Hypersurfaces have no embedded points (Vakil 5.5.I) Here's a question from Vakil's FOAG. 
If $f\in k[x_1,\ldots,x_n]$ is non-zero, show that $A:=k[x_1,\ldots,x_n]/(f)$ has no embedded points. Hint: suppose $\bar{g}\in A$ is a zero-divisor, and choose a lift $g\in k[x_1,\ldots,x_n]$ of $\bar{g}$. Show that $g$ has a common factor with $f$.
The hint is easily proven: if $\bar{g}\bar{h} = 0$ with $\bar{h}\neq 0$ in $A$, then $gh\in (f)$ in $k[x_1,\ldots,x_n]$ for some lift $h$ of $\bar{h}$. $h\notin (f)$ since $\bar{h}\neq 0$, so $g$ has a common factor with $f$ since $k[x_1,\ldots,x_n]$ is a UFD.
Unfortunately, I do not see how the hint is related to embedded points. Presumably it has something to do with the following property of associated primes which is assumed to be true:
(C) An element $f$ of a Noetherian ring $A$ is a zero-divisor of the finitely generated $A$-module $M$ (i.e., there exists $m \neq 0$ with $fm = 0$) if and only if it vanishes at some associated point of $M$ (i.e., is contained in some associated prime of $M$).
I feel like I'm missing something obvious here. Could someone help me out please? Note that Vakil is adopting a geometric approach here: associated points are defined to be the generic points of irreducible components of the support of some element, and primary decomposition has not been developed, so it would be great if an answer could be given from this geometric perspective.
 A: Let $R:=k[x_1,\ldots,x_n]$ just for ease of notation, let $\pi:R\to R/(f)$ be the projection and let $P\subseteq R/(f)$ an associated prime ideal, i.e. there is some $g\in R$ such that $P=\operatorname{ann}(g)=\{ h\in R/(f) \mid gh=0 \}$. Then, $Q:=\pi^{-1}(P)=\{ h\in R \mid gh\in (f)\}$, so you know that $g$ and $f$ have a (largest) common divisor $d$, say $f=ad$ and $g=bd$. Then, $bdh=gh\in(f)$ is equivalent to $a$ being a factor of $h$, so $Q=(a)\supseteq(f)$. Since $Q$ is a minimal prime ideal over $(f)$, the ideal $P=(\pi(a))$ is a minimal prime ideal of $R/(f)$ - hence a generic point.
A: Let $A = k[x_1,\dots,x_n]/(f)$ and $X = \mathrm{Spec}(A)$. Let $q$ be an associated point. Now let $g$ be the function s.t $\overline{\{q\}}$ is $\mathrm{Supp}(g)$. Then suppose that there is a point $p$ s.t $\overline{\{p\}}$ is an irreducible component of $X$ and $g_p = 0$. Then by $\mathbf{[C]}$ we have that $g$ is a zero divisor. Thus we infer that $\mathrm{gcd}(g,f) = \lambda$ which is non trivial.
Let $x \in X$ be a point.
Suppose that $g_x = 0$. Then there exists an $s \in A \setminus x$ s.t $sg = 0$. But then, by U.F.D stuff, $f/\lambda \mid s$. Thus $f/\lambda \in A\setminus x$.
If $f/\lambda \in A\setminus\{x\}$ then since $(f/\lambda)g = 0$ we have that $g_x = 0$
We conclude that $x \in \mathrm{Supp}(g)$ iff $f/\lambda \in x$. Thus $V(f/\lambda) = V(q)$ and so $\mathrm{Rad}((f/\lambda)) = q$. Now since $(f/\lambda) \supset (f)$ and $k[x_1,\dots,x_n]$ is a U.F.D, we have that $\mathrm{Rad}((f/\lambda)) = \pi(\mathrm{Rad}((f/\lambda))) = \pi((a)) = (\pi(a))$, where $\pi$ is the quotient map from $k[x_1,\dots,x_n]$ to $A$. Thus $q = (a)$.
Now, if a prime is principal then it is minimal and so $q$ belongs to the irreducible component of $X$ given by $\overline{\{p\}}$ we must have that $p = q$. Thus $q$ is not embedded.
Edit: Removed comment about support of functions
