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Is there an example which shows that the functor $B\otimes_R(-)$ is not left-exact, given a ring $R$ and a right $R$-module $B$?

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1 Answer 1

Consider the one-to-one function of $\mathbb{Z}$-modules $$\begin{align*} f\colon\mathbb{Z}&\longrightarrow\mathbb{Z}\\ a&\longmapsto 2a \end{align*}$$

If we tensor with $\mathbb{Z}/2\mathbb{Z}$, the map $f$ induces a map $$\mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}\ \longrightarrow\ \mathbb{Z}/2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}\cong\mathbb{Z}/2\mathbb{Z}$$ that sends $\overline{1}\otimes 1$ to $$\overline{1}\otimes f(1) = \overline{1}\otimes 2 = \overline{1}\otimes (1+1) = \overline{1}\otimes 1 + \overline{1}\otimes 1= \mathbf{0}.$$ So the map induced by the one-to-one map $f$ is not one-to-one.

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