Computing $\operatorname{Ext}^1_R(R/x,M)$ How to compute $\operatorname{Ext}^1_R(R/(x),M)$ where $R$ is a commutative ring with unit, $x$ is a nonzerodivisor and $M$ an $R$-module?
Thanks.
 A: $0 \to R \xrightarrow{~x} R \to R/(x) \to 0$ is a projective resolution. Applying $\hom(-,M)$ gives the complex $0 \to \hom(R/(x),M) \to \hom(R,M) \xrightarrow{x} \hom(R,M) \to 0$, which identifies with $0 \to \{m \in M : xm=0\} \to M \xrightarrow{x} M \to 0$. It follows 
$$ \mathrm{Ext}^n(R/(x),M) = \left\{\begin{array}{l} \{m \in M : xm=0\} & n =0 \\ M/xM & n=1 \\ 0 & n >1 \end{array}\right.$$
A: There is an alternative way to doing this problem than taking a projective resolution. Consider the ses of $R$ - modules
$$0 \to R \stackrel{x}{\to} R \to R/(x) \to 0$$
where the multiplication by $x$ map is injective because it is not a zero divisor in $R$. Now we recall a general fact from homological algebra that says any SES of $R$ - modules gives rise to an LES in Ext. We need only to care about the part
$$0 \to \textrm{Hom}_R(R/(x),M) \to \textrm{Hom}_R(R,M) \stackrel{f}{\to} \textrm{Hom}_R(R,M) \to \textrm{Ext}^1_R(R/(x),M) \to 0 \to 0\ldots $$
where the zeros appear  because $R$ as a module over itself is free (and hence projective) so that $\textrm{Ext}^1_R(R,M) = 0$. Now we recall that $\textrm{Hom}(R,M) \cong M$ because any homomorphism from $R$ to $M$ is completely determined by the image of $1$. It is easily seen now that under this identification, $\textrm{im} f \cong xM$ so that 
$$\textrm{Ext}^1_R(R/(x),M) \cong M/xM.$$
