This question is part II of my proof reading of Lemma of devissage from Mumford's & Oda's Algebraic Geometry II, findable on page 81; Theorem 6.12:
Theorem 6.12 (“Lemma of devissage”). Let $K$ be the abelian category of coherent $\mathcal{O}_X$ modules on a noetherian scheme $X$, and K′ $\subset$ Ob(K) an exact subset. We have K′ $=$ Ob(K), if for any closed irreducible subset $Y ⊂ X$ with generic point $y$ there exists an coherent $\mathcal{O}_X$ module $\mathcal{G} ∈ $ K′ with support $Y$ such that $\mathcal{G}_y$ is a one-dimensional $k(y)$-vector space.
Few comments on exact subset. Let K be an abelian category, and denote by Ob(K) the set of its objects. A subset K′ $\subset$ Ob(K) is called exact if $0 ∈$ K′ and if the following is satisfied: in an exact sequence $0 → A′ → A → A′′ → 0$ in K, if two among $A, A′$ and $A′′$ belong to K′, then the third also belongs to K′.
We continue with the proof:
Proof. For simplicity, a closed subset $Y ⊂ X$ is said to have property P($Y$) if any $\mathcal{S} ∈$ Ob(K) with $Supp(\mathcal{S}) ⊂ Y$ satisfies $\mathcal{S} ∈ $ K′. We need to show that X has property P($X$). By noetherian induction, it suffices to show that a closed subset $Y ⊂ X$ has property P($Y$) if any closed subset $Y′ ⫋ Y$ has property P($Y'$). Thus we now show $\mathcal{F} ∈ $ Ob(K) satisfies $\mathcal{F} ∈ $ K′ if $Supp(\mathcal{F}) ⊂ Y$. Endow $Y$ with the unique structure of closed reduced subscheme $Y_{red}$ of $X$ with the ideal sheaf $\mathcal{F}$ . Since $\mathcal{J} = \sqrt{\mathcal{J}} \subset \sqrt{Ann(\mathcal{F})}$, there exists $n > 0$ such that $\mathcal{J}^n\mathcal{F} = (0)$. Looking at successive quotients in the filtration $$ \mathcal{F} ⊃ \mathcal{J} \mathcal{F} ⊃ \mathcal{J}^2 \mathcal{F} ⊃ ... ⊃ \mathcal{J}^{n-1} \mathcal{F} ⊃ \mathcal{J}^n \mathcal{F} = (0) $$ we may assume $n = 1$, that is, $\mathcal{J} \mathcal{F} = (0)$, in view of the exactness of K′. (???) Let $j : Y → X$ be the closed immersion so that $\mathcal{F} = j_*j^*\mathcal{F}$.
Q: Why does the exactness of K′ allow us to reduce to the case $n = 1$, i.e. that $\mathcal{J} \mathcal{F} = (0)$? What is the argument which allows to perform this reduction?
In addition I would like to remark that I asked this question some time ago in MO.
Assume you have proven the case $n=1$. The authors claim that by induction on $n$, one may prove that every $\mathcal{G}$ in $\mathbf{K}$ with support in $Y$ and verifying $\mathcal{J}^n\mathcal{G}=0$ is in $\mathbf{K}'$.
The first step of the induction is given by the $n=1$ case. Then, assume that the step $n-1$ has been proven and take a $\mathcal{G}$ as above. From step $n-1$ you get that $\mathcal{JG}$ is in $\mathbf{K}'$ and since $\mathcal{F}/\mathcal{JF}$ is killed by $\mathcal{J}$, the step $n=1$ tells you that $\mathcal{F}/\mathcal{JF}\in\mathbf{K}'$. Thus, you have an exact sequence $0\rightarrow\mathcal{JF}\rightarrow \mathcal{F}\rightarrow\mathcal{F}/\mathcal{JF}\rightarrow0$ where the kernel and the cokernel are in $\mathbf{K}'$: the exactness of $\mathbf{K}'$ tells you that $\mathcal{F}$ is also in $\mathbf{K}'$.