I'm not very sure about this argument. Let $\mathscr{I},\mathscr{J}$ two ideal sheaves (you can think about ideal sheaves over a projective variety or even the projective space itself) and assume that $\mathscr{J}$ is a subsheaf of $\mathscr{I}$ with $\mathrm{supp}(\mathscr{I})=\mathrm{supp}(\mathscr {J})$.
What happens to the support of $\mathscr{I}/\mathscr{J}$?
I imagined that $\mathrm{supp}(\mathscr{I}/\mathscr J) = \varnothing$ but I'm not sure how to prove it. Can it be made switching to local rings, namely considering two ideals $I\supseteq J$ on a local (noetherian) ring $A$ with the same support as $A$-modules? That seems reasonable to me, being $$\mathrm{supp}(\mathscr{I})=\{ \text {points }p\text { such that } \mathscr{I}_p\neq 0\}$$
If the latter is true, then I'll only have to prove that for ideals as forementioned $\mathrm{supp}(I/J)=\varnothing$ holds (does this follow by some trick about radicals and factor ideals?).
Thank you very much for your help.
Take $A=\mathbb Z, I=\mathbb Z, J=2\mathbb Z$. $\operatorname{supp} I = \operatorname{supp} J = \operatorname{Spec} \mathbb Z$.
$\operatorname{supp} I/J = \operatorname{supp} \mathbb Z/2\mathbb Z = \{(2) \} \neq \emptyset$.
Furthermore the support can have arbitrary large dimension: Take $R$ to be the polynomial ring in $n$ variables, $I$ a maximal ideal and $J$ be a prime ideal of height $1$, contained in $I$. Then $\operatorname{supp} I/J = \mathcal V(J) = \operatorname{Spec} R/J$ is $(n-1)$-dimensional.