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Let $X$ and $Y$ be schemes of finite type over a field $k$. Assume that $X$ is geometrically reduced. If two morphisms $f,g$ from $X$ to $Y$ agree as maps on $X(\bar{k}) \rightarrow Y(\bar{k})$ , I want to show that they are the same maps topologically. Here $\bar{k}$ denotes the algebraic closure. I can see they agree on closed points, but why do they agree on other points?

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up vote 3 down vote accepted

If $Y$ is a separated scheme, then the subset $K\subset X$ of points where $f$ and $g$ agree is closed in $X$.
If this set contains the dense subset $X^{cl}\subset X$ of closed points of $X$, then $K=X$.

Note that density (and even very density) of $X^{cl}$ is a general property of schemes locally of finite type over a field:cf. EGA IV, Troisième Partie, §10.

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Thank you for your answer . However, the book I am using has mot defined separated yet, which makes me curious if tere is another way. – Heidar Svan Feb 20 '13 at 15:34

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