# Theorem 4.2.21 in Liu: Regular closed point in a geometrically reduced algebraic variety

I dont't understant the beggining of the proof of theorem 4.2.21 in Liu.

Theorem: A geometricaly reduced algebraic variety X contain always a regular closed point.

The proof begin in saying that we can suppose that X is affine and integral and I don't understand why.

I explain my ideas, wich can be totaly out of the right way...

For integral I had the idea to take an irreducible compponent Y of X such that $\operatorname{dim}X=\operatorname{dim}Y$. Then Y is integral because irreducible and reduced (geometricaly reduced imply reduced). If Y has a regular closed point $y$ then $y$ is closed in $X$ but why is it regular in $X$? As $y$ is closed then (2.5.24) $\operatorname{dim}\mathcal{O}_{Y,y}=\operatorname{dim}Y=\operatorname{dim}X$ but we don't have $\operatorname{dim}T_{X,y}=\operatorname{dim}T_{Y,y}$ but only $\operatorname{dim}T_{X,y}\geqslant\operatorname{dim}T_{Y,y}$ wich proove that $\operatorname{dim}\mathcal{O}_{X,y}\leqslant\operatorname{dim}T_{X,y}$ wich is always right. I don't see how obtain $\operatorname{dim}\mathcal{O}_{X,y}=\operatorname{dim}T_{X,y}$. But maybe I'm not in the right way.

For affine I had the idea to take a minimal (finite) affine covering $(U_i)_i$. Let $i_0$ with $\operatorname{dim}U_{i_0}=\operatorname{dim}X$. With minimality we have $F:=X\setminus\cup_{i\neq i_0}U_i\neq\emptyset$ and $F\subseteq U_{i_0}$. As $F$ is closed in $U_{i_0}$ then it is affine. If $F$ has a regular closed point $x$ then $x$ is closed in $X$, but the problem is to show the regularity in $X$. My idea was regular in $F\Rightarrow$ regular in $U_{i_0}\Rightarrow$ regular in $X$. I see wy the second is write but not the first: I don't have information about $\operatorname{dim}F$ nor $\mathcal{O}_{F,x}$. I think I'm not in the right way, other I should be more specific on $F$?

-

If the components of $X$ are $X_0,\ldots, X_n$, with $\dim X_0 = \dim X$, then take $Y = X - \bigcup_{i=1}^n X_i$, which is open in $X$, open dense in $X_0$, and thus has the same dimension as $X$. By shrinking $Y$ a little you can even assume it's affine. The point is that because $Y$ is open in $X$, you get $\dim T_{X,y} = \dim T_{Y,y}$ for all $y\in Y$.
Edit To see why $y$ closed in $Y$ implies $y$ closed in $X$: with $y$ as above, note that if $x\in \overline{\{y\}}$ (closure in $X$), then $x\in X_0$ as $X_0$ is closed and contains $y$. If $U\subseteq X_0$ is an open affine containing $x$, then $y\in U$ as well -- otherwise the closed set $X_0 - U$ contains $y$ but not $x$. As $\overline{\{y\}}\cap U$ is the closure of $y$ in $U$ and contains $x$, it suffices to show that $y$ corresponds to a maximal ideal in $U$. This follows from the fact that by our choice of $y$, $\dim \mathcal{O}_{X,y} = \dim X = \dim X_0 = \dim U$.
Ok, but if I get $y\in Y$ that is closed and regular in $Y$, I don't see why $y$ stay closed in $X$. As $Y$ is open in $X$ there can be point of $X$ (of $X_0$ in fact) that could adhere to $y$? –  Soranzo Nov 18 '13 at 21:29