2

$\newcommand{\Spec}{\operatorname{Spec}}\newcommand{\P}{\mathbb P}\newcommand{\C}{\mathbb C}\newcommand{\A}{\mathbb A}$ Let $\A^n_\C=\Spec \C[x_1,\dots,x_n]$, and let $f:\A^n_\C\rightarrow \Spec \C$ be the obvious morphism induced by the inclusion $\C\hookrightarrow \C[x_1,\dots,x_n]$. This morphism is clearly separated, closed, and of finite type. I am trying to show that it's not proper, which means I am going to have to find a scheme $X$ so that $\pi_X:\A^n_\C\times X\rightarrow X$ is not closed.

I was thinking of trying $X=\P^n_\C$ but I don't have a good intuition for what the closed sets of $\A^n_\C\times_\C\P^n_\C$ should be. Any affine scheme I try, which I do seem to have a good intuition for what the closed sets should be, does not seem to actually not be closed, so I am little lost on what to do.

Edit: With regards to Mummy the turkey's comment, let $V=\mathbb V(xy-1)\subset \mathbb A^2$, and let $\mathbb A^2\rightarrow \mathbb A^1$ be given by $\mathbb C[x]\hookrightarrow \C[x,y]$, $x\mapsto x$. Then $\mathbb V(xy-1)$ contains the prime ideal $\langle xy-1\rangle$ which gets mapped to the generic point under the projection, so the image of $\mathbb V(xy-1)$ cannot be closed.

To generalize this to $\mathbb A^n$, I suppose we just do the same thing, but it isn't much harder to see that $x_1\cdots x_n\cdot x_{n+1}-1$ is an irreducible polynomial than it is to see that $xy-1$ is?

Chris
  • 5,420
  • Why not try first with $\mathbb{A}^1$ and projecting the hyperbola in $\mathbb A^2 \cong \mathbb{A}^1 \times \mathbb{A}^1$. Can you then generalise that? – Mummy the turkey May 27 '24 at 18:50
  • @Mummytheturkey please see my edit. – Chris May 27 '24 at 18:58
  • Not Mummy the turkey, but as far as why $X_{1} \cdots X_{n} \cdot X_{n+1} - 1$ is irreducible, I think it should be the case that $\mathbb{C}[X_{1}, X_{2}, \ldots, X_{n+1}]/\langle X_{1} \cdots X_{n} \cdot X_{n+1} - 1 \rangle \cong \mathbb{C}[X_{1}, X_{1}^{-1}, X_{2}, X_{2}^{-1}, \ldots, X_{n}, X_{n}^{-1}]$. Can probably use the universal property of localization and universal property of the quotient to write down inverse maps. The idea is that both of these rings should represent the same scheme, namely $\mathbb{G}_{m, \mathbb{C}}^{n}$, because they have the same functor of points. – Alex Wertheim May 27 '24 at 19:15
  • @AlexWertheim this makes sense to me I think – Chris May 27 '24 at 19:17
  • 1
    @AlexWertheim this is indeed true and very easy to prove, thank you. – Chris May 27 '24 at 19:29
  • Alternatively, use Gauss's lemma inductively. @Chris, I suggest you write this up as an answer :) – Mummy the turkey May 27 '24 at 20:00
  • @Mummytheturkey I have written it up thank you for your help :) – Chris May 27 '24 at 21:51

2 Answers2

3

$\newcommand{\A}{\mathbb A}\newcommand{\P}{\mathbb P}\newcommand{\C}{\mathbb C}\newcommand{\Spec}{\operatorname{Spec}}\newcommand{\V}{\mathbb V}$

Clearly the map $\A^n_\C\rightarrow \Spec \C$ is closed, separated, and of finite type. We need to show that this morphism is not universally closed. Consider the scheme morphism $\pi:\A^n_\C\times_\C \A^1_\C\rightarrow \A^1_\C$. This morphism comes from (up to isomorphism) the ring homomorphism $\C[y]\hookrightarrow \C[x_1,\dots,x_{n+1}]$. Consider the closed subset $\V(x_1\cdots x_{n+1}-1)$, we claim that: \begin{align*} \C[x_1,\dots,x_{n+1}]/\langle x_1\cdots x_{n+1}-1\rangle\cong \C[x_1,\dots, x_{n}]_{x_1\cdots x_{n}} \end{align*} which is an integral domain. Indeed, note that there is a map: \begin{align*} \C[x_1,\dots,x_{n+1}]\longrightarrow \C[x_1,\dots, x_{n}]_{x_1\cdots x_{n}} \end{align*} given by $x_i\mapsto x_i$ for $i\leq n$, and $x_{n+1}\mapsto 1/(x_1\dots x_n)$. This map clearly factors through the quotient hence we have well defined map: \begin{align*} \phi:\C[x_1,\dots,x_{n+1}]/\langle x_1\cdots x_{n+1}-1\rangle\rightarrow \C[x_1,\dots, x_n]_{x_1\cdots x_{n}} \end{align*} Now note that there is map: \begin{align*} \C[x_1,\dots, x_{n}]\longrightarrow \C[x_1,\dots,x_{n+1}]/\langle x_1\cdots x_{n+1}-1\rangle \end{align*} given by the composition of the inclusion map with the map with the projection map. We have that $[x_1\cdots x_n]$ is invertible in $\C[x_1,\dots,x_{n+1}]/\langle x_1\cdots x_{n+1}-1\rangle$ so the there is a well defined map: \begin{align*} \psi: \C[x_1,\dots, x_{n}]_{x_1\cdots x_n}\longrightarrow \C[x_1,\dots,x_{n+1}]/\langle x_1\cdots x_{n+1}-1\rangle \end{align*} These are then clearly inverses of one another, so we have that the two rings are isomorphic. As the localization of an integral domain is an integral domain, it follows that $x_1\cdots x_{n+1}-1$ is irreducible.

The induced projection map then takes $\langle x_1\cdots x_{n+1}-1\rangle \in \V(x_1\cdots x_{n+1}-1)$ to the zero ideal, which is the generic point in $\A^1_\C$. It follows that $\pi(\V(x_1\cdots x_{n_1}-1))$ cannot be closed, so the map $\A^n_\C\rightarrow \Spec \C$ is not universally closed.

Chris
  • 5,420
3

This might be a bit overkill (and I think it is interesting to ask how exactly universal closedness fails), but there is a noteworthy general principle[1]:

A morphism is affine and proper if and only if it is finite.

Since morphisms between affine schemes are always affine, it is sufficient to show that your morphism is not finite. But clearly the polynomial ring $\mathbb C[x_1, \dotsc, x_n]$ is not a finite module (i.e. finite-dimensional vectorspace) over $\mathbb C$.

[1] https://mathoverflow.net/questions/125740/why-is-a-proper-affine-morphism-finite

red_trumpet
  • 11,169