A point $P$ in a variety $X$ (affine,quasi-affine,projetive,quasi-projective)is closed if the closure $\overline{\{P\}}=\{P\}$.Will someone be kind enough to give me some hints on this?Thank you very much!
Are points in a variety always closed?
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4@user14242: From the string of questions you have been asking, I'm beginning to get the impression you're trying to run before you can stand, to stretch a metaphor. Anyway. Your question is not clear. What do you mean by a variety—a locally ringed space in the sense of Hartshorne Chapter I, or an integral scheme of finite type over $\operatorname{Spec} k$, for some algebraically closed $k$? – 2011-08-12
1 Answers
Lets assume that you are working over an algebraic closed set. I'll do the affine case for you: We have that a set $A\subset k^n$ its algebraic if its of the form of $V(I)$ for some ideal $I\subset k[x_1,\dots,x_n]$ so we have to show that every point of $k^n$ is of that form. We have by the Nullstellensatz that there is a one-to-one inclusion-reversing correspondence between algebraic sets in $\mathbb{A}^n$ and radical ideals. Furthermore it can be shown that an algebraic set is irreducible if and only if its ideal is a prime ideal. So a maximal ideal $\mathfrak{m}\subset k[x_1,\dots,x_n]$ corresponds to a minimal irreduble closed subset, wich must be a point. We have then the following
Theorem. Every maximal ideal of $k[x_1,\dots,x_n]$ is of the form $\mathfrak{m} = (x_1 - a_1,\dots,x_n - a_n)$ for some $a_1,\dots,a_n\in k$.
Its worth mentioning that this results do not hold if you are not working over an algebraic closed set.