Let $X$ be a noetherian and separated scheme, and let $\mathscr{F}$ be a coherent sheaf on $X.$ Then $Z = \operatorname{Supp}\mathscr{F}$ is closed. We can put a closed subscheme structure on $Z.$ Choose one, and let $i \colon Z \to Z$ be the closed immersion and let $\mathscr{I}$ be the sheaf of ideals defining $Z,$ i.e. we have a short exact sequence $$ 0 \to \mathscr{I} \to \mathscr{O}_X \to i_*\mathscr{O}_Z \to 0. $$ I'm wondering if it is true that $\mathscr{I}\mathscr{F} = 0.$
We know that a sheaf is zero if and only if its stalks are all zero. So it suffices to show that $(\mathscr{I}\mathscr{F})_p = 0.$
My first question is the following. Is it true that $(\mathscr{I}\mathscr{F})_p = \mathscr{I}_p\mathscr{F}_p?$ I'm not sure how to show that direct limits preserve the action of $\mathscr{I}(U)$ on $\mathscr{F}(U).$
But suppose the statement in the above paragraph is true. If $p \notin Z,$ then $\mathscr{F}_p = 0$ by the definition of the support, so $\mathscr{I}_p\mathscr{F}_p = 0.$ So suppose $p \in Z.$ We have an exact sequence $$ 0 \to \mathscr{I}_p \to \mathscr{O}_{X,p} \to (i_*\mathscr{O}_Z)_p \to 0. $$ We know that for $p\in Z,$ $$(i_*\mathscr{O}_Z)_p = \mathscr{O}_{Z,p} = \mathscr{O}_{Y,p}/\mathscr{I}_p.$$ But I'm not sure if we can use this to show that $\mathscr{I}_p$ annihiliates $\mathscr{F}_p.$
Any help would be much appreciated.
The scheme structure you choose matters. Here's an example: take $X=\operatorname{Spec} k[x]$ and $\mathcal{F}=\widetilde{k[x]/(x^n)}$. If you take $\mathcal{I}=\widetilde{(x^m)}$ for $m<n$, then $\mathcal{I}$ does not annihilate $\mathcal{F}$, while if you take $\mathcal{I}=\widetilde{(x^n)}$, then everything works like you want. To have $\mathcal{I}\mathcal{F}=0$, one must take $\mathcal{I}$ contained in the annihilator of $\mathcal{F}$. Indeed when people define the support of a sheaf on a scheme they usually take the subscheme cut out by the annihilator, as Mohan mentions in the comments, which is more data than just the underlying set.