Let $X$ be an noetherian scheme, $Y$ an irreducible closed subscheme of $X$ with generic point $y$ and $\mathscr G$ a coherent sheaf of $\mathscr O_X$-modules. Consider the following statement:
If $\mathscr G_y\ne 0,$ then $\operatorname{Supp}(\mathscr G)=Y.$
I know this holds for $X=Y$ but I think it isn't correct otherwise because I can take $Y=y$ to be a closed point and $\mathscr G = \mathscr O_X$ and violate the statement.
But unless I am reading it wrong, the proof of the "lemme de dévissage", Theorem 3.1.2 in EGA III, and even more explicitly the statement of Corollary 3.1.3 which follows it, both claim just the above statement. I am confused; could someone please explain how I should understand what's written in EGA (or is it perhaps a mistake and if so, how should one correct it)? Thanks a lot!
Edit: Let me elaborate on what bugs me. Theorem 3.1.2 only contains the condition that for every irreducible closed subset $Y\subset X$ with a generic point $y$ there should exist a sheaf $\mathscr G\in \mathbf K'$ such that $\mathscr G_y$ is a one-dimensional vector space over $k(y).$ But then in the proof when $\mathscr G$ is called upon (second paragraph on p. 116) there is a passage:
Par hypothèse, il y a un $\mathscr O_X$-Module cohérent $\mathscr G$ (nécessairement de support $Y$) tel que $\mathscr G_y$ soint un $k(y)$-espace vectoriel de dimension $1.$
Then in the course of the proof this is used in the proof and the fact that $\mathscr G^m$ has support inside $Y$ is reiterated. Then finally the statement of Corollary 3.1.3 literally contains the following:
... est remplacée par $\mathscr G_y\ne 0$ (ce qui équivaut à $\operatorname{Supp}(\mathscr G) = Y$).
From all of that it's hard not to come to the conclusion that the statement I wrote above is being implied.
Let $X$ be a Noetherian scheme; one can prove that $X$ has finitely many irreducible components and they are in bijection with the generic points of $X$.
Let $Y$ be an irreducible components of $X$, and let $\eta$ be the generic point of $Y$, I recall that: \begin{equation} Y=\overline{\{\eta\}}^X. \end{equation} Let $\mathcal{G}$ be a coherent $\mathcal{O}_X$-module, or more in general $\mathcal{G}$ is locally of finite type; by definition: \begin{equation} \mathrm{Supp}\mathcal{G}=\{x\in X\mid\mathcal{G}_x\neq0\}, \end{equation} considering the zero morphism of $\mathcal{O}_X$-modules \begin{equation} 0\to\mathcal{G} \end{equation} passing to stalks: $0_x\to\mathcal{G}_x$ is an epimorphism of $\mathcal{O}_{X,x}$-modules if and only if $\mathcal{G}_x=0_x$.
Let $x\notin\mathrm{Supp}\mathcal{G}$, by hypothesis there exists an open affine neighbourhood $U$ of $x$ such that $\mathcal{G}_{|U}$ is associated to a finite $\mathcal{O}_X(U)$-module $M$, let $\{s_1,\dots,s_n\}$ a generator system of $M$; by hypothesis: \begin{equation} \forall k\in\{1,\dots,n\},\,\exists V_k\subseteq U\,\text{open}\mid s_{k|V_k}=0\Rightarrow x\in V=\bigcap_{k=1}^nV_k\,\text{is open and}\,\mathcal{G}_{|V}=\widetilde{0}. \end{equation} From all this, for any $x\notin\mathrm{Supp}\mathcal{G}$ there exists an open neighbourhood $V_x$ such that $\mathcal{G}_{|V_x}=\widetilde{0}$; then \begin{equation} V=\bigcup_{x\notin\mathrm{Supp}\mathcal{G}}V_x\,\text{is open and}\,\mathcal{G}_{|V}=0 \end{equation} and therefore \begin{equation} \mathrm{Supp}\mathcal{G}=X\setminus V\,\text{is a closed subset of}\,X; \end{equation} by hypothesis: \begin{equation} \eta\in\mathrm{Supp}\mathcal{G}\cap Y=\mathrm{Supp}_Y\mathcal{G}\Rightarrow Y=\mathrm{Supp}_Y\mathcal{G}. \end{equation} Remark: I had use only the hypothesis that $Y$ is a scheme and $\mathcal{G}$ is locally of finite type.
Update: Grothendieck and Dieudonné affirm that we can reduce the proof to the case that $\mathcal{IF}=0$, where $\mathcal{I}$ is the ideal sheaf associated to $Y$ then: \begin{equation} \mathcal{F}\cong i_{*}\left(i^{*}\mathcal{F}\right); \end{equation} my idea is: by hypothesis, we can assume that there exists a coherent $\mathcal{O}_Y$-module $\mathcal{G}^{\prime}$ such that $\mathcal{G}^{\prime}_{\eta}$ is a $\kappa(\eta)$-vector space of dimension $1$; then we can get \begin{equation} \mathcal{G}=i_{*}\mathcal{G}^{\prime}. \end{equation} Because $Y$ is a closed (reduced) subscheme of $X$, then the inclusion $i:Y\to X$ is (quasi-)separated ([B], corollary 7.4.10) and quasi-compact ([B], lemma 7.3.8 and proposition 7.3.9) and therefore $\mathcal{G}$ is a coherent $\mathcal{O}_X$-module ([B], proposition 6.9.9), and in particular: \begin{equation} \mathrm{Supp}\mathcal{G}=Y\,\text{([B], exercise 6.9.5)}. \end{equation} [B] Bosch S., Algebraic Geometry and Commutative Algebra