The answer to your second question is pleasantly general and simple.

Given a completely general scheme $X$ (no noetherian, separation, ...hypothesis) and an arbitrary closed subspace $F\subset |X|$ of its underlying topological space, there is a *unique* closed reduced subscheme $Y\subset X$ whose underlying set is $|Y|=F$. Here is the proof:

i) If $X=Spec A$ is affine, $Y$ is given by the reduced ideal $I=\bigcap_{x\in F} j_x \;$

[as usual, for $x\in SpecA, j_x \subset A$ denotes the ideal corresponding to the point $x$],

ii) If $X$ is not affine, the reduced scheme $Y=V_{sch}(\mathcal I)$ is obtained by the unique ideal sheaf $\mathcal I\subset \mathcal O_X$ restricting on each open affine $U=Spec A$ to the ideal sheaf $\tilde I$ associated to the $I$ above.

**Reference** EGA 1, Chap.1 , §5, *Proposition* (5.2.1)

**Addendum: the scheme structure on the support of a sheaf.**

For reference purpose, let me describe the schematic structure on the support of a sheaf in a fairly general setting.

The situation is that we have a completely arbitrary scheme $X$ (no noetherian assumption) and a quasi-coherent sheaf $\mathcal F$ of $\mathcal O_X$-Modules of finite type on $X$. ($\mathcal F$ needn't be coherent and so this applies to those strange schemes where $\mathcal O_X$ is not coherent!)

Then there exists a smallest closed subscheme $i:Y\hookrightarrow X$ with underlying set $|Y|=supp(\mathcal F)$ and a sheaf of finite type $\mathcal F'$ of $\mathcal O_Y$-Modules with support $|Y|$ such that $i_* \mathcal F'=\mathcal F$.

Of course if $X=SpecA$ then $\mathcal F=\tilde M$ for some finitely generated $A$-module $M$, then we have $Y=V_{sch}(annM)$ and $\mathcal F'=\widetilde {M^\prime}$ , where $M^\prime$ is $M$ seen as an $A/annM$-module.

Although no coherence is requested of $\mathcal F$, some finiteness condition is necessary, else $supp M$ wouldn't even be closed: just look at the $\mathbb Z$-module $\mathbb Q$ whose support is the non-closed generic point of $Spec(\mathbb Z)$

wouldbe satisfied with, it's hard to know what else to say. $\endgroup$