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.



Injective Change of Rings
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 $\text{id}\_R(A)\leq 1$.


