$X \to Y$ flat $\Rightarrow$ the image of a closed point is also a closed point?

Solution 1:

This is just a consequence of Nullstellensatz (the flatness is useless). If $x\in X$ is a closed point, then $k(x)$ is a finite extension of $k$ and so is $k(f(x))$ (being a subextension of $k(x)$). This is enough to show that $f(x)$ is a closed point.

If $X, Y$ are not finite type over a field, this is false even if $f$ is finite type (and faithfully flat if you like): consider a DVR $R$ with uniformizing element $\pi$, and let $f : X=\mathbb A^1_R \to Y=\mathrm{Spec}(R)$ be the canonical morphism. It is finite type and faithfully flat. The polynomial $1-\pi T\in R[T]$ defines a closed point of $X$ whose image by $f$ is the generic point of $Y$.