Injective modules in a short exact sequence

Question

Let $0→A→B→C→0$ be an exact sequence in the category of $R$ modules, where $R$ is commutative with $1$, and $B$ be injective. In a text book it is said that all three modules are injective, or the injective dimension of $A$ is one more than that of $C$. I could not understand why. Can anybody explain this to me? Thanks.

Answer 1

Fact: inj dim $M\leq i$ if and only if $\mathrm{Ext}^{i+1}(N,M)=0$ for every (cyclic) module $N$.

Using the long exact homology sequence for $\mathrm{Ext}$ we get: $$0\to \mathrm{Hom}(N,A)\to \mathrm{Hom}(N,B)\to \mathrm{Hom}(N,C)\to \mathrm{Ext}^1(N,A)\to 0\to \mathrm{Ext}^1(N,C)\to \mathrm{Ext}^2(N,A)\to 0\to \mathrm{Ext}^2(N,C)\to \cdots$$

This shows that $\mathrm{Ext}^i(N,C)\simeq \mathrm{Ext}^{i+1}(N,A)$, for all $i\ge 1$

Suppose that $\mathrm{id}(C)=n$; then $\mathrm{Ext}^{n+1}(N,C)=0$, and then $\mathrm{Ext}^{n+2}(N,A)=0$, so $\mathrm{id}(A)\le n+1$. If $\mathrm{id}(A)\le n$ we get $\mathrm{Ext}^{n+1}(N,A)=0$, so $\mathrm{Ext}^{n}(N,C)=0$, and thus $\mathrm{id}(C)<n$, a contradiction.