# Local ring of regular functions

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.

-
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, –  Matt E May 10 '12 at 3:29

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 $\varphi(X)$.
Dear Dylan, I think you mean something like "$f^{-1}(0)$ is closed and contains (a dense subset of, and hence all of) $Y$." Regards, –  Matt E May 10 '12 at 4:38
@MattE Woops, I thought $f$ was the morphism as well. Gunctional notation. Thanks! –  Dylan Moreland May 10 '12 at 4:52