# The set of rational points of a scheme of finite type over a field.

In this book http://ukcatalogue.oup.com/product/9780199202492.do#.UYDnvZNk1bA (Liu's Algebraic Geometry book), we can find the next proposition;

Proposition 3.2.20. Let $X$ be a geometrically reduced algebraic variety over a field $k$. Let $k^s$ be the separable closure of $k$. Then $X(k^s)$ is not empty.

,and the proof of this Proposition starts with assuming $k=k^s$, $X$ is affine and integral by replacing $X$ by an irreducible open affine subset. But I cannot understand how we can make second assumption.(Throughout the book an algebraic variety is defined by a of finite type $k$-scheme)

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It is not true that any finite type k-scheme has k-rational points. Consider: $\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$ as a $\mathbb{R}$-scheme. –  Dedalus May 1 '13 at 10:57
@Dedalus Thanks! –  User0829 May 1 '13 at 11:09

Like any variety, $X$ has a decomposition as the finite union of its irreducible components: $X=\bigcup_{i=1}^r X_i$.
Each $X_i$ is an integral closed subvariety $X_i\subset X$ of $X$, so that a rational point of $X_i$ is also a rational point of $X$ and so it suffices to prove the result for $X_1$ (say).
In other words we may suppose $X$ integral.
Then our integral $X$ has (by definition of an algebraic variety) a covering $X=\bigcup U_j$ by affine open subvarietes $U_j$ and the key point is that those $U_j$'s are automatically integral too.
To conclude, just notice that a rational point of $U_j$ is a rational point of $X$ so that we have indeed reduced (through replacing $X$ by any $U_j$) to the case where $X$ is an affine integral variety.