I'm a bit stuck on one of the 'unimportant' exercises in section §8.3 of Vakil's notes on scheme theoretic images. The question states
If $\pi: X \to Y$ is a locally closed embedding into a locally Noetherian scheme (so X is also locally Noetherian), then the associated points of the scheme-theoretic closure are (naturally in bijection with) the associated points of X. (Hint: Exercise 8.3.B.) Informally, we get no nonreduced structure on the scheme-theoretic closure not “forced by” that on X.
Now the Exercise 8.3.B referenced here is simply proving that if $\pi : X \to Y$ is quasicompact then we can show that associated points of the image subscheme are the images of associated points in $X$ — assuming we can show that a locally closed embedding from locally Noetherian to locally Noetherian is quasicompact, then this would give us one direction of the bijective correspondence. However, I'm not quite sure why we must have that $\pi$ is quasicompact in general, nor do I understand why we should have
$x$ is an associated point of $X$ $\Longrightarrow$ $\pi(x)$ is an associated point of $(\operatorname{Im} \pi)_{sch} \hookrightarrow Y$
(i.e. the reverse direction holds under the additional assumption of locally closed embedding).
Best Answer
Claim: Any locally closed embedding $f:X\to Y$ with target a locally noetherian scheme is quasi-compact.
Proof. First, note that it suffices to check that a morphism $f:X\to Y$ is quasi-compact on open affines - that is, $f$ is quasi-compact iff $f^{-1}(U)$ is quasi-compact for every affine open $U\subset Y$. This is because any quasi-compact open subset of $Y$ is covered by finitely many affine opens and a finite union of quasi-compact spaces is again quasi compact: cover an arbitrary quasi-compact open in $Y$ by finitely many affine opens, note that their preimages are quasi-compact, and then note that the union of these preimages is the preimage of the unions.
If $Y$ is locally noetherian, then every affine open $\operatorname{Spec} R\subset Y$ is the spectrum of a noetherian ring. Hence every subset of $\operatorname{Spec} R$ is quasi-compact, and so $f^{-1}(\operatorname{Spec} R)$ is quasi-compact, proving the claim. $\blacksquare$
To proceed with the rest of the question, note that after showing $\pi$ is quasi-compact, we know that the underlying topological space of the scheme-theoretic image is precisely the closure of $\pi(X)$ (corollary 8.3.5). Thus $X$ is an open subscheme of $\operatorname{im}(\pi)$ (exercises 8.3.C/8.1.M) and by exercise 5.5.F, the associated points can be calculated locally. This is enough to show a bijection, since open immersions preserve stalks.