TITLE: Base step of induction in Hartshorne III 9.5
QUESTION [7 upvotes]: 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 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-dimensionsional 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?

REPLY [5 votes]: 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.