# Closed, irreducible subset of an affine variety

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.

• Subvarieties of $X$ are $\{(0,y)\mid y\in K\}$ and $\{(x,0)\mid x\in K\}$. Commented Sep 17, 2018 at 12:54
• @Wuestenfux Thanks. I can see this by inspection, but how do we prove there's no other subvarieties?
– scd
Commented Sep 17, 2018 at 13:24
• Correspondence with ideals? Commented Sep 17, 2018 at 13:25
• @Wuestenfux That's the question. In the book, he used the conclusion about varieties to draw conclusion on the corresponding ideals, not the other way around. That's how he conclude the height of the corresponding maximal ideal to one point has height 1.
– scd
Commented Sep 17, 2018 at 13:30
• The minimal primes are $(x_1)$ and $(x_2)$ -- they consist of zero-divisors, after all. A height-$1$ prime contains one of these and so corresponds to a height-$1$ prime in a polynomial ring (PID). Commented Sep 17, 2018 at 18:28

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.