Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\phi: Y\to X$ be an affine (finite & dominant) morphism of (smooth) $\Bbbk$-varieties. Let $Y_P$ be the scheme-theoretic fiber of $P\in X$, i.e. $Y_P=Y\times_X\mathrm{Spec}(\Bbbk(P))$. I was told that $Y_P=\mathrm{Spec}((\phi_\ast\mathcal{O}_Y)_P)$, but I do not see how to prove it.

(I assume that the conditions in brackets can be omitted, but you may assume them if necessary)

Clearly, we can do this locally and assume $Y=\mathrm{Spec}(B)$ as well as $X=\mathrm{Spec}(A)$. Now, I would have said that $Y_P=\mathrm{Spec}(B\otimes_A\Bbbk(P))$ but $(\phi_\ast\mathcal{O}_Y)_P=B\otimes_A A_P$, which is why I am confused.

share|improve this question
    
I would phrase your last paragraph differently: clearly we can do this locally on $X$, so assume $X=\operatorname{Spec}A$, and, since $\phi$ is affine, that also $Y$ is affine. –  Mariano Suárez-Alvarez Jan 21 '12 at 23:36

1 Answer 1

up vote 7 down vote accepted

"I was told that..." They lied !

More seriously, everything you write is perfectly correct, including the reduction to the case where both schemes are affine. And indeed the conditions in brackets are irrelevant.

The morphism $i: Spec (A_P) \to X=Spec (A)$ however is not uninteresting.
It is technically not an immersion but strongly resembles one. Its image is exactly the intersection of all neighbourhoods of $P \in X$, and may be thought of as some sort of germ of $X$ at $P$.

Coming back to $\phi:Y\to X$, the canonical morphism (= projection) of the fibered product $Y\times_X Spec (A_P)=Spec( B\otimes_A A_P) $ into $Y \;$ is injective and its image $Y_{(P)}\subset Y$ can be thought of as some thickening of the genuine fiber $Y_P=Spec (B\otimes_A\kappa(P))$ .
Summing-up, we have $Y \supset Y_{(P)}=\phi^{-1}(Spec (A_P))\supset Y_P=\phi^{-1}(Spec (\kappa(P)))$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.