Hypersurfaces have no Embedded Points

In this post, we assume the following properties of associated points of an $A$ module $M$ (for some Noetherian ring $A$).

1. Associated points of $M$ are precisely the generic points of irreducible components of support of some elements in $M$ (viewed as global section on $X = \text{Spec} A$).
2. $M$ has finitely many associated points
3. An element of $A$ annihilates some non-zero element of $M$ iff it vanishes at some associated point of $M$. 

Notice that $m_p = 0$ iff $x m = 0$ for some $x \not \in p$, i.e. $\text{ann}(m) \not \subset p$. In other words, $p \in \text{Supp}(m)$ iff $\text{ann}(m) \subset p$. 

So $\text{Supp}(m)$ consists of exactly the prime ideals of $A$ containing $\text{ann}(m)$. Thus its irreducible components should correspond to the minimal such prime ideals. In other words, the associated primes of $A$ lying inside support of $m$ are exactly minimal prime ideals containing $\text{ann}(m)$. Thus from (1) and (2), we already can deduce that if $f \in A$ annihilates some nonzero $m \in M$ then $f$ lies inside some associated point of $M$ (namely those lying inside $\text{Supp}(m)$).

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Lemma. Let $A = k[x_1, \ldots, x_n]/(f)$. Then $A$ has no embedded points.

Proof. Let $p$ be an associated prime. Then $p$ is a minimal prime ideal containing $\text{ann}(g)$ for some $g \in A^*$. We claim that $p$ is a minimal prime ideal containing $f$. Indeed, if $\text{ann}(g) \neq 0$, then $f \mid hg$ for some $h$ but $f \not \mid g$, so by unique factorization, $f$ and $g$ must have some greatest common divisor $1 \neq f_1 \neq f$. Suppose $f = f_1 f_2$. Then $\text{ann}(g) =  (f_2).$ Thus $p$ is a minimal prime ideal containing $f_2$, and must also be a minimal prime ideal containing $f$.