**Proof.**
The implications (1) implies (2) implies (3) implies (4) are all trivial. Thus we prove (4) implies (1). Suppose that $N_1 \to N_2 \to N_3$ is exact. Let $K = \mathop{\mathrm{Ker}}(N_2 \to N_3)$ and $Q = \mathop{\mathrm{Im}}(N_2 \to N_3)$. Then we get maps

\[ N_1 \otimes _ R M \to K \otimes _ R M \to N_2 \otimes _ R M \to Q \otimes _ R M \to N_3 \otimes _ R M \]

Observe that the first and third arrows are surjective. Thus if we show that the second and fourth arrows are injective, then we are done^{1}. Hence it suffices to show that $- \otimes _ R M$ transforms injective $R$-module maps into injective $R$-module maps.

Assume $K \to N$ is an injective $R$-module map and let $x \in \mathop{\mathrm{Ker}}(K \otimes _ R M \to N \otimes _ R M)$. We have to show that $x$ is zero. The $R$-module $K$ is the union of its finite $R$-submodules; hence, $K \otimes _ R M$ is the colimit of $R$-modules of the form $K_ i \otimes _ R M$ where $K_ i$ runs over all finite $R$-submodules of $K$ (because tensor product commutes with colimits). Thus, for some $i$ our $x$ comes from an element $x_ i \in K_ i \otimes _ R M$. Thus we may assume that $K$ is a finite $R$-module. Assume this. We regard the injection $K \to N$ as an inclusion, so that $K \subset N$.

The $R$-module $N$ is the union of its finite $R$-submodules that contain $K$. Hence, $N \otimes _ R M$ is the colimit of $R$-modules of the form $N_ i \otimes _ R M$ where $N_ i$ runs over all finite $R$-submodules of $N$ that contain $K$ (again since tensor product commutes with colimits). Notice that this is a colimit over a directed system (since the sum of two finite submodules of $N$ is again finite). Hence, (by Lemma 10.8.4) the element $x \in K \otimes _ R M$ maps to zero in at least one of these $R$-modules $N_ i \otimes _ R M$ (since $x$ maps to zero in $N \otimes _ R M$). Thus we may assume $N$ is a finite $R$-module.

Assume $N$ is a finite $R$-module. Write $N = R^{\oplus n}/L$ and $K = L'/L$ for some $L \subset L' \subset R^{\oplus n}$. For any $R$-submodule $G \subset R^{\oplus n}$, we have a canonical map $G \otimes _ R M \to M^{\oplus n}$ obtained by composing $G \otimes _ R M \to R^ n \otimes _ R M = M^{\oplus n}$. It suffices to prove that $L \otimes _ R M \to M^{\oplus n}$ and $L' \otimes _ R M \to M^{\oplus n}$ are injective. Namely, if so, then we see that $K \otimes _ R M = L' \otimes _ R M/L \otimes _ R M \to M^{\oplus n}/L \otimes _ R M$ is injective too^{2}.

Thus it suffices to show that $L \otimes _ R M \to M^{\oplus n}$ is injective when $L \subset R^{\oplus n}$ is an $R$-submodule. We do this by induction on $n$. The base case $n = 1$ we handle below. For the induction step assume $n > 1$ and set $L' = L \cap R \oplus 0^{\oplus n - 1}$. Then $L'' = L/L'$ is a submodule of $R^{\oplus n - 1}$. We obtain a diagram

\[ \xymatrix{ & L' \otimes _ R M \ar[r] \ar[d] & L \otimes _ R M \ar[r] \ar[d] & L'' \otimes _ R M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M \ar[r] & M^{\oplus n} \ar[r] & M^{\oplus n - 1} \ar[r] & 0 } \]

By induction hypothesis and the base case the left and right vertical arrows are injective. The rows are exact. It follows that the middle vertical arrow is injective too.

The base case of the induction above is when $L \subset R$ is an ideal. In other words, we have to show that $I \otimes _ R M \to M$ is injective for any ideal $I$ of $R$. We know this is true when $I$ is finitely generated. However, $I = \bigcup I_\alpha $ is the union of the finitely generated ideals $I_\alpha $ contained in it. In other words, $I = \mathop{\mathrm{colim}}\nolimits I_\alpha $. Since $\otimes $ commutes with colimits we see that $I \otimes _ R M = \mathop{\mathrm{colim}}\nolimits I_\alpha \otimes _ R M$ and since all the morphisms $I_\alpha \otimes _ R M \to M$ are injective by assumption, the same is true for $I \otimes _ R M \to M$.
$\square$

## Comments (2)

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