Proving second injective change of rings theorem

Question

I am trying to prove the second injective change of rings theorem:$\DeclareMathOperator{\id}{id}$

Let $R$ be a ring, $A$ be an $R$-module, $x\in R$ a central non-unit non-zerodivisor such that $xa\neq 0$ for all nonzero $a\in A$. Then either $A$ is injective, or $$\id_R(A)\geqslant 1+\id_{R/xR}(A/xA)$$ where $\id_{R}$ denotes the injective dimension as a module over $R.$

This is Exercise 4.3.3 of Weibel's Introduction to homological algebra. I am trying to mimic the proof of the corresponding theorem for projective dimension (theorem 4.3.5 in the book), which uses induction on $n=\id_{R/x}A$.

What I have right now:

Assume $A$ is not injective. Then we have an exact sequence

$$0\longrightarrow A\longrightarrow Q\longrightarrow E\longrightarrow 0 $$ with $Q$ injective. Applying $\text{Hom}(R/xR,-)$ (over $R$), we get an exact sequence $$0\longrightarrow \text{Hom}(R/xR,A)\longrightarrow \text{Hom}(R/xR,Q)\longrightarrow \text{Hom}(R/xR,E)\longrightarrow \text{Ext}^1_R(R/xR,A)\longrightarrow 0 $$ This is actually a sequence of $R/xR$-modules. Now, since $x$ is a nonzerodivisor on $A$, $\text{Hom}(R/xR,A)=0$. Also, $\text{Ext}^1_R(R/xR,A)=A/xA$. I was going to use that to connect $\id_{R/xR}(A/xA)$ to $\id_{R/xR}(E/xE)$ and use induction. Everything would work if I knew how to show something like $\id_{R/xR}(\text{Hom}(R/xR,E))\leqslant \id_R(E)$, but I am stuck at this step (and not sure whether this is the correct way to go).

Answer 1

Assume that $A$ is not injective with $\operatorname{id}_RA<\infty$. Let $0\longrightarrow A\longrightarrow Q\longrightarrow E\longrightarrow 0 $ be exact with $Q$ the injective hull of $A$. Since $x$ is a nonzero divisor on $A$, and $\operatorname{ker}(Q\stackrel{x}{\longrightarrow}Q)\cap A=0$, so $Q\stackrel{x}{\longrightarrow}Q$ is a monomorphism. Thus $Q\stackrel{x}{\longrightarrow} xQ$ is an isomorphism. Since $Q$ is an injective $R$-module, one can easily show that $Q=xQ$, then the previous isomorphism induces an isomorphism $E=Q/A\cong xQ/xA= Q/xA$.

On the other hand, notice $\operatorname{Hom}_R(R/(x),E)\cong \operatorname{Hom}_R(R/(x),Q/xA)=A/xA$. Suppose $0\longrightarrow E\longrightarrow E^0\longrightarrow E^1\longrightarrow...\longrightarrow E^n\longrightarrow0$ is an injective resolution of $E$. Applying $\operatorname{Hom}_R(R/(x),-)$ to this resolution, we get an injective resolution of $\operatorname{Hom}_R(R/(x),E)=A/xA$ over $R/(x)$. All these together solve the problem.