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

Let $A$ be a commutative (unital) ring, and $A[x_1,\ldots,x_n]$ a polynomial ring over it in some finite number of variables. The inclusion $i\colon A \hookrightarrow A[x_1,\ldots,x_n]$ induces (by contraction) a continuous surjection $\mathrm{Spec}(i)\colon \mathrm{Spec}(A[x_1,\ldots,x_n]) \twoheadrightarrow \mathrm{Spec}(A)$ on the prime spectra. Is $\mathrm{Spec}(i)$ a closed map of topological spaces? Does this become the case if $A$ is assumed to be Noetherian and/or an integral domain or a field?

If it's not closed, (under whatever assumptions on $A$), could someone provide a simple counterexample?

I realize this is probably a very stupid question. It seems like the map should be obviously be closed or obviously not be, but I've vacillated as to which. I seem finally to have devised a proof it is closed, but I am suspicious of this quasi-proof, because it seems to make an exercise I've been working on easier than the hint provided would indicate, and also fails to use some of the hypotheses granted for the exercise. Also, if it were true, I would expect to have seen some mention of it on the Internet or in some text, and so far I haven't.

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Incidentally, it is true that the map is open, since it is flat and of finite presentation. –  Akhil Mathew Feb 15 '11 at 13:59

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.

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Thank you very much. This is exactly what I was looking for. –  jdc Feb 15 '11 at 13:45

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.

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My god, what a disaster. Thank you for this. I would vote this answer up, along with Matt E's, but (quite rightly, as it turns out) I have no reputation. –  jdc Feb 15 '11 at 13:46