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Let $\alpha:R\to S$ be a map of unital rings, and let $M$ be an $R$-module. We have a canonical map of $R$-modules:

$$\begin{array}{rcl}i:M&\longrightarrow&S\otimes_RM\\[.05in]m&\longmapsto&1\otimes m\end{array}$$

where the $S$-module $S\otimes_RM$ obtained by extension of scalars is considered an $R$-module via restriction along $\alpha$. Contrary to appearances, this map is not necessarily injective, as seen in the following counterexample:

Let $R=\mathbb{Z}$, $S=\mathbb{Z}/2\mathbb{Z}$, and let $M$ be the $\mathbb{Z}$-module = abelian group of order $3$ generated by $t\in M$. Then $2\cdot t=t^2\ne 0$, but $$i(2\cdot t)=1\otimes (2\cdot t)=\alpha(2)\otimes t=0\otimes t=0_{S\otimes_RM}$$

What conditions should we require of $R,S,\alpha,$ and $M$ so that $i$ is an injective map of $R$-modules? It would be nice to find equivalent conditions, but I'd appreciate sufficient or necessary conditions alone as well.

The above example shows that a necessary condition is that $\ker\alpha\subset \operatorname{Ann}_R(M)$. A sufficient condition is that $\alpha$ is injective and $M$ is a flat $R$-module. Any other results in this direction?

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    $\begingroup$ Hint: Look at the vast literature on "pure homomorphisms" of commutative rings. $\endgroup$ – Martin Brandenburg Oct 28 '13 at 8:48
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    $\begingroup$ For instance you might try Section 3.3.1 of Rotman's book on homological algebra. $\endgroup$ – user2055 Feb 3 '14 at 1:55

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