Stabilizer of two short exact sequences at the same time

Question

For two short exact sequences of say, finitely generated modules of some ring, $0\rightarrow N\xrightarrow{a} R\xrightarrow{b} M\rightarrow0, 0\rightarrow K\xrightarrow{a'}R\xrightarrow{b'}L\rightarrow0.$ Let $\phi \in Aut(R)$ and $\phi$ acting on these two short exact sequences by $\phi.a=\phi\circ a, \phi.b=b\circ\phi^{-1},\phi.a'=\phi\circ a', \phi.b'=b'\circ\phi^{-1}.$ My question is that the book I am reading says that by some diagram chasing, the cardinality of stabilizer of this action of $Aut(R)$ is the same as the cardinality of $Hom(Coker\space b'a, Ker\space b'a).$ I have no idea how to do explain it, can somebody help me?

Answer 1

First, I make a remark on why should $Coker (b'\circ a)$ and $Ker(b'\circ a)$ enter the picture.

Let $C=Coker (b'\circ a)$ and $D=Ker(b'\circ a)$. Let also slightly abuse notation to identify $N$ with $a(N)$ and $K$ with $a'(K)$. Let $Stab$ denote the set of all $\phi\in Aut(R)$ that stabilize both sequences. Notice that $\phi\in Stab\ $ if and only if

1. $\phi$ is the identity on $N$ and $K$;
2. $\phi(r)-r\in N\cap K$.

Therefore $C$ and $D$ are naturally relevant to the situation, because we can think $C\cong R/(N+K)$ and $D\cong N\cap K$.

Now I answer your question.

Let $\psi: Hom(C,D)\to End(R)$ be given by $h\mapsto \phi=d\circ h\circ c + id_R$, where $c: R\to L\to C$ and $d: D\to N\to R$. I claim that $\psi$ provides the bijection between $Hom(C,D)$ and $Stab$.

In other words, $\phi=id+h$, all language abuses being made.

What follows is tedious verification

Let $h\in Hom(C,D)$, $\phi=\psi(h)$ and $r\in R$.

1. $\phi(r)=r$ if $r\in K$. Indeed $R\to L$ is the kernel of $K\to R$, so $c(r)=0$;
2. $\phi(r)=r$ if $r\in N$. Indeed $L\to C$ is the kernel of $N\to L$, so $c(r)=0$;
3. $\phi(r)-r\in N\cap K$. Indeed $\phi(r)-r$ is in the range of $d$;
4. $\phi:R\to R$ is invertible because $\phi-id_R$ is nilpotent of order two (so the inverse of $\psi(h)$ is $\psi(-h)$).

Thus we actually have $\psi: Hom(C,D)\to Stab$.

Given $\phi\in Stab$, we reconstruct $h=\psi^{-1}(\phi)$ as follows.

Let $s\in C$, choose $r\in R$ with $c(r)=s$, let $h(s)=d^{-1}(\phi(r)-r)$.

1. $h$ is well-defined. Indeed if $s=c(r')$ then $r'=r+k+n$ for some $k\in a'(K)$ and $n\in a(N)$. But then $\phi(k)=k$ and $\phi(n)=n$, so $\phi(r')-r'=\phi(r)-r$.
2. $\psi(h)=\phi$. Indeed $d\circ h\circ c (r) = \phi(r)-r$ for all $r\in R$;
3. If $\phi=\psi(k)$, then $h=k$. Indeed $c:R\to C$ is surjective and $h(c(r)) = d^{-1}\circ d\circ k\circ c(r)$ for all $r\in R$.