Prove that $x \otimes x' \in M \otimes_R M'$ is nonzero.

Question

Let $R$ be an integral domain and let $M, M'$ be free (left) $R$-modules. Prove that for any nonzero $x \in M$ and $x' \in M'$, the element $x \otimes x' \in M \otimes_R M'$ is nonzero. Prove that this statement does not hold if $R$ is not an integral domain.

My attempt: Let $x \in M$ and $x' \in M'$ be arbitrary and nonzero. Since $M$ and $M'$ are free $R$-modules, they have bases $B$ and $C$. Thus, there exist $\lambda_i \in R$, $b_i \in B$ and $\mu_j \in R$, $c_j \in C$, such that $x = \lambda_1 b_1 + \cdots + \lambda_n b_n$ and $x' = \mu_1 c_1 + \cdots + \mu_m c_m$. Then, $$ x \otimes x' = (\lambda_1 b_1 + \cdots + \lambda_n b_n) \otimes x' = \lambda_1 (b_1 \otimes x') + \cdots + \lambda_n (b_n \otimes x'). $$ What can I do now? From here it didn't know how to continue.

Now, if $R$ is not an integral domain, I took $M$ and $M'$ as $\mathbb{Z}_4$-module $\mathbb{Z}_4$, which are free. Then $\overline{2}\in M,M'$ is nonzero and $\overline{2}\otimes \overline{2}=\overline{1}\otimes {(2\cdot \overline{2})}=\overline{1}\otimes \overline{4}=\overline{1}\otimes \overline{0}=\overline{0}$. I think this counterexample should be ok, but please check also this. Thanks for helping.

Note: My post is different than this post: When the tensor product of two module elements is nonzero. As here I wanna know WHY $x\otimes x'$ is nonzero (and here it only says "It works for nonzero elements in free modules over an integral domain.").

Also, I want to solve this exercise WITHOUT creating new linear maps. Can this be possible?

Answer 1

Since $x \neq 0$, there is a linear map $\omega : M \to R$ with $\omega(x) \neq 0$. (Consider a basis and project to some basis vector whose coefficient is non-zero). Similarly, there is a linear map $\omega' : M' \to R$ with $\omega'(x') \neq 0$. Consider the linear map $\omega \otimes \omega' : M \otimes M' \to R \otimes R \cong R$ (which exists by functoriality of the tensor product). It maps $x \otimes x'$ to $\omega(x) \omega'(x')$. This is non-zero since $R$ is an integral domain and the factors are non-zero. Hence, $x \otimes x' \neq 0$.

If you want to work without linear maps and only with bases (it's really the same argument): If $B$ is a basis of $M$ and $B'$ is a basis of $M'$, then $(b \otimes b')_{b \in B, b' \in B'}$ is a basis of $M \otimes M'$. Let us write $x = \sum_{b} x(b) \cdot b$ and $x' = \sum_{b'} x'(b') \cdot b'$ with coefficients in $R$, almost all of which are zero. Then $x \otimes x' = \sum_{b,b'} x(b) x'(b') \cdot (b \otimes b')$. By assumption, at least one $x(b)$ is non-zero. Likewise, at least one $x'(b')$ is non-zero. Then the coefficient of the basis element $b \otimes b'$ in $x \otimes x'$ is non-zero, hence $x \otimes x' \neq 0$.

Your counterexample is fine (but please write $\mathbb{Z}/4\mathbb{Z}$ for the quotient group, the notation $\mathbb{Z}_4$ means something else) and works in a much more general setting. Let $R$ be any ring with elements $a,b$ with $ab = 0$. Then $a \otimes b = 0$ in $R \otimes R$. This actually shows:

A commutative ring $R$ has the property that tensor products of two non-zero elements in free modules remain non-zero if and only if $R$ has no non-zero zero divisors. (That is, either $R = 0$ or $R$ is an integral domain.)