Morphisms of Affine Sets and Morphisms of Corresponding Coordinate Rings. I stumbled across something that I really couldn't really figure out.
So suppose you have a morphism of affine algebraic sets: $f: X \rightarrow Y$ and the corresponding coordinate ring morphisms: $f': k[Y] \rightarrow k[X]$
Why is it equivalent to saying that $f$ is closed if any only $f'$ is going up?
I've been wondering this for the past couple of days and I just can't figure it out. Is there an explanation to this at all?
 A: The map on coordinate rings will be denoted by $f^*$ instead of $f'$ .
Exercise 5.10 of Atiyah and Macdonald states that $f^*$ is going-up iff $f(V(\mathfrak{p}))=V(f^{*-1}(\mathfrak{p}))$ for all primes $\mathfrak{p}$.
This will be the main ingredient.
Assume $f^*$ to have the going up property.
Let $W\subset X$ be closed. 
Then $W=V(I)$ for some radical ideal $I\subset K[X]$.
It has a finite number of irreducible components $V(I)=\cup_{i=0}^n V(\mathfrak{p}_i)$. This gives
$$f(V(I))=f(\cup_{i=0}^n V(\mathfrak{p}_i))=\cup_{i=0}^n f(V(\mathfrak{p}_i))=\cup_{i=0}^n V(f^{*-1}(\mathfrak{p}_i)).$$
Hence $f(V)$ is closed as a finite union of closed subsets.
Let $\mathfrak{p}$ be any prime in $k[X]$. We always have $f(V(\mathfrak{p})) \subset V(f^{-1}(\mathfrak{p}))$.
Let $f(V(\mathfrak{p}))\subset V(I)$ then $I\subset \cap_{\mathfrak{p}\subset\mathfrak{m}} f^{*-1}(\mathfrak{m})= f^{*-1}(\mathfrak{p})$ so $V(f^{-1}(\mathfrak{p})) \subset V(I)$. 
Assume that $f$ is closed, then $f(V(\mathfrak{p})) = V(f^{-1}(\mathfrak{p}))$ by the previous. The result from A&M implies that $f^*$ has the going-up property.
