[proofplan] The implications $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4)$ follow by restricting to progressively smaller classes of exact sequences. The substantive work is in the converses: $(3) \Rightarrow (2)$ follows because $T_M$ is already right exact, so preserving injectivity upgrades it to preserving short exact sequences. $(2) \Rightarrow (1)$ decomposes an arbitrary exact sequence into short exact sequences and traces images and kernels through the tensored pieces. $(4) \Rightarrow (3)$ uses a finitary reduction: any element of $\ker(\mathrm{id}_M \otimes f)$ involves only finitely many generators, so the vanishing can be witnessed in finitely generated submodules where condition (4) applies. [/proofplan] [step:Prove the immediate implications $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4)$] $(1) \Rightarrow (2)$: A short exact sequence $0 \to N' \to N \to N'' \to 0$ is in particular an exact sequence. If $T_M$ preserves exactness of all exact sequences, it preserves exactness of short exact sequences. $(2) \Rightarrow (3)$: Let $f: N' \to N$ be an injective $R$-module homomorphism. We must show $\mathrm{id}_M \otimes f: M \otimes_R N' \to M \otimes_R N$ is injective. Define the cokernel $N'' := N / \operatorname{im}(f)$ and let $\pi: N \to N''$ be the canonical projection. Since $f$ is injective and $\pi$ is the cokernel of $f$, the sequence \begin{align*} 0 \to N' \xrightarrow{f} N \xrightarrow{\pi} N'' \to 0 \end{align*} is a short exact sequence. By (2), the tensored sequence \begin{align*} 0 \to M \otimes_R N' \xrightarrow{\mathrm{id}_M \otimes f} M \otimes_R N \xrightarrow{\mathrm{id}_M \otimes \pi} M \otimes_R N'' \to 0 \end{align*} is exact. In particular $\mathrm{id}_M \otimes f$ is injective. $(3) \Rightarrow (4)$: Condition (4) restricts condition (3) to injections between finitely generated modules. Since (3) applies to all injections, it applies to those between finitely generated modules. [/step] [step:Prove $(3) \Rightarrow (2)$: flatness plus right exactness gives preservation of short exact sequences] Let $0 \to N' \xrightarrow{f} N \xrightarrow{g} N'' \to 0$ be a short exact sequence of $R$-modules. By the [Right Exactness of Tensoring](/theorems/2838) applied to $N' \xrightarrow{f} N \xrightarrow{g} N'' \to 0$, the tensored sequence \begin{align*} M \otimes_R N' \xrightarrow{\mathrm{id}_M \otimes f} M \otimes_R N \xrightarrow{\mathrm{id}_M \otimes g} M \otimes_R N'' \to 0 \end{align*} is exact (i.e., exact at $M \otimes_R N$ and $M \otimes_R N''$). It remains to show injectivity of $\mathrm{id}_M \otimes f$. Since $f$ is injective (exactness of the original sequence at $N'$), condition (3) gives $\mathrm{id}_M \otimes f$ injective. Combining, the sequence \begin{align*} 0 \to M \otimes_R N' \xrightarrow{\mathrm{id}_M \otimes f} M \otimes_R N \xrightarrow{\mathrm{id}_M \otimes g} M \otimes_R N'' \to 0 \end{align*} is exact. [/step] [step:Prove $(2) \Rightarrow (1)$: decompose an arbitrary exact sequence into short exact sequences] Let $A \xrightarrow{f} B \xrightarrow{g} C$ be an exact sequence, i.e., $\operatorname{im}(f) = \ker(g)$. We must