$\operatorname{Spec}(k[x])$ has infinite points.

Question

Let $k$ be a field. I have to prove that $\operatorname{Spec}(k[x])$ has infinite points.

If $k$ is infinite it is obvious: in fact there are infinite maximal ideals $(x-\alpha_i)$, with $a_i \in k$. But if $k$ is a finite field, how can I prove my claim? Is there a general argument like the proof of the infinitude of primes of $\mathbb{Z}$?

See http://math.stackexchange.com/questions/585201/ for a generalization. — Martin Brandenburg, Nov 29 '13 at 07:16

score 10 · Accepted Answer · 2013-09-17T14:55:58.113

10

The maximal ideals in $k[x]$ are principal, generated by irreducible polynomials. Then you have to prove that there are infinitely many (non-associate) irreducible polynomials. In order to do this use the Euclid's trick for proving that there are infinitely many primes.

edited Sep 17 '13 at 14:55

answered Sep 17 '13 at 08:56

5

+1. Euclid's trick works for $k[x]$ because we know the units of $k[x]$. (It doesn't apply for a DVR for instance). – Cantlog Sep 17 '13 at 10:48

Martin Brandenburg · Answer 2 · 2013-09-17T10:09:52.023

5

You can reduce the finite case to the infinite case.

Consider the morphism $\mathrm{Spec}(\overline{k}[x]) \to \mathrm{Spec}(k[x])$. The fiber of $(0)$ is $\mathrm{Spec}(\overline{k}(x))$, just one point, and the fiber of $(f)$ with $f \in k[x]$ irreducible is $\mathrm{Spec}(\overline{k}[x]/(f))$, which is a finite set. It follows that if $\mathrm{Spec}(\overline{k}[x])$ is infinite, the same is true for $\mathrm{Spec}(k[x])$, and they have the same cardinality.

Actually this "geometric" proof can be made more "algebraic": There are infinitely many elements in $\overline{k}$. Since every polynomial over $k$ has only finitely many roots in $\overline{k}$, there are also infinitely many elements in $\overline{k}$ which are not conjugate to each other (pairwise). Hence, their minimal polynomials generate pairwise distinct maximal ideals in $k[x]$.

edited Sep 17 '13 at 10:09

answered Sep 17 '13 at 10:04

Martin Brandenburg

163,620

2

+1 In the present case it may be necessary to also prove the known fact that $\overline{k}$ is infinite even though $k$ is finite. This has been done e.g. here. – Jyrki Lahtonen Sep 17 '13 at 10:58
1

Yes, I assumed the trivial fact that algebraically closed fields are infinite. – Martin Brandenburg Sep 17 '13 at 12:44
@MartinBrandenburg But what's is wrong if I use the same method if the field is finite? Why I have to pass to $\bar{k}[x]$? – ArthurStuart Sep 19 '13 at 20:27

$\operatorname{Spec}(k[x])$ has infinite points.

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