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Let $A, B$ be commutative rings with identity, $f: A \longrightarrow B$ a ring morphism, $M$ an $A$-module. Given $b\in B, x\in M$, does the following statement hold?

$b\otimes x=0$ in $B \otimes_A M$ iff there exist finitely many $a_1,\dots, a_n\in\operatorname{Ann}(x)\subset A, b_1,\dots, b_n\in B$ such that $b=b_1f(a_1)+\cdots+b_nf(a_n)\in B$. (Here $\operatorname{Ann}(x)=\{a\in A: ax=0\}.$)

One direction is obvious.

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No, this is false. Consider the element $1 \otimes 2 \in \mathbf{Z}/2\mathbf{Z} \otimes_{\mathbf{Z}} \mathbf{Z}$. (We have $1 \otimes 2 = 2 \otimes 1 = 0 \otimes 1 = 0$, but the only element in the annihilator of $2 \in \mathbf{Z}$ is $0$.)

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  • $\begingroup$ Oh, thank you very much! $\endgroup$
    – Lao-tzu
    Jan 17 '14 at 3:45

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