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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3Do you know the definition of the Zariski topology? If you understand the definition, then you shouldn't have any problems with the question. If you don't, then that's what you should be asking about. – 2011-08-12
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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.