Is the space of $k$-symmetric tensors generated by the "diagonal elements"? Let $A$ be a commutative ring with 1, and $M$ a free $A$-module of rank $n$.
Let $Sym^k(M)$ denote the subspace of $M^{\otimes k}$ generated by the symmetric tensors.
Is this space spanned by the diagonal elements $m^{\otimes k}$ for $m\in M$?
 A: $Sym(k)(M)$ is free with basis $\bar e_I =\sum_{\sigma \in S_k/Fix(I)}e_{\sigma(i_1)}\otimes\cdots \otimes e_{\sigma(i_k)}$ where $I={i_1\le \ldots \le i_k}$ all possible multisubsets of $\Lambda$ of cardinality $k$   . 
One  can check the general equality:
$$\sum_{\sigma \in S_k} v_{\sigma(1)}\otimes \cdot \otimes v_{\sigma(k)}= \sum_{J\subset I}(-1)^{|I|-|J|}(\sum_{j \in J} v_j)^{\otimes k}$$
Therefore, if $I$ is an ordinary subset then $\bar e_I$ is generated by diagonal elements. If $k!$ is invertible ( or even a weaker condition) this is true for all $I$ , since we can divide by the order of the group $Fix(I)$. 
I see a problem in the case $k=3$, $M=\mathbb{Z}^2$ free over $\mathbb{Z}$, with basis $e_1$, $e_2$. Then the basic element corresponding to the multisubset $(1^2,2)$ is $e_1\otimes e_1 \otimes e_2+ e_1\otimes e_2 \otimes e_1 + e_2\otimes e_1 \otimes e_1$. In fact, we can show that $Sym^3(\mathbb{Z}^2)$ is not generated by elements of form $(\alpha e_1 + \beta e_2)^{\otimes 3}$. Indeed, expanding we get 
$$(\alpha e_1 + \beta e_2)^{\otimes 3}=\alpha^3 e_1\otimes e_1 \otimes e_1 + \alpha^2 \beta (e_1\otimes e_1 \otimes e_2+ e_1\otimes e_2 \otimes e_1 + e_2\otimes e_1 \otimes e_1) + \alpha \beta^2 (e_1\otimes e_2 \otimes e_2+ e_2\otimes e_1 \otimes e_2 + e_2\otimes e_2 \otimes e_1) + \beta^3 e_2\otimes e_2 \otimes e_2$$
Denoting by $u=e_1\otimes e_1 \otimes e_2+ e_1\otimes e_2 \otimes e_1 + e_2\otimes e_1 \otimes e_1$, $v=e_1\otimes e_2 \otimes e_2+ e_2\otimes e_1 \otimes e_2 + e_2\otimes e_2 \otimes e_1$, the question is whether any $p u + q v$ is in the span of $\alpha^2 \beta u+ \alpha \beta^2 v$.  But it's easy to see that $\alpha^2 \beta + \alpha \beta^2$ is always even, so the answer is negative. 
I thank @Dap: for the correction regarding the basis of $Sym^k(M)$. 
