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This is from Qing Liu's Algebraic Geometry and Arithmetic Curves, page 131: enter image description here

Why can we assume that $X$ is integral? Can we prove that an irreducible component of a geometrically reduced algebraic variety given with the reduced closed subscheme structure is geometrically reduced?

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  • $\begingroup$ Having a nilpotent function on a irreducible component means that you have a nilpotent function on your whole variety. $\endgroup$ – iou May 8 '20 at 15:56
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Liu's definition of an algebraic variety over a field $k$ is a scheme of finite type over $k$. In particular, such a scheme $X$ is noetherian and has finitely many irreducible components $X_1\cup\cdots\cup X_n$. Then $X_1\setminus (X_2\cup\cdots\cup X_n)$ is an open irreducible subscheme, and so we may pick an affine open irreducible subscheme $U\subset X_1\setminus (X_2\cup\cdots\cup X_n)\subset X$.

As geometrically reduced implies reduced and any open subscheme of a reduced scheme is reduced, $U$ is reduced. As open immersions are preserved under base change, we have that $U_{\overline{k}}$ is an open subscheme of $X_{\overline{k}}$, which implies $U_{\overline{k}}$ is reduced by the same logic as in the previous sentence. So $U$ is an affine, irreducible, geometrically reduced subscheme. In particular, $U$ is an integral affine scheme. Further, any regular closed point of $U$ is a regular closed point of $X$, so if $U$ has a regular closed point, then $X$ must have a regular closed point.

It should be pointed out that it's much easier to work with irreducible opens for this reduction rather than (closed) irreducible components. It's automatic that any open subscheme of a reduced scheme is reduced, but one needs to add more conditions when speaking of closed subschemes. So one might as well make one's life easier by just jumping straight to an irreducible open affine.

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