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Let $f\colon X \to Y$ be a flat morphism of schemes of finite type over a field k. For any point $x\in X$, let $y = f(x)$.

Assume that $\operatorname{dim} Y = 0$.

Hartshorne now claims that $X_y$ is defined by a nilpotent ideal in $X$. Why?

I made the following attempt:

If $Y$ is of dimension zero, then, in the classical case, we have a variety $X$ mapping to a bunch of isolated points. Since $y = f(x)$, if we further assume that $X$ is connected, then the whole of $X$ maps to $y$. So, indeed, $X_y$ is the whole of $X$, and the whole of $X$ is defined by a nilpotent ideal.

But how does it work in the general case? How does one prove, in the scheme setting, that if $Y$ is zero-dimensional then every component of $X$ maps to a (obviously closed) point of $Y$? And how does one prove first that $X_y$ is defined by a nilpotent ideal, instead of deducing it from the fact that it is the whole space?

cos_dm_math21
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Rodrigo
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1 Answers1

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As you point out we may assume $Y=Spec(A)$ where $(A, \mathfrak{m})$ is a local Artinian $k$-algebra and that $y$ corresponds to $\frak{m}$. In this situation $X_y=X\otimes_A A/\frak{m}$ (I'll use this shorthand for $X\times_{Spec(A)}Spec(A/\mathfrak{m})$).

Now we just want to figure out the ideal sheaf defining $X_y\hookrightarrow X$. Consider the short exact sequence $0\to \mathfrak{m}\to A\to A/\mathfrak{m} \to 0$.

By flatness, we still get exactness $0\to \mathfrak{m}\mathcal{O}_X \to \mathcal{O}_X \to \mathcal{O}_X\otimes_A A/\mathfrak{m}\to 0$. But $\mathcal{O}_X\otimes_A A/\mathfrak{m}=\mathcal{O}_{X_y}$ and hence the ideal sheaf is $\mathfrak{m}\mathcal{O}_X$. This is nilpotent because $A$ is Artinian.

Matt
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  • Thank you Matt! I think it remained to say that because Y is of finite type over k, it is noetherian. Then by, say 8.5 Atiyah, Noetherian + dim 0 <=> Artinian. I'll not accept the question just yet, because I would like to see an answer with a more geometric explanation, although your exposition was very clear and helpful :) – Rodrigo Jun 27 '13 at 21:41
  • @Rodrigo Geometrically the explanation is this. Your base is 0-dimensional and you only care a certain fiber, so it allows you to restrict to that component. But that component is a purely infinitesimal thing (one closed point plus nilpotent fuzzy other stuff). There is a fiber over the closed point (sometimes called "the closed fiber" and sometimes called "the special fiber"). You start with a flat map, so this could be thought of as a "family" containing the closed fiber or "deformation" of the closed fiber. – Matt Jun 28 '13 at 16:16
  • (cont) In either case, the fact that the base is infinitesimal just forces the total family over the base to be a single "nice" closed fiber (finite type over a field, closed in the family) with some infinitesimal other stuff. This is usually called an "infinitesimal thickening" of the scheme $X_y$. It is just $X_y$ plus some extra fuzzy stuff. But this is geometrically just what it means to have nilpotent ideal sheaf. An infinitesimal thickening has the same underlying topological space (just like $Spec(A)$ had the same top space as $Spec(A/m)$). – Matt Jun 28 '13 at 16:20