Injective Change of Rings

Question

Sorry if this is too elementary, but when I was going to ask this question on math.stackexchange, I saw the same question with three up-votes and no answer. So I decided to post it here.

I am doing the following problem:

Let $R$ be a ring, $x\in R$ a central non-unit non-zerodivisor. If $A\neq 0$ is an $R/xR$ module with $id_{R/xR}A$ finite, then $$id_R(A)=1+id_{R/xR}A$$ where $id_{R}（id_{R/xR}）$ means injective dimension as a module over $R(R/xR)$

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 4.3.3 in the book), which uses induction on $n=pd_{R/x}A$. The problem is that I cannot prove the base case, which says that:

If $A$ is an injective $R/xR$ module, then $id_A=1$.

I can prove that $A$ is not an injective $R$ module, so $id_RA\geq1$, but I cannot prove the other inequality. Can anyone give some hints to me? Thanks in Advance.

I don't know if it's against the rules to copy-paste another question, if so, please tell me.

Answer 1

The underlying idea of the below argument comes from the technique of spectral sequence presented in Chapter five of Weibel's book.

Let $A$ be an injective $R/xR$-module. Let $M$ be an arbitrary $R$-module. Let $P_*\to M$ be a projective resolution of $M$. One has $$ \hom_{R}(P_*,A) \cong \hom_R(P_*, \hom_{R/xR}(R/xR, A))\cong \hom_{R/xR}(P_*\otimes_{R}R/xR, A). $$ Note that $H_i(P_*\otimes_RR/xR) = \text{Tor}^R_i(M,R/xR) =0$ for $i\geq 2$. Since $A$ is injective, $$ H^i(\hom_{R/xR}(P_*\otimes_{R}R/xR, A)) =0 $$ for $i\geq 2$. Thus $\text{Ext}_R^i(M,A)=0$ for $i\geq 2$. This shows that $\text{id}_R(A)\leq 1$.