For a (noncommutative) ring $R$, I will denote by $\langle r_1,\ldots,r_n\rangle_{\mathrm{sym}}$ the set of elements $p$ of $R$ generated by the elements $r_1,\ldots,r_n\in R$, such that $p$ viewed as a "polynomial" $p(r_1,\ldots,r_n)$ is invariant under permutations $(r_1,\ldots,r_n)\mapsto (r_{\pi(1)},\ldots,r_{\pi(n)})$, $\pi\in S_n$.
Conjecture: For a (free?) ring $R$, $\langle r_1,\ldots,r_n\rangle_{\mathrm{sym}}=\langle \sum_{i=1}^n r_i, \sum_{i=1}^n r_i^2,\ldots,\sum_{i=1}^n r_i^k,\ldots\rangle$.
In practice, I am only interested in the special case of tensors. Consider a set of vectors $v_1,\ldots,v_n\in V$ and the algebra $\langle v_1,\ldots,v_n\rangle$ of all tensors generated by them. Similarly one can define permutation-invariant tensors $\langle v_1,\ldots,v_n\rangle_{\mathrm{sym}}$ (note that this is not the same as symmetric tensors: for example $v_1\otimes v_1\otimes v_2+v_2\otimes v_2\otimes v_1$ is permutation-invariant but not symmetric in the traditional sense).
I claim that all such permutation-invariant tensors can be expressed as linear combinations of terms that are tensor products of power-sums of the form $\sum_{i=1}^n v_i^{\otimes k}$.
For example, $v_1\otimes v_1\otimes v_2+v_2\otimes v_2\otimes v_1=(v_1^{\otimes 2}+v_2^{\otimes 2})\otimes (v_1+v_2)-(v_1^{\otimes 3}+v_2^{\otimes 3})$.
I would like to find out whether such a theorem is known, or if it is incorrect.
Best Answer
Your claim happens to be correct for $n=2$, but is false for $n \geq 3$, as I show below. For convenience, I omit the "tensor product" symbol in my computations ; you just need to remember that multiplication is not commutative in my notation.
Denote by $B$ your set of $n$ vectors $v_1,\ldots,v_n$. Recall that a pure tensor is a tensor of the form $b_1b_2\ldots b_k$ where each $b_i$ is in $B$. In general, a tensor will not be pure but will be a linear combination of pure tensors. The action of $S_n$ on all tensors restricts to an action on pure tensors ; for a pure tensor $p$, denote by $Orb(p)$ the orbit of $p$ under this action (thus $Orb(p)$ is a finite set with cardinality dividing $|S_n|=n!$) and let $s(p)=\sum_{q\in Orb(p)}q$. Then, the $s(p)$ form a basis of the subspace of permutation-invariant tensors. The question is therefore whether every $s(p)$ is in your subspace defined by power sums (let us call this subspace $T$).
Let $p$ be a pure tensor, we can uniquely write $p=b_1^{e_1}b_2^{e_2}\ldots b_l^{e_l}$ where each $b_i$ is in $B$, $b_{i+1}\neq b_i$ ; we call $l$ the complexity of $p$. By extension, the complexity of a linear combination of pure tensors is the largest complexity of the component tensors.
For $n=2$, your claim is true because of
Lemma. Suppose $n=2$ and let $p=b_1^{e_1}b_2^{e_2}\ldots b_l^{e_l}$ as above, $q=b_1^{e_1}b_2^{e_2}\ldots b_{l-1}^{e_{l-1}}$ and $r=b_l^{e_l}$ (so that $p=qr$). Then $s(p)-s(q)s(r)$ has complexity $\lt l$.
Proof of lemma. Let $\tau=(1,2)$ be the unique non-identity element of $S_2$. Then $s(p)=p+\tau(p)$ for every $p$. Hence
$$s(p)-s(q)s(r)=(p+\tau(p))-(q+\tau(q))(r+\tau(r))=q\tau(r)+\tau(q)r=s(q\tau(r))$$
and $q\tau(r)=b_1^{e_1}b_2^{e_2}\ldots b_{l-1}^{e_{l-1}+e_l}$ has complexity $l-1$. QED
Once we have the lemma, it is easy to deduce that every $s(p)$ is in $T$ by induction on the complexity of $p$.
When $n\geq 3$ however, there is a dimension mismatch : for tensor products of three elements, the permutation-invariant subspace has dimension $5$ with basis $s(v_1^3),s(v_1^2v_2),s(v_1v_2^2),s(v_1v_2v_1),s(v_1v_2v_3)$. The corresponding subspace of $T$ has dimension $4$ with basis $t_1^3,t_1t_2,t_2t_1,t_3$.