A doubt in the proof of Prop. 1.10 of Hartshorne's Algebraic Geometry

I have a doubt in the proof of Proposition 1.10 of Hartshorne's book Algebraic Geometry, which states that if $Y$ is a quasi-affine variety, then its dimension is the dimension of its closure.

In this proof the author picks a maximal chain $Z_0 \subset Z_1 \subset \dots \subset Z_n$ of irreducible closed subsets of $Y$, and states that then also the "closure" chain $\overline{Z_0} \subset \overline{Z_1} \subset \dots \subset \overline{Z_n}$ of irreducible closed subsets of $\overline{Y}$ is maximal, referring to the fact that an open non-empty subset of an irreducible set is irreducible and dense. I tried to prove it by contradiction: if the closure chain were not maximal, then there would exist an index $i\in \{0,\dots,n-1\}$ and another irreducible closed subset $W$ of $\overline{Y}$ (which is then closed in all the affine $n$-space) such that $\overline{Z_i} \subset W \subset \overline{Z_{i+1}}$, or $W$ would satisfy $\overline{Z_n} \subset W$. Then my idea was to look at the intersection of these sets with the quasi-affine variety $Y$. But since $Y$ is quasi-affine, $Y=U\cap X$ with $U$ open in the Zariski topology of the affine $n$-space, and $X$ an affine variety, i.e. an irreducible Zariski-closed set. So I have that $Y\cap W=U\cap (X\cap W)$ which is an open subset of the closed set $W\cap X$. But now how can I use the above fact, since I don't know if the set $W\cap X$ is irreducible? Is the intersection of irreducible sets always irreducible?

• Dear Diogenes, dare I ask if you are Indian? – Georges Elencwajg Apr 22 '14 at 11:33
• Dear Georges, no... I am European. – Diogenes Apr 22 '14 at 13:47
• Dear Diogenes: thanks for answering my indiscreet question. I am passionately interested in linguistics and I have noticed that Indians sometimes use "doubt" where Americans (and other English speaking populations) would use "question". But of course the American sense of "doubt" makes perfect sense in your question ... – Georges Elencwajg Apr 22 '14 at 17:57
• @ Georges: That's why you asked that question! I was wondering whether the reason was that my written English is awful (it is understood that I am not saying that Indians speak English worse than Europeans, in general!) :) – Diogenes Apr 22 '14 at 18:20

Use the fact that $Y\cap W$ is open in the irreducible set $W$ to conclude that $Y\cap W$ is an irreducible topological space. It is also closed in $Y$ since $W$ is closed in $X$, so this completes your proof by contradiction.
• I don't see that $Y\cap W$ is open in $W$. I only see that $Y\cap W=U\cap (W\cap X)$ is open in $W\cap X$. – Diogenes Apr 22 '14 at 10:59
• @Diogenes $W\subseteq\bar Y\subseteq X$ since $X$ is necessarily a closed set in the ambient space. This means $Y\cap W=U\cap(W\cap X)=U\cap W$. – user714630 Apr 22 '14 at 11:01
• Ok, thanks, so by taking intersections with $Y$ I obtain a chain of irreducible closed subsets of $Y$ which then must be strict otherwise from the density of the opens I'll obtain equality in the closure chain. Is it right? But now I have another doubt: to construct the closure chain we take closures wrt the whole affine space, while to use density we take closures relative to the subspace topologies of the different irreducible subsets, so perhaps the previous reasoning is not right, because we have to take the closures wrt the same subspace to deduce equality... – Diogenes Apr 22 '14 at 11:19
• The closure of $Y\cap W$ is $W$ respect to both the subspace topology on $W$ and to the topology on affine space since $W$ is closed in affine space itself. – Kevin Carlson Apr 22 '14 at 11:30