Does a non-empty locally closed subset of a $k$-scheme of finite type always contain a closed point?

Question

Let $k$ be a field. Let $X$ be a scheme of finite type over $k$. We denote by $X_0$ the set of closed points of $X$. Let $Y$ be a locally closed subset of $X$. Is $Y \cap X_0$ non-empty whenever $Y$ is non-empty?

Motivation See my comment to Avi Steiner's answer to this question.

score 2 · Answer 1 · answered Dec 22 '13 at 11:56

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Write $Y=Z \cap U$ with $Z \subseteq X$ closed and $U \subseteq X$ open. Now we may view $Y$ as an open subset of the closed subset $Z$ of $X$. We make $Z$ into a closed subscheme of $X$. Then it is again of finite type over $k$, and we have $Z_0 = X_0 \cap Z$. Thereby we reduce the problem to open subsets. But then it follows from a corollary of Hilbert's Nullstellensatz, stating that the closed points are dense.

answered Dec 22 '13 at 11:56

Martin Brandenburg

181,922

How do you prove that $X_0$ is dense in $X$ when $X$ is not affine? – Makoto Kato Dec 22 '13 at 14:59
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Denseness is a local property. Also remember that $X_0 \cap U = U_0$ for every open $U$ in $X$, since $x$ is closed iff $k(x)$ is finite over $k$. – Martin Brandenburg Dec 22 '13 at 15:02
How do you prove that $x$ is closed iff $k(x)$ is finite over $k$? – Makoto Kato Dec 22 '13 at 15:26
I've just posted a question on this subject. http://math.stackexchange.com/questions/615709/is-the-set-of-closed-points-of-a-k-scheme-of-finite-type-dense – Makoto Kato Dec 22 '13 at 16:52
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If $x$ is closed, then $k(x)$ is finite over $k$ by Zariski's Lemma (http://en.wikipedia.org/wiki/Zariski%27s_lemma). Conversely, assume that $k(x)$ is finite over $k$. Consider an open affine neighborhood $\mathrm{Spec}(A)$, so that $A$ is a finitely generated $k$-algebra, and $x$ corresponds to a prime ideal $\mathfrak{p}$. Then $\mathrm{Quot}(A/\mathfrak{p})$ is finite over $k$, so that also $A/\mathfrak{p}$ is integral over $k$. Since $k$ is a field, it then follows that the integral domain $A/\mathfrak{p}$ is a field, i.e. $\mathfrak{p}$ is maximal. Hence, $x$ is closed. – Martin Brandenburg Dec 22 '13 at 18:07
How do you prove that a closed point of an affine open subset of $k$-scheme $X$ of finite type is a closed point of $X$? – Makoto Kato Dec 23 '13 at 03:14

score -1 · Answer 2 · answered Dec 22 '13 at 08:24

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I believe any one dimensional local ring is a counter example: The generic point is open and not closed.

answered Dec 22 '13 at 08:24

alpacahaircut

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Could you please give an explicit example: finite type over $k$ seems difficult to achieve. – Georges Elencwajg Dec 22 '13 at 08:54
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Sure: $k[t]_{(t)}$. – alpacahaircut Dec 22 '13 at 08:55
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That is not of finite type over $k$. – Georges Elencwajg Dec 22 '13 at 08:57
Ahh yes, you are correct. My geometry is a little rusty. Suppose $k$ is finite though, then $k[t]_{(t)}$ is of finite type. Am I right? – alpacahaircut Dec 22 '13 at 09:22
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Nope. If a local ring is of finite type over $k$, its maximal ideal must be nilpotent. By the way, you better write @hisname in the head of your comment to ping him. Otherwise he might not notice it. – Makoto Kato Dec 22 '13 at 09:51
Nice comment about nilpotency, dear @Makoto. I'd prove it by noticing that the local ring is Jacobson. I'm sure you have a better proof ... – Georges Elencwajg Dec 22 '13 at 10:16
@GeorgesElencwajg [I'm sure you have a better proof] Not at all. It's almost the same thing as Hilbert's Nullstellensatz. – Makoto Kato Dec 22 '13 at 15:07

score -2 · Answer 3 · edited Apr 13 '17 at 12:19

-2

As Martin Brandenburg's answer shows that we may suppose that $Y$ is a non-empty open subset of $X$. Then $Y \cap X_0$ is non-empty because $X_0$ is dense in $X$ by the proposition of my answer to this question.

edited Apr 13 '17 at 12:19

Community

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answered Dec 23 '13 at 03:25

Makoto Kato

44,216

Does a non-empty locally closed subset of a $k$-scheme of finite type always contain a closed point?

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