I was told that closed points are Zariski dense in the variety by my teacher. But I don't understand what does it mean. Every single point in a variety is closed in Zariski topology because it's the zero locus of a maximal ideal. So how could I understand this statement?
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3Possible duplicate of Closed points are dense in $\operatorname{Spec} A$ – Crostul Dec 17 '15 at 12:31
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@Crostul I'm not sure it's very helpful to just close this question as a duplicate -- there's a fundamental misunderstanding in the question (not all points of a variety are closed). – Najib Idrissi Dec 17 '15 at 12:33
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It's not true that "every single point in a variety is closed in the Zariski topology". Look at the variety $\mathbb{A}^1 = \operatorname{Spec} \Bbbk[x]$. The generic point corresponding to the zero ideal $(0) \subset \Bbbk[x]$ is not closed. However the set of all closed points of $\mathbb{A}^1$ is dense for the Zariski topology (meaning that its closure is all of $\mathbb{A}^1$). See e.g. this question for a reference (an algebraic variety is in particular of finite type).
Najib Idrissi
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So that means the set of all closed points is dense. But how could we prove it? – Intoks Liobein Dec 17 '15 at 12:35
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@IntoksLiobein See the link I provided (I updated my answer to include a better one). – Najib Idrissi Dec 17 '15 at 12:36
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I see, the union of all closed point correspond to the intersection of all maximal ideals. – Intoks Liobein Dec 17 '15 at 12:39
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What I mean is affine variety which is a closed set in zariski.topology, that's actually all what I know. – Intoks Liobein Dec 17 '15 at 12:48
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@IntoksLiobein So you mean a closed set inside $\operatorname{Spec} \Bbbk[x_1, \dots, x_n]$ for some field $\Bbbk$ and some integer $n$? – Najib Idrissi Dec 17 '15 at 12:49