Let $f\colon Y \to X$ be a quasicompact morphism of schemes and suppose that $f(Y)$ is stable under specialization. Then $f(Y)$ is closed.
I'm trying to follow the proof given here http://math.ucsd.edu/~kedlaya/math203b/projective-proper.pdf and also in Hartshorne. (They are on the similar lines though )
I failed to understand the following things even after trying a lot! But I guess there are some simple facts I am not being able to connect.
1) How can we assume that closure of $f(Y)$ is $X$? I don't understand how do we "replace $X$ by the reduced induced structure in $\overline{f(Y)}$"?
2) I don't understand the finishing. I think we have to prove that if $\varphi: A \to B$ is an injective homomorphism and $\mathfrak{p}$ is a minimal prime ideal of $A$, then it is the contraction of some prime ideal in $B$. Can anyone please supply a proof of this?
Thanks a lot!
The morphism $f$ factors as $Y \to V(I) \hookrightarrow X$, where $I$ is the kernel of $f^\#$. Observe that $f' : Y \to V(I)$ enjoys the same properties as $f$ and that the conclusion for $f'$ gives the conclusion for $f$.
Well, $A_{\mathfrak{p}} \to B_{\mathfrak{p}}$ is injective, hence $B_{\mathfrak{p}} \neq 0$, so it has some prime ideal, but this corresponds to a prime ideal of $B$ which contracts to $\mathfrak{p}$ (because of minimality).