$0 \rightarrow Q \rightarrow M \rightarrow P \rightarrow 0$ is isomorphic to some $0 \rightarrow N \rightarrow M \rightarrow M/N \longrightarrow 0$

Question

Let $A$ be a ring and $M, Q, P$ be $R$-modules. Let $N$ be a submodule of $M$. Then any short exact sequence of the form $0 \longrightarrow Q \longrightarrow M \longrightarrow P \longrightarrow 0$ is isomorphic to the exact sequence an exact sequence $0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0$. Why is that the case? I imagine we somehow consider the diagram \begin{array}{ccccccccc} 0 &\longrightarrow & Q & \longrightarrow & M & \longrightarrow & P & \longrightarrow & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 &\longrightarrow & N & \longrightarrow & M & \longrightarrow & M/N & \longrightarrow & 0 \end{array} and want to construct some isomorphism between $P$ and $M/N$, and between $Q$ and $N$, but I'm having a hard time seeing how we could do so.

Answer 1

What it means for $0 \longrightarrow Q \longrightarrow M \longrightarrow P \longrightarrow 0$ to be a short exact sequence is that

(1) $Q \rightarrow M$ is injective.

(2) $\text{im}(Q \rightarrow M) = \ker(M \rightarrow P)$

(3) $M \rightarrow P$ is surjective

Informally, we get the sequence $0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0$ by conceiving of $Q$ as a submodule of $M$ and $P$ as a projection of $M$.

To be precise, we're just going to apply the first isomorphism theorem for modules twice. As you suspected, this produces canonical isomorphisms $Q \cong N$ and $P \cong M/N$. After establishing the isomorphisms, it will just remain to check that the sequences themselves are isomorphic, which amounts to checking that the diagram you gave above is commutative.

Let's set $N = \text{im}(Q \rightarrow M)$. Then the first isomorphism theorem guarantees $Q / \ker(Q \rightarrow M) \cong \text{im}(Q \rightarrow M) = N$, and by $(1)$, the kernel is trivial so that $Q \cong N$.

Similarly the first isomorphism theorem guarantees $M / \ker(M \rightarrow P) \cong \text{im}(M \rightarrow P)$. Applying both $(3)$ $\big(\text{im}(M \rightarrow P) =P \big)$, and $(2)$ $\big(\ker(M \rightarrow P) = N\big)$ we have $M / N \cong P$.

Answer 2

Hint: The diagram you are looking for is $\require{AMScd}$ \begin{CD} 0 @>>> Q @>i>> M @>p>> P @>>> 0\\ @. @ViVV @V\operatorname{id}_MVV @V\varphi VV \\ 0 @>>> \operatorname{im} i @>>> M @>>> M/\operatorname{im} i @>>> 0 \end{CD}

Where $\varphi$ is the unique map such that $(\varphi\circ p)(m) = m + \operatorname{im}i$. To prove that such $\varphi$ exists (and is unique), use that $p$ is epimorphism and $\operatorname{im} i = \ker p$.