TITLE: If $X$ is affine reduced, show that $f\neq 0 \Rightarrow \overline {D(f)} = \operatorname {Supp} f$
QUESTION [5 upvotes]: If $\operatorname {Spec}A$ is reduced, show that $f\neq 0 \Rightarrow \overline {D(f)} = \operatorname {Supp} f$

Attempt at a solution:
Clearly $D(f) \subset \operatorname{Supp} f$. Since the latter is closed, we can take closure on both sides to obtain one of the containments.
The nontrivial part is to show the reverse containment, which I'd get if I manage to show that the Rightarrow below is actually an iff:
$z\in \overline {D(f)} \iff $ every neighborhood of $z$ intersects $D(f)$ $\Rightarrow fA_z\neq 0 \iff z\in \operatorname{Supp} f$. 
But I am stuck here and I don't know how to use reducedness of A.

REPLY [7 votes]: Assume that some open neighborhood of $z$, say $D(g)$, does not intersect $D(f)$. This means $D(fg)=\emptyset$, i.e. $fg$ is nilpotent. Since $A$ is reduced, it follows $fg=0$. We have $g \notin z$, hence this equation shows $f=0$ in $A_z$.