# Stabilizer of two short exact sequences at the same time

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?

## 1 Answer

### 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 $$r\in R$$.