Closed, irreducible subset of an affine variety

Question

This question rised when I'm reading Kemper's 'A course in commutative algebra' chapter 6 and 7. let $K$ be a algebraically closed field and $X=\{(x_1,x_2)\in K^2|x_1x_2=0\}$ is an affine variety. The book claimed the height of the maximal ideal in the coordinate ring $K[X]$ is 1, since the maximal ideal corresponding to one-point sub-variety $\{x_0\}$ of $X$, and the chain starting with $\{x_0\}$ of irreducible subsets of X has length 1. I'm confused about the last part. Could anyone offer some help about why the chain starting with $\{x_0\}$ of irreducible subsets of X has length 1? Thanks in advance.

Answer 1

Since $K$ is an integral domain, $x_1x_2=0$ implies $x_1=0$ or $x_2=0$. So $X=\{(x_1,0), (0,x_2)\mid x_1,x_2\in K\}$. Then e.g.\ $Y_1=\{(x_1,0)\mid x_1\in K\}$ is a subvariety and so we have the chain $Y_0=\{(0,0)\}\subset Y_1 \subset X$ of length 1.