Example 10.12.12. Consider the injective map $2 : \mathbf{Z}\to \mathbf{Z}$ viewed as a map of $\mathbf{Z}$-modules. Let $N = \mathbf{Z}/2$. Then the induced map $\mathbf{Z} \otimes \mathbf{Z}/2 \to \mathbf{Z} \otimes \mathbf{Z}/2$ is NOT injective. This is because for $x \otimes y\in \mathbf{Z} \otimes \mathbf{Z}/2$,

$(2 \otimes 1)(x \otimes y) = 2x \otimes y = x \otimes 2y = x \otimes 0 = 0$

Therefore the induced map is the zero map while $\mathbf{Z} \otimes N\neq 0$.

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