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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?


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up vote 9 down vote accepted

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

$$\textrm{Ext}^1_R(R/(x),M) \cong M/xM.$$

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Who downvoted my answer? Why is it wrong? –  fpqc Nov 21 '12 at 9:57
I made a typo in writing $\textrm{Ext}(R,R)$ instead of $\textrm{Ext}(R,M)$. Still the fact remains that the latter $R$ - module is zero. –  fpqc Nov 21 '12 at 10:01
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$0 \to R \xrightarrow{~x} R \to R/(x) \to 0$ 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

$$ \mathrm{Ext}^n(R/(x),M) = \left\{\begin{array}{l} \{m \in M : xm=0\} & n =0 \\ M/xM & n=1 \\ 0 & n >1 \end{array}\right.$$

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