Reduced Schemes have no Embedded Points

In this Post, we only talk about associated points of an affine scheme $X = \text{Spec} A$.

We only use the following axiom of associated points: associated points are generic points of irreducible components of the support of sections.
Note that $X$ is itself is always a possible support (of the function $1$, for example), so the generic points of its irreducible components are associated points. The other associated points are called embedded points.

Example: Let  $A= k[x,y]/(xy, y^2)$.

 Possible supports of sections:
$\emptyset$ = $\text{Supp}(0). $
$X = \text{Supp} f$, for $f \not \in (y)$.
$\{(x,y)\} = \text{Supp}(f)$, $ f \in (y) \backslash (y^2, xy)$.

Thus the associated points of $X$ are $(y)$ and $(x,y)$. Among these, only $(x,y)$ is an embedded point.

Notice that  $A$  has an embedded point corresponding exactly to the unique non-reduced point

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Reduced Schemes have no Embedded Points

Let $A$ be an integral domain. Then as the only possible support of sections of $A$ is $\text{Spec} A$,  the the generic point is the only associate point of $\text{Spec}A$.
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Lemma. If $A$ is reduced then $X = \text{Spec}A$ has no embedded points.

Proof. First consider the case when $\text{Spec} A$ is irreducible, i.e. $A$ is a domain. Then $(0)$ is the only associated point, but it is also the generic point of $X$, so there is no embedded point.

 Now let $X =\text{Spec}A$ be any reduced scheme. Let $f \in A^*$. Then $\text{Supp} (f) = \overline{D(f)}$.We claim that its irreducible components are also irreducible components of $X$ (thus all associated points are generic pts of irreducible components of $X$ and hence there is no embedded points).

Note that the irreducible components of a closed subspace $Y \subset X$ are always closed and irreducible in $X$, but might not be maximal.

Claim: the irreducible components of $\overline{D(f)}$ are exactly the irreducible components of $X$ that meet $D(f)$.

Clearly all irreducible components of $\overline{D(f)}$ are contained in some irreducible components of $X$ that meet $D(f)$. Thus it suffices to show that if $\mathbb{V}(\mathfrak{p})  \cap D(f) \neq \emptyset$ then $$\overline{\{\mathfrak{p}\}} =\mathbb{V}(\mathfrak{p})  \subset \overline{D(f)}, \text{i.e.} $$
$$\mathfrak{p} \in \overline{D(f)}.$$
Suppose $\mathbb{V}(\mathfrak{p})  \cap D(f) \neq \emptyset$, i.e. for some $\mathfrak{q} \supset \mathfrak{p}, \mathfrak{q} \not \ni f$. Then in particular, $\mathfrak{p} \not \ni f$. (Or we can see that $\mathbb{V}(\mathfrak{p}) \not \subset \mathbb{V}(f)$.)
Thus $\mathfrak{p} \in D(f)$.
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Another way to phrase it is that if $D(f) \cap \mathbb{V}(\mathfrak)(p)$ is not empty then it is dense in $\mathbb{V}(\mathfrak{p})$ so its closure in $\mathbb{V}(\mathfrak)(p)$ must be $\mathbb{V}(\mathfrak)(p)$. Hence $\overline{D(f)} \supset \mathbb{V}(\mathfrak)(p)$
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