One knows that the support $S$ of a coherent sheaf on a noetherian scheme is closed. E.g. on an affine scheme $X=Spec(A)$ and $F$ corresponding to a finitely generated $A$-module $M$, then the closed subset which corresponds to $S$ is just $V(Ann(M))$.

One often says that $S$ is endowed with the structure of a closed subscheme by taking the sheaf of ideals $Ann(F)$ defined as the kernel of $\mathcal O_X \rightarrow Hom_{\mathcal O_X}(F,F)$.

Now my question: this is not the (unique) reduced subscheme structure, isn't it? Can one anyhow describe the reduced structure?


  • 2
    $\begingroup$ Could you explain a bit about how you came to this question? As is, it's not clear what constitutes a description. It sounds like you wouldn't be satisfied with "it's the subscheme structure corresponding to the radical of $Ann(F)$" but without knowing what sort of answer you would be satisfied with, it's hard to know what else to say. $\endgroup$ Aug 25, 2011 at 6:39
  • $\begingroup$ My problem was to know whether the structure induced by $Ann(F)$ is already reduced. If you read the first answer of mathoverflow.net/questions/34621/… it looks as if. That was confusing me a bit. $\endgroup$
    – Descartes
    Aug 25, 2011 at 7:37

2 Answers 2


No it isn't the reduced induced closed subscheme structure in general. For example, let $A={\bf Z}$, $M={\bf Z}/4{\bf Z}$. Then ${\rm Ann}(M)=4{\bf Z}$ and the prime ideal defining $S$ (with its reduced structure) is $2{\bf Z}={\rm rad}({\rm Ann}(M))$. So if $S$ is endowed with the reduced structure, it is isomorphic to ${\rm Spec}({\bf Z}/2{\bf Z})$ and if it is endowed with the structure given by the annihilator then it is isomorphic to ${\rm Spec}({\bf Z}/4{\bf Z})$.


The answer to your second question is pleasantly general and simple.

Given a completely general scheme $X$ (no noetherian, separation, ...hypothesis) and an arbitrary closed subspace $F\subset |X|$ of its underlying topological space, there is a unique closed reduced subscheme $Y\subset X$ whose underlying set is $|Y|=F$. Here is the proof:
i) If $X=Spec A$ is affine, $Y$ is given by the reduced ideal $I=\bigcap_{x\in F} j_x \;$
[as usual, for $x\in SpecA, j_x \subset A$ denotes the ideal corresponding to the point $x$],
ii) If $X$ is not affine, the reduced scheme $Y=V_{sch}(\mathcal I)$ is obtained by the unique ideal sheaf $\mathcal I\subset \mathcal O_X$ restricting on each open affine $U=Spec A$ to the ideal sheaf $\tilde I$ associated to the $I$ above.

Reference EGA 1, Chap.1 , §5, Proposition (5.2.1)

Addendum: the scheme structure on the support of a sheaf.
For reference purpose, let me describe the schematic structure on the support of a sheaf in a fairly general setting.
The situation is that we have a completely arbitrary scheme $X$ (no noetherian assumption) and a quasi-coherent sheaf $\mathcal F$ of $\mathcal O_X$-Modules of finite type on $X$. ($\mathcal F$ needn't be coherent and so this applies to those strange schemes where $\mathcal O_X$ is not coherent!)
Then there exists a smallest closed subscheme $i:Y\hookrightarrow X$ with underlying set $|Y|=supp(\mathcal F)$ and a sheaf of finite type $\mathcal F'$ of $\mathcal O_Y$-Modules with support $|Y|$ such that $i_* \mathcal F'=\mathcal F$.
Of course if $X=SpecA$ then $\mathcal F=\tilde M$ for some finitely generated $A$-module $M$, then we have $Y=V_{sch}(annM)$ and $\mathcal F'=\widetilde {M^\prime}$ , where $M^\prime$ is $M$ seen as an $A/annM$-module.
Although no coherence is requested of $\mathcal F$, some finiteness condition is necessary, else $supp M$ wouldn't even be closed: just look at the $\mathbb Z$-module $\mathbb Q$ whose support is the non-closed generic point of $Spec(\mathbb Z)$

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    $\begingroup$ I guess you don' mean that $F'$ is of finite type (but rather quasi-coherent, not more). $\endgroup$ Sep 16, 2011 at 15:20

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