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This is probably a trivial question.

While reading the paper

R. Elkik, Singularites rationnelles et deformations, Invent. Math. 47 Ž1978., 139147.

I came across the following short exact sequence. Consider a flat morphism $f:X\rightarrow S=Spec(R)$ of k-schemes of finite type, $X$ normal + CM, and pick a regular parameter $t\in R$. If $X_t=X\times_{S} Spec R/tR$ is the fiber over $t$, then multiplication by $t$ induces a short exact sequence $$0 \rightarrow \omega_X \stackrel{t}{\rightarrow} \omega_X \rightarrow \omega_{X_t} \rightarrow 0$$

Is it straightforward to derive such a sequence?

Thanks in advance for any suggestion.

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Sure.

Note first that we have a short exact sequence

$$ 0 \to O_X \xrightarrow{t} O_X \to O_{X_t} \to 0 $$ If $D$ is the divisor corresponding to $t = 0$ on $X$, you can also view this as $$0 \to O_X(-D) \to O_X \to O_D \to 0.$$

Anyways, now apply the functor $\mathcal{H}om_{O_X}(\bullet, \omega_X)$. You get $$ 0 \to \mathcal{H}om_{O_X}(O_{X,t}, \omega_{X}) \to \mathcal{H}om_{O_X}(O_X, \omega_X) \xrightarrow{t} \mathcal{H}om_{O_X}(O_X, \omega_X) \to \mathcal{E}xt^1(O_{X_t}, \omega_X) \to \mathcal{E}xt^1_{O_X}(O_X, \omega_X) $$ Ok, the first term is zero (homing torsion into non-torsion), the last term is also zero since you are homing from a free module. Finally, $\mathcal{E}xt^1(O_{X_t}, \omega_X) = \omega_{X_t}$ by basic properties of dualizing/canonical modules, and the result follows.

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  • $\begingroup$ I suppose $\omega_X$ really means $\omega_{X/R}$ (not the same thing if $R$ isn't Gorenstein.) The fact that $\omega_X$ is $R$-flat rests on the fact that $f$ automatically has CM fibers (since it is flat and $X$ is CM). Once this $R$-flatness property is noted, the question amounts to the formation of $\omega_{X/R}$ commuting with base change on $R$ along the quotient map $R \rightarrow R/(t)$, but as you know, in the CM-morphism case it commutes with any base change at all. $\endgroup$ – user36938 Jul 31 '13 at 4:03
  • $\begingroup$ You are making this too hard. I mean $\omega_X$ not $\omega_{X/R}$ (although in Elkik's case, you can assume $R$ is a DVR if I recall correctly). You don't need the flatness at all, $\omega_X$ is S2, so a regular element for $R$ acts as a regular element on $\omega_X$. There is no need for any base change for canonical sheaves here. $\endgroup$ – Karl Schwede Jul 31 '13 at 4:07
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    $\begingroup$ @Karl Schwede: OK, not having the paper in front of me, I only saw what is in the question, where nothing is said about $R$ or what $\omega_X$ is supposed to mean, so I had to guess. If $R$ is CM but not Gorenstein then $\omega_X$ and $\omega_{X/R}$ are quite different things, so are you saying Elkik uses $\omega_X$ and not $\omega_{X/R}$? Also, $R$-flatness tells us that $t$ is regular on $X$, but the S2 condition merely concerns the existence of some sequence regular for the module, not that a specific element of the ring acts injectively. Why does $t$ act injectively on $\omega_X$? $\endgroup$ – user36938 Jul 31 '13 at 4:45
  • $\begingroup$ In Elkik's paper, all schemes are finite-type $\mathbb{C}$-schemes. So $\omega_X$ makes perfect sense. Elkik really wants to prove something about $\omega_X$ and not the relative term (although you can prove something along the line he wants with the relative term). You are right though, without a little more information you can't know that $\omega_X$ even makes sense. Let me respond to the other momentarily. $\endgroup$ – Karl Schwede Jul 31 '13 at 11:51
  • $\begingroup$ Since $X \to \text{Spec R}$ is flat, that fact that $t$ is a non-zero divisor on $R$ implies $t$ is a non-zero divisor on $X$. Since it's a non-zero divisor on $X$ it is a non-zero divisor on $\omega_X$. There isn't anything fancy to see this last statement, you can use the exact sequence I wrote in my answer. Note that the first $\mathcal{H}om$ is still zero by dimension reasons. $\endgroup$ – Karl Schwede Jul 31 '13 at 12:00

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