)\subset R$$. Note that $$xy\otimes z=x\otimes yz=xz\otimes y=z\otimes xy$$ in $$M^{\otimes 2}$$, so $$xy\otimes z\in (M^{\otimes 2})^{\Sigma_2}$$. However, I claim that $$xy\otimes z$$ is not in the span of the image of $$\operatorname{Sym}:M^2\to M^{\otimes 2}$$.
To prove this, it suffices to show $$xy\otimes z$$ is not in the span of the degree $$\leq 3$$ parts of tensors of the form $$a\otimes a$$ or $$a\otimes b+b\otimes a$$, for $$a,b\in M$$. Moreover, $$a$$ and $$b$$ can be assumed to be monomials (if they are sums of monomials, then $$a\otimes a$$ and $$a\otimes b+b\otimes a$$ expand out into a sum of such terms formed by monomials). Note also that if $$a$$ is a monomial of degree $$2$$ and $$b$$ is a monomial of degree $$1$$, then $$a\otimes b=b\otimes a$$ by a manipulation similar to the one used above to show $$xy\otimes z=z\otimes xy$$. Since $$k$$ has characteristic $$2$$, this means $$a\otimes b+b\otimes a$$ will be $$0$$ in that case. Similar reasoning also shows that if $$a$$ and $$b$$ have degree $$1$$, then any degree $$3$$ multiple of $$a\otimes b+b\otimes a$$ will be $$0$$. So, we only need consider the case $$a\otimes a$$ where $$a$$ has degree $$1$$; that is, we must show that $$xy\otimes z$$ is not in the submodule generated by $$x\otimes x,y\otimes y,$$ and $$z\otimes z$$. This is easy: there is a homomorphism $$M^{\otimes 2}\to R^{\otimes 2}\cong R$$ which sends $$xy\otimes z$$ to $$xyz$$, which is not in the submodule generated by $$x^2,y^2,$$ and $$z^2$$.
On the other hand, it is true if $$M$$ is projective. First, it is easy to see it is true if $$M$$ is free by picking a basis and considering the induced basis on $$M^{\otimes n}$$, which is permuted by the action of $$\Sigma_n$$. Now suppose $$M$$ is projective, so it is a direct summand of a free module; say $$F$$ is free and $$F=M\oplus N$$. Note that the action of $$\Sigma_n$$ respects the direct sum decomposition $$F^{\otimes n}=M^{\otimes n}\oplus K$$ where $$K$$ is all the other terms you get by distributing the tensor product over the direct sum in $$F=M\oplus N$$. Also, if $$v=(x,y)\in F^n$$ with $$x\in M^n$$ and $$y\in N^n$$, then the $$M^{\otimes n}$$ component of $$\operatorname{Sym}(v)$$ is an integer multiple of $$\operatorname{Sym}(x)$$ (the multiple being because $$\operatorname{Stab}(x)$$ might be larger than $$\operatorname{Stab}(v)$$). It follows that the intersection of the span of the image of $$\operatorname{Sym}:F^n\to F^{\otimes n}$$ with $$M^{\otimes n}$$ coincides with the span of the image of $$\operatorname{Sym}:M^n\to M^{\otimes n}$$. But the span of the image of the span of the image of $$\operatorname{Sym}:F^n\to F^{\otimes n}$$ is all of $$(F^{\otimes n})^{\Sigma_n}$$ since $$F$$ is free, and so in particular it contains all of $$(M^{\otimes n})^{\Sigma_n}$$.
It is also true if $$n!$$ is a unit in $$R$$. In that case, given $$x\in (M^{\otimes n})^{\Sigma_n}$$, you can write $$x$$ as a sum of simple tensors $$x=\sum_i t_i$$, and then $$x=\frac{1}{n!}\sum_{\sigma\in\Sigma_n}\sigma(x)=\frac{1}{n!}\sum_i\sum_{\sigma\in\Sigma_n} \sigma(t_i)$$ and $$\sum_{\sigma\in\Sigma_n} \sigma(t_i)$$ is an integer multiple of an element of the image of $$\operatorname{Sym}$$ for each $$i$$.