In Mumford's & Oda's Algebraic Geometry II, on page 81 the authors give a proof for 6.12: Lemma of dévissage. After a careful reading of the proof, I failed to understand an argument:
Theorem 6.12 (“Lemma of dévissage”). Let $\mathbf K$ be the abelian category of coherent $\mathcal{O}_X$-modules on a noetherian scheme $X$, and $\mathbf K’\subset\operatorname{Ob}(\mathbf K)$ an exact subset. We have $\mathbf K’=\operatorname{Ob}(\mathbf K)$, if for any closed irreducible subset $Y\subset X$ with generic point $y$ there exists an coherent $\mathcal{O}_X$-module $\mathcal{G}\in\mathbf K’$ with support $Y$ such that $\mathcal{G}_y$ is a one-dimensional $\mathbb k(y)$-vector space.
About exact subset: Let $\mathbf K$ be an abelian category, and denote by $\operatorname{Ob}(\mathbf K)$ the set of its objects. A subset $\mathbf K’\subset\operatorname{Ob}(\mathbf K)$ is said to be exact if $0\in\mathbf K’$ and if the following is satisfied: In an exact sequence $0\to A’\to A\to A’’\to 0$ in $\mathbf K$, if two among $A$, $A’$ and $A’’$ belong to $\mathbf K’$, then the third also belongs to $\mathbf K’$.
We continue with the proof:
Proof. For simplicity, a closed subset $Y\subset X$ is said to have property $\mathbf P(Y)$ if any $\mathcal{S}\in\operatorname{Ob}(\mathbf K)$ with $\operatorname{Supp}(\mathcal{S})\subset Y$ satisfies $\mathcal{S}\in\mathbf K’$. We need to show that $X$ has property $\mathbf P(X)$. By noetherian induction, it suffices to show that a closed subset $Y\subset X$ has property $\mathbf P(Y)$ if any closed subset $Y’\subsetneqq X$ has property $\mathbf P(Y’)$. Thus we now show $\mathcal{F} ∈ $ Ob(K) satisfies $\mathcal{F}\in\mathbf K’$ if $\operatorname{Supp}(\mathcal{F})\subset Y$. Endow $Y$ with the unique structure of closed reduced subscheme $Y_{\mathrm{red}}$ of $X$ with the ideal sheaf $\mathcal{J}$. Since $\mathcal{J}\supset\operatorname{Ann}(\mathcal{F})$ (???), there exists $n > 0$ such that $\mathcal{J}^n\mathcal{F} = (0)$. Looking at successive quotients in the filtration ...
Q. Why $\mathcal{J}\supset\operatorname{Ann}(\mathcal{F})$? I don't understand the logic behind the argument.
Recall the definitions. $\operatorname{Supp}(\mathcal{F})= \{x \in X\mid\mathcal{F}_x \neq 0 \}$ and $\operatorname{Ann}(\mathcal{F})_x = \{r \in \mathcal{O}_{X,x} \mid \forall f \in \mathcal{F}_x,\, r \cdot f =0 \}$.
We need to show that $\forall x \in X$ we have an inclusion $\mathcal{J}_x\supset \operatorname{Ann}(\mathcal{F})_x$ of $\mathcal{O}_{X,x}$-modules. If $x \not \in Y$ it's trivial because in that case $\mathcal{J}_x= \mathcal{O}_{X,x}$.
The case $ x \in Y$ I don't understand. Assume for example that $\mathcal{F}_x=0$. This means that $\operatorname{Ann}(\mathcal{F})_x= \mathcal{O}_{X,x}$. Otherwise because $ x \in Y$ we obtain $\mathcal{J}_x \neq \mathcal{O}_{X,x}$, a contradiction.
Do I miss any additional assumption on $Y$?