Relation of $\operatorname{Ext}$ and projective dimension I have some problem to understand the proof of proposition 8.38, page 473 from An Introduction to Homological Algebra by Rotman.
Proposition: Let $x\in Z(R)$ be an element which is not a zero-divisor and let $R^*=R/(x)$. Moreover, let $M$ be a left $R$-module with $x$ regular in $M$. If $\DeclareMathOperator{\pd}{pd}\pd_R(M)=n\lt\infty$, then $\pd_{R^*}(M/xM)\le{n-1}$.

For the proof I have to use the fact that if $\pd_R(M)=n\lt\infty$, we have that $\operatorname{Ext}_R^{n+1}(L,M)=\{0\}$ for all left  $R$-modules $L$. I don't know why. Can you give me a hint? 

Thanks!
 A: I have no idea how Rotman gets from  $\DeclareMathOperator{\pd}{pd}\pd_R(M)=n\lt\infty$ that $\operatorname{Ext}_R^{n+1}(L,M)=0$ for all $R$-modules $L$. (In my opinion this is a mistake.) Instead I can show you an alternative proof:

Let $x\in Z(R)$ be an element which is not a zero-divisor and let $R^*=R/(x)$. Moreover, let $M$ be a left $R$-module with $x$ regular in $M$. 
  If $\pd_R(M)\lt\infty$, then $\pd_{R^*}(M/xM)\lt\infty$.

Let $0\to F_n\to F_{n-1}\to\cdots\to F_1\to F_0\to M\to 0$ be a projective (free) resolution of $M$. One can split it into short exact sequences $0\to K_i\to F_i\to K_{i-1}\to 0$, $0\le i\le n-1$. (Here $K_{-1}=M$.) Now notice that $x$ is regular on $K_i$ for all $i\ge 0$ since $x$ is regular on $R$, and therefore on every free $R$-module. 
Start with $$0\to K_0\to F_0\to M\to 0$$ and tensor it by $R/(x)$. We get $$0\to K_0/xK_0\to F_0/xF_0\to M/xM\to 0.$$ This follows from $\operatorname{Tor}_1^R(R/(x),M)=0$ which at its turn follows from the short exact sequence $0\to R\stackrel{x\cdot}\to R\to R/(x)\to 0$ after tensorizing it by $M$ and taking into account that $x$ is $M$-regular.
Then $$0\to K_i/xK_i\to F_i/xF_i\to K_{i-1}/xK_{i-1}\to 0$$ are short exact sequences for $0\le i\le n-1$. This shows us that $0\to F_n/xF_n\to F_{n-1}/xF_{n-1}\to\cdots\to F_1/xF_1\to F_0/xF_0\to M/xM\to 0$ is a projective (free) resolution of $M/xM$, so $\pd_{R^*}(M/xM)\lt\infty$. $\quad\blacksquare$
Now set $k=\pd_{R^*}(M/xM)$. From Rotman, Proposition 8.38(ii) (whose proof is pretty clear!) we get $k+1\le\pd_R(M)\Leftrightarrow k\le\pd_R(M)-1$.
