Lemma: $\text{Supp} s$ is closed
Proof. $s_p = 0$ iff $s$ is $0$ (as a section) on a neighborhood of $p$. Thus the complement of $\text{Supp} s$ in $X$ is open.
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In general, for an $A$-module $M$ and $p \in \text{Spec} A$, we see that $p \in \text{Supp}(m)$ iff $m \neq 0 $ in $M_p$ $\iff$ $\text{ann}(m) \subset p$.
Claim: If $X = \text{Spec} A$ for a integral domain $A$ and $f \in A^*$ then $\text{Supp} f = X.$
Proof. For every $f \in A$, if $f \neq 0$ then $\text{ann}(f) = 0$ so $\text{Supp}(f) = \text{Spec} A$.
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Notice that if $s_p = 0$ iff $s$ vanishes on a neighborhood of $p$, so $s(p) = 0$. In other words, if $s(p) \neq 0$ then $p \in \text{Supp}(s)$, i.e.
$$\overline{D(s)} \subset \text{Supp}(f).$$
However, the other direction is not true in general. For $X = \text{Spec} k[x,y]/(y^2, xy)$, the global section $s = y$ is $0$ at $p=(x,y)$ but $s_p$ is not zero. (Because there are extra differential operators at $p$).
Claim: If $X = \text{Spec} A$ for a reduced ring $A$ then $\text{Supp} f = \overline{D(f)}.$
Proof. Clearly if $p \in D(f)$ then $f(p) \neq 0$ so $f_p \neq 0$ and thus $\text{Supp} f \supset D(f)$, and thus $\text{Supp} f \supset \overline{D(f)}.$
(Another way to state the same thing: Suppose $p \in \overline{D(f)}$. Then every neighborhood of $p$ contains points of $D(f)$ so $f$ cannot vanish on any neighborhood of $p$.)
Conversely, suppose $p \in \text{Supp} f$. Then $f_p \neq 0$. Thus for all (sufficiently small) neighborhood $W \ni p$, we have $f|_W \neq 0$.
Warning: The common mistake here is to deduce from this that $W \cap D(f) \neq \emptyset$, by arguing that as $f|_W$ is not equal to the zero function, $f$ cannot vanish on all of $W$, i.e, $W \not \supset \mathbb{V}(f)$. However in general schemes, functions are NOT determined by their values at points. That is why we need to reduced hypothesis. (The difference between two functions that equal pointwise lies in the nilradical, so if the nilradical is $0$, functions are determined by their values at points.)
As $W$ is itself a reduced scheme, and $f|_W$ is uniquely determined by its values at points of $W$, we do have that $W \not \subset \mathbb{V}(f)$ and thus $W \cap D(f) \neq \emptyset$.
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Example. Let $X = \text{Spec} k[x]/(x^2)$ be the fat point. Let $p = (x)$ be the only point of $X$. Then
$Supp f = \emptyset$ if $f \in (x^2)$ (i.e. if $f \sim 0$ as elements of $k[x]/(x^2)$).
$Supp f = X$, otherwise.
Indeed, if $f$ does not vanish at $p$ to begin with, then $f_p \neq 0$. On the other hand, if $f(p) = 0$ and $f \neq 0$ then $f $ is a constant multiple of $x$. Now $x_p \neq 0$. This can be checked algebraically, but geometrically it is because a neighborhood of $p$ contains the operator: $f \mapsto $ the linear term of $f$. Notice that $x$ does not vanish under this operator.
See What does the Stalk Perceive for more explanation.
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Let $X = \text{Spec} A$. Let $f \in A$. Then $p$ is in $\text{Supp} f$ iff $f$ is not in the kernel of
$A \to A_p$.
Notice $\text{Supp} (fg) = \text{Supp}(f) \cap \text{Supp} (g)$.
Example. Let $A = \frac{k[x,y]}{(y^2, xy)}$ and $X = \text{Spec} A$. Let $f \in A$. We claim that the support of $f$ is either $\emptyset, (x,y)$ or $X$.
Proof. Note the elements of $\text{Spec} A$ are prime ideals of $k[x,y]$ containing $(y)$. In other words, $(y)$ is the generic point of $X$.
Case 1. $f \not \in (y)$. Then $D(s)\ni (y)$ so $\text{Supp}(s) \supset \overline{D(s)} = X$.
Case 2. $f \in (y)$. Then $f$ vanishes at all points of $X$ since all prime ideals of $A$ contains $(y)$. (Another way to see this is to notice that $D(f)$ is an open subset of $X$ not containing the generic point so must be empty).
For all $p \neq (x,y)$, $p$ is a reduced point of $X$. As $f$ vanishes on a neighborhood of $p$ and $p$ is reduced, $f$ must actually be equal to the zero section on a neighborhood of $p$, i.e. $f_p = 0$. So $p \not \in \text{Supp}(f)$.
Thus we only need to check if $f_{(x, y)} = 0$
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Some other random reasonings that achieve the same thing:
Suppose $p \neq (x, y)$, i.e. $p \in D(x)$.
We have $s_p$ iff $s$ is $0$ on some neighborhood of $p$. Without loss of generality, assume that neighborhood is $D(xf) \subset D(x)$ for some $f$.
We have a morphism $A \to O_X(D(x)) = A_x = k[x] \to O(D(xf))$ (given by $y \mapsto 0$). As $k[x]$ is an integral domain, the localization map $O_X(D(x)) \to O_X(D(xf))$ is injective, so $s \in A$ restricts to $0$ in $D(xf)$ iff it restricts to $0$ on $D(x)$, i.e. iff it is in the kernel of
$$A \mapsto A_x$$
i.e.
$$k[x,y]/(y^2, xy) \to k[x]$$ given by $y \mapsto 0$ and $x \mapsto 0$.
The kernel of this map is $(y)$.
So for all $s \in (y)$, $s_p = 0$ at all $p \in D(x)$. We cannot conclude yet for all such $s$, that $\text{Supp} (s) = X - D(x) = (x, y)$. We need to check if $s_{(x,y)} = 0$.
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Let $p =(x,y)$. At the stalk of $X$ at $p$ there are extra differential vectors generated by the $x, y, x^2$. Thus $s_p = 0$ iff the $y$ term, $x$ term and $x^2$ term of $s$ is zero. A way to rephrase it is to think of the ``$y$-operator'', as a point $\text{Spec}k[y]/(y^2) \to X$. As its image is the point $(x, y)$, we can pullback maps on neighborhoods of $(x,y)$ to maps on neighborhood of $(y)$. In other words, we have a local homomorphism of stalks $A_p \to (k[y]/(y^2))_{(y)}$. Thus $s $ is $0$ in $(O_X)_p$ iff its pullback is $0$ in $k[y]/(y^2))_{(y)}$. Among the element $s \in (y)$, only those in $y^2$ pullsback to $0$. So if $s$ is a multiple of $y$, then $s_{(x,y)} \neq 0$, and thus $\text{Supp} (s) = (x,y)$.
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Lemma. Let $M$ be an $A$-module. Then the natural map
$$M \to \prod_{\text{associated }$p$} M_p$$
is an injection.
$$M \to \prod_{\text{associated }$p$} M_p$$
is an injection.
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