I am studying tensor products and the following question came up to me, and dummit and foote does not address an answer to this. Is the following statement true?
Let us consider $M^{\otimes 3}$, where $R$ is a commutative ring and $M$ is a module over $R$. Let $x,y,z \in M$. If $x \otimes y \otimes z=0$ then $x=ry+r'z$ for some $r,r' \in R$.
Would it change if we assume $M$ to be free?
Thanks for your help!
False. In $({\Bbb Z}_9)^{\otimes 3}$ over $\Bbb Z$ we have $x \otimes 3 \otimes 3 = 0$ for all $x$, but since $\langle 3\rangle = \{0,3,6\}$, any $x\not\in \langle 3\rangle$ gives a counter-example.
Suppose $M$ is free with basis $\{e_i\}_{i\in I}$. Then $M^{\otimes 3}$ is also free with basis $\{e_i\otimes e_j \otimes e_k\}_{(i,j,k)\in I^3}$. Write $x = \sum_{i\in I} x_i e_i$, and similarly for $y$ and $z$ (finite sums). Assume $y,z\neq 0$. Then $$0 = x \otimes y \otimes z = \sum_{i,j,k\in I} x_i y_j z_k e_i \otimes e_j \otimes e_k,$$and so $x_iy_jz_k = 0$ for all choices of $i,j$ and $k$. There are $j_0,k_0\in I$ such that $y_{j_0},z_{k_0}\neq 0$. Then if $R$ is a domain, $x_iy_{j_0}z_{k_0}=0$ gives $x_i=0$. But since $i$ was arbitrary, $x=0$.
This clearly adapts to $M^{\otimes n}$, in the same setting. Only more boring to write with all indices and such.