show $\operatorname{im}(\mathrm{id}_M \otimes f) = \ker(\mathrm{id}_M \otimes g)$. Define the following submodules and quotients: \begin{align*} K &:= \ker(f), & I &:= \operatorname{im}(f) = \ker(g), & Q &:= \operatorname{im}(g). \end{align*} The exactness $\operatorname{im}(f) = \ker(g)$ yields the following short exact sequences: \begin{align*} 0 &\to K \xrightarrow{\iota} A \xrightarrow{\bar{f}} I \to 0, \\ 0 &\to I \xrightarrow{j} B \xrightarrow{\bar{g}} Q \to 0, \end{align*} where $\iota: K \hookrightarrow A$ is the inclusion, $\bar{f}: A \to I$ is $f$ with codomain restricted to $\operatorname{im}(f)$, $j: I \hookrightarrow B$ is the inclusion, and $\bar{g}: B \to Q$ is $g$ with codomain restricted to $\operatorname{im}(g)$. By (2), tensoring each short exact sequence with $M$ preserves exactness: \begin{align*} 0 &\to M \otimes_R K \xrightarrow{\mathrm{id}_M \otimes \iota} M \otimes_R A \xrightarrow{\mathrm{id}_M \otimes \bar{f}} M \otimes_R I \to 0, \\ 0 &\to M \otimes_R I \xrightarrow{\mathrm{id}_M \otimes j} M \otimes_R B \xrightarrow{\mathrm{id}_M \otimes \bar{g}} M \otimes_R Q \to 0. \end{align*} From the first tensored sequence, $\mathrm{id}_M \otimes \bar{f}$ is surjective, so $\operatorname{im}(\mathrm{id}_M \otimes \bar{f}) = M \otimes_R I$. From the second tensored sequence, $\mathrm{id}_M \otimes j$ is injective and $\ker(\mathrm{id}_M \otimes \bar{g}) = \operatorname{im}(\mathrm{id}_M \otimes j)$. Now observe that $f = j \circ \bar{f}$ (as maps $A \to B$) and $g = \iota_Q \circ \bar{g}$ where $\iota_Q: Q \hookrightarrow C$ is the inclusion. Hence $\mathrm{id}_M \otimes f = (\mathrm{id}_M \otimes j) \circ (\mathrm{id}_M \otimes \bar{f})$ and $\mathrm{id}_M \otimes g = (\mathrm{id}_M \otimes \iota_Q) \circ (\mathrm{id}_M \otimes \bar{g})$. We compute: \begin{align*} \operatorname{im}(\mathrm{id}_M \otimes f) &= \operatorname{im}((\mathrm{id}_M \otimes j) \circ (\mathrm{id}_M \otimes \bar{f})) \\ &= (\mathrm{id}_M \otimes j)(\operatorname{im}(\mathrm{id}_M \otimes \bar{f})) \\ &= (\mathrm{id}_M \otimes j)(M \otimes_R I) \\ &= \operatorname{im}(\mathrm{id}_M \otimes j). \end{align*} The third equality uses surjectivity of $\mathrm{id}_M \otimes \bar{f}$. Next: \begin{align*} \ker(\mathrm{id}_M \otimes g) &= \ker((\mathrm{id}_M \otimes \iota_Q) \circ (\mathrm{id}_M \otimes \bar{g})). \end{align*} Since $\iota_Q: Q \hookrightarrow C$ is injective, condition (3) (which follows from (2) by the implication already proved) gives $\mathrm{id}_M \otimes \iota_Q$ injective. Therefore $\ker((\mathrm{id}_M \otimes \iota_Q) \circ (\mathrm{id}_M \otimes \bar{g})) = \ker(\mathrm{id}_M \otimes \bar{g})$. By the second tensored short exact sequence, $\ker(\mathrm{id}_M \otimes \bar{g}) = \operatorname{im}(\mathrm{id}_M \otimes j)$. Combining: $\operatorname{im}(\mathrm{id}_M \otimes f) = \operatorname{im}(\mathrm{id}_M \otimes j) = \ker(\mathrm{id}_M \otimes g)$. [guided] The strategy is to decompose the exactness relation $\operatorname{im}(f) = \ker(g)$ into two short exact sequences, tensor each (preserving exactness by hypothesis (2)), and reassemble. The factorisation $f = j \circ \bar{f}$ separates the "surjective onto the image" part ($\bar{f}$) from the "inclusion of the image into $B$" part ($j$). After tensoring: - $\mathrm{id}_M \otimes \bar{f}$ is surjective (from the first tensored SES), so $\operatorname{im}(\mathrm{id}_M \otimes f) = (\mathrm{id}_M \otimes j)(M \otimes_R I)$. - $\mathrm{id}_M \otimes j$ is injective (from the second tensored SES), so $(\mathrm{id}_M \otimes j)(M \otimes_R I) = \operatorname{im}(\mathrm{id}_M \otimes j)$. For the kernel side, $g = \iota_Q \circ \bar{g}$. Since $\iota_Q$ is injective and (2) implies (3), $\mathrm{id}_M \otimes \iota_Q$ is injective. The injectivity of $\mathrm{id}_M \otimes \iota_Q$ ensures $\ker(\mathrm{id}_M \otimes g) = \ker(\mathrm{id}_M \otimes \bar{g})$, and the second tensored SES gives $\ker(\mathrm{id}_M \otimes \bar{g}) = \operatorname{im}(\mathrm{id}_M \otimes j)$. Why could this fail without flatness? Without the injectivity of $\mathrm{id}_M \otimes j$ (which is the flatness input), the image of $\mathrm{id}_M \otimes f$ could be strictly smaller than the kernel of $\mathrm{id}_M \otimes g$: the tensor product could "create" new kernel elements that do not come from $M \otimes_R A$. [/guided] [/step] [step:Prove $(4) \Rightarrow (3)$: reduce to finitely generated submodules] Let $f: N' \to N$ be an injective $R$-module homomorphism. We must show $\mathrm{id}_M \otimes f: M \otimes_R N' \to M \otimes_R N$ is injective. Let $\xi \in \ker(\mathrm{id}_M \otimes f)$. Write $\xi$ as a finite sum of simple tensors: \begin{align*} \xi = \sum_{k=1}^{r} m_k \otimes n'_k \end{align*} for some $m_1, \ldots, m_r \in M$ and $n'_1, \ldots, n'_r \in N'$. The condition $(\mathrm{id}_M \otimes f)(\xi) = 0$ means \begin{align*} \sum_{k=1}^{r} m_k \otimes f(n'_k) = 0 \quad \text{in } M \otimes_R N. \end{align*} Let $N'_0$ denote the submodule of $N'$ generated by $\{n'_1, \ldots, n'_r\}$. This is a finitely generated submodule of $N'$. Since $f$ is injective, the restriction $f|_{N'_0}: N'_0 \to N$ is injective. Let $N_0$ denote the submodule of $N$ generated by $\{f(n'_1), \ldots, f(n'_r)\}$. Then $f(N'_0) \subseteq N_0$, and both $N'_0$ and $N_0$ are finitely generated. The relation $\sum_{k=1}^{r} m_k \otimes f(n'_k) = 0$ holds in $M \otimes_R N$. Since the $f(n'_k)$ all lie in $N_0$, this relation also holds in $M \otimes_R N_0$ (the canonical map $M \otimes_R N_0 \to M \otimes_R N$ induced by the inclusion $N_0 \hookrightarrow N$ sends $\sum m_k \otimes f(n'_k)$ in $M \otimes_R N_0$ to $\sum m_k \otimes f(n'_k)$ in $M \otimes_R N$; but we need to be careful since elements can be zero in $M \otimes_R N$ without being zero in $M \otimes_R N_0$). To handle this, we use the following finitary observation: a relation $\sum_k m_k \otimes x_k = 0$ in $M \otimes_R N$ that involves only finitely many elements $x_k$ from a directed system of finitely generated submodules of $N$ is already witnessed in some finitely generated submodule. More precisely, since $N = \varinjlim_\lambda N_\lambda$ as the directed colimit of its finitely generated submodules, and tensor products commute with directed colimits ($M \otimes_R N = \varinjlim_\lambda (M \otimes_R N_\lambda)$), the element $\sum_k m_k \otimes f(n'_k)$ mapping to zero in $M \otimes_R N$ means it maps to zero in $M \otimes_R N_\lambda$ for some finitely generated submodule $N_\lambda$ of $N$ containing all the $f(n'_k)$. Fix such an $N_\lambda$. Define $N'_\lambda := f^{-1}(N_\lambda) \cap N'_0$. Since $f(N'_0) \subseteq N_\lambda$ (as all $f(n'_k) \in N_\lambda$), in fact $N'_\lambda = N'_0$. The restriction $f|_{N'_0}: N'_0 \to N_\lambda$ is injective (as $f$ is injective globally). Both $N'_0$ and $N_\lambda$ are finitely generated. By condition (4), the map $\mathrm{id}_M \otimes (f|_{N'_0}): M \otimes_R N'_0 \to M \otimes_R N_\lambda$ is injective. Since $\sum_k m_k \otimes f(n'_k) = 0$ in $M \otimes_R N_\lambda$ and this is the image of $\sum_k m_k \otimes n'_k$ under $\mathrm{id}_M \otimes (f|_{N'_0})$, injectivity gives $\sum_k m_k \otimes n'_k = 0$ in $M \otimes_R N'_0$. The canonical map $M \otimes_R N'_0 \to M \otimes_R N'$ (induced by the inclusion $N'_0 \hookrightarrow N'$) sends $\sum_k m_k \otimes n'_k$ to $\xi$. Since $\sum_k m_k \otimes n'_k = 0$ in $M \otimes_R N'_0$, we get $\xi = 0$ in $M \otimes_R N'$. Hence $\ker(\mathrm{id}_M \otimes f) = 0$, so $\mathrm{id}_M \otimes f$ is injective. [guided] The key difficulty in $(4) \Rightarrow (3)$ is that condition (4) only handles finitely generated modules, but $N'$ and $N$ may be arbitrary. The resolution uses the fact that tensor products commute with directed colimits. Any $R$-module $N$ is the directed colimit of its finitely generated submodules: $N = \varinjlim_\lambda N_\lambda$ where the index set is the collection of all finitely generated submodules ordered by inclusion. The tensor product commutes with this colimit: $M \otimes_R N \cong \varinjlim_\lambda (M \otimes_R N_\lambda)$. What does this mean concretely? An element $\sum_k m_k \otimes x_k \in M \otimes_R N$ is zero if and only if it is zero in $M \otimes_R N_\lambda$ for some finitely generated $N_\lambda$ containing all the $x_k$. This is because the colimit is a filtered colimit of abelian groups: an element maps to zero in the colimit iff it maps to zero at some finite stage. Armed with this, the argument is: take $\xi = \sum_k m_k \otimes n'_k$ in $\ker(\mathrm{id}_M \otimes f)$. The elements $n'_1, \ldots, n'_r$ generate a finitely generated submodule $N'_0 \subseteq N'$. The relation $\sum_k m_k \otimes f(n'_k) = 0$ in $M \otimes_R N$ is witnessed in some finitely generated submodule $N_\lambda \subseteq N$ containing $f(N'_0)$. The restricted injection $f|_{N'_0}: N'_0 \hookrightarrow N_\lambda$ has finitely generated source and target, so (4) gives $\mathrm{id}_M \otimes (f|_{N'_0})$ injective, forcing $\xi = 0$. What would go wrong if we tried to apply (4) directly without the colimit argument? The relation $\sum_k m_k \otimes f(n'_k) = 0$ in $M \otimes_R N$ does not immediately imply the same relation in $M \otimes_R N_0$ for the finitely generated submodule $N_0$ generated by the $f(n'_k)$: the tensor relation might involve elements of $N$ outside $N_0$. The directed colimit machinery handles exactly this subtlety. [/guided] [/step]