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I'm reading page 10, exercise 3 part (c) of:

http://www.mit.edu/~ssam/alggeom-I.pdf

in order to claim that if f is zero on a dense subset of $Y$ then f is zero everywhere don't we need that the codomain, i.e $k$ is a Hausdorff space? or we don't? in this case $k$ is an algebraically closed field and $f: Y \rightarrow k$ is a continuous map where $k$ has the Zariski topology so $k$ is non-Hausdorff.

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    Dear user10, You are correct that an argument via Hausdorffness is not valid, but nevertheless the statement is true. (Just not by that particular argument.) Regards,2012-05-10

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But $\{0\}$ is still closed in the Zariski topology and, as you said, $f$ is continuous, so $f^{-1}(0)$ is a closed subset of $Y$ which contains $\varphi(X)$.

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    @MattE Woops, I thought $f$ was the morphism as well. Gunctional notation. Thanks!2012-05-10