This is related to Weibel Exercise 4.3.3 Third Theorem for injective change of rings of Chpt 4, Sec 3. $id_R(-)$ denotes injective dimension of $(-)$ module over ring $R$.
Third Theorem: Suppose $R$ is commutative noetherian local ring and $A$ is f.g. over $R$. Say maximal ideal of $R$ is $m$. Let $x\in m$ s.t. $x\cdot:A\to A$ and $x\cdot:R\to R$ are injections.(i.e. $x$ multiplication is injective.) Then $id_R(A)=id_R(A/xA)$.
I have done two theorems before this one. $id_R(A)\geq id_R(A/xA)$ is instantaneous from the theorem before. I have no idea how to prove $id_R(A)\leq id_R(A/xA)$.
$\textbf{Q:}$ How do I proceed to prove the inequality? I considered induction on $id_R(A/xA)$. I could not even get the base case working. However, the least number it can start with is $id_R(A/xA)\geq 1$ by $id_R(A/xA)=id_{R/x}(A/xA)+1$ if $A/xA\neq 0$. In other words, if I have $id_R(A/xA)=0$, I would have $A/xA=0$ but over noetherian local ring, I would have $A=xA\implies A=0$ by $A$ f.g. So the base case for induction is $1$. The annoying part is given a short exact sequence over $R$ of $A$ as the following $0\to A\to I\to C\to 0$ with $I$ injective over $R$. If I tensor the sequence with $R/x$, I may get non-trivial contribution from $Tor^1(I,R/x)$. This term is normally turned off by projectivity and one has to show $Tor^1(A,R/x)=0$ to get the proof working.
The other option is to consider $0\to A\xrightarrow{x\cdot}A\to A/xA\to 0$ exact sequence. Take any $P_\cdot\to M$ projective resolution of $M$ over $R$. Consider $0\to Hom(P_\cdot, A)\to Hom(P_\cdot, A)\to Hom(P_\cdot, A/xA)\to 0$ inducing long exact sequence. The last $Hom_R(P_\cdot, A/xA)=Hom_R(P_\cdot,Hom_{R/x}(R/x,A/xA))=Hom_{R/x}(P_\cdot\otimes R/x,A/xA)$. So there will be $Tor^1(M,R/x)$ "showing up" but $Hom_{R/x}(-,A/xA)$ is no longer exact functor. It is not something like $Hom(Tor^1(M,R/x),A/xA)$.