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?
Thanks in advance! 🙂
Best Answer
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.