# Irreducible component of a geometrically reduced algebraic variety

This is from Qing Liu's Algebraic Geometry and Arithmetic Curves, page 131: 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?

• Having a nilpotent function on a irreducible component means that you have a nilpotent function on your whole variety. – iou May 8 '20 at 15:56

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.