This is 8.3.B in Vakil's Foundations of Algebraic Geometry.
Let $f : X \rightarrow Y$ be a quasicompact morphism of locally Noetherian schemes. Then, the associated points of the scheme-theoretic image of $f$ are a subset of the image of the associated points of the associated points of $X$.
I'm not exactly sure what this means. I think that if $i : Z \rightarrow Y$ is the scheme-theoretic image, and $z \in Z$ is an associated point, then there is some associated point $x \in X$ such that $f(x) = i(z)$.
I managed to prove it in the case where $X$ and $Y$ are both affine. Then, from quasicompactness, I know that the scheme-theoretic image can be computed affine-locally. But the thing I'm having trouble with is how to reduce to $X$ is affine?
Let $U = Spec(R) \subseteq Y$ be affine, and then $V_i = Spec(S_i)$ be an affine cover of $f^{-1}(U)$. then, $f$ induces ring homomorphism $\phi_i : R \rightarrow S_i$. Let $I_i$ be the kernel of $\phi_i$. Then, the scheme-theoretic image is cut out by $\cap_i I_i$. I can't find a way to relate the associated points of $R / I_i$ with the associated points of $R/I$. How can I do this?
Best Answer
I will use your notation. Let $z\in Z$ be an associated point of the image $Z$.
Let $z\in U=Spec R\subset Y$ be an affine open neighbourhood of $z$. Note that $z$ is an associated point of $U$.
Now you cover $f^{-1}(U)$ by affine opens $V_i$'s.
Consider the new map $$g:=\coprod_i f_i: \coprod_i V_i\to U.$$ Now the domain is an affine scheme. Observe that the scheme theoretic image of $g$ is the scheme theoretic image of the original morphism $f:f^{-1}(U)\to U$. Apply the affine-to-affine case to this map, so we are done.