Does inclusion of a ring into a polynomial ring induce a closed map on prime spectra?

Solution 1:

Another good example to think about, more geometric and less arithmetic than that of Qiaochu, is obained by taking $A = k[x]$ ($k$ an algebraically closed field) and considering $k[x] \to k[x,y]$. The induced map on Spectra is the map $\mathbb A^2 \to \mathbb A^1$ from the affine plane to the affine line given by the projection $(x,y) \mapsto x$.

This is a (perhaps the most!) famous example of a non-closed map in algebraic geometry, which motivates the defintion of projective spaces, properness, complete varieties, and so on.

To see that it is non-closed, consider the hyperbola $xy = 1$, which is a closed subset of $\mathbb A^2$. Its image is the subset $x \neq 0$ of $\mathbb A^1$, which is not closed.

Here is the geometric picture: to see if a given point $x_0$ of $\mathbb A^1$ is in the image of this map, we have to take the vertical line $x = x_0$ and intersect with hyperbola $x y = 1$, and ask whether or not this intersection is non-empty. What we see is that the intersection is non-empty if $x_0 \neq 0$, but as we pass to the limit $x_0 = 0$, the intersection suddenly becomes empty.

This illustrates the general phenomenon that in affine varieties, there is no "conservation of intersection number" when we make continuous deformations of the varieties being intersected. Rectifying this problem is one of the main motivations for introducing projective spaces, or, more generally, complete varieties.

Technically, if you look at the basic definitions related to projective varieties, or more generally, complete varieties, you will see that "conservation of intersection number" is not explicitly mentioned, but that the property of properness (which has to do with the closedness of certain maps) is what looms large. This may seem a little mysterious, but in fact it turns out that the failure of certain maps to be closed is more or less equivalent to the failure of conservation of intersection number. The example of the map $\mathbb A^2 \to \mathbb A^1$ above illustrates how the two issues are connected, and this is one reason why it is worth thinking about this example very carefully.

Solution 2:

Nope. Let $A = \mathbb{Z}_{(p)}, n = 1$. Then $\text{Spec } A[x]$ has a closed point given by the morphism $\phi : A[x] \to \mathbb{Q}$ sending $x$ to $\frac{1}{p}$, and this closed point maps to the generic point in $\text{Spec } A$, which is not closed.

$A$ is Noetherian and an integral domain, so neither of those hypotheses help. If $A$ is a field then $\text{Spec } A$ consists of a single point.