Closed immersions and stalks I am a bit confused about the following situation:
if you have a closed immersion $i: Y\rightarrow X$ of schemes and a coherent sheaf $F$ on $X$ which has stalks only in $Y$ (i.e. $F_x$=0 for points $x$ not in $Y$), can I then consider $F$ as a module over $Y$ only by taking the inverse image $i^{-1}F$, and not $i^*F$?
Or are both constructions here the same? Or do I have to take the pullback $i^*$ if I want to consider it as a $\mathcal O_Y$-module?
Thanks for your help!
 A: No, you may not replace $i^*F$ by $i^{-1}F$, because the latter sheaf has no natural structure of $\mathcal O_Y$-Module.
A counterexample.
Consider the affine line  $X=\mathbb A^1_{\mathbb C} =Spec(\mathbb C[T])$ and its closed subscheme consisting of just the origin:
$Y=V_{sch}(T)= Spec ( \mathbb C[T]/(T))$ .
Then  for the $\mathbb C[T]$-module  $M=\mathbb C[T]/(T^2)$, the  associated sheaf   $F=\tilde M$ has support $Y$ but cannot be considered a sheaf of $\mathcal O_Y$-Modules, because  $M=\mathbb C[T]/(T^2)$is not  a $\mathbb C[T]/(T)$-module.
The correct restriction is the sheaf $i^*F$ on $Y$ associated to the $\mathbb C[T]/(T)$-module
 $$\mathbb C[T]/(T)) \otimes _{\mathbb C[T]} \mathbb C[T]/(T^2)=\mathbb C[T]/(T)$$
In other words $i^*F=\mathcal O_Y\;$ !
Edit
On a more optimistic note, what you ask is almost true (which explains why my example looks rather contrived):
Any coherent sheaf $F$ on a noetherian scheme $X$ with support contained in a closed subscheme $Y\subset X$ has a filtration by coherent sheaves $F=F_0 \supset F_1\supset  \ldots  \supset F_N=0$ on $X$ whose quotients do satisfy $i^{-1} 
({F_k/F_{k+1}})=i^*({F_k/F_{k+1}})$.
The key to this proposition is the analysis done by Matt in the affine case  of the precise condition for $i^{-1}$ to coincide with $i^*$. 
A: It may be easier to think first in the affine case (which in fact is less special than it may first appear, since any closed immersion is an affine morphism).
So suppose that $A \to B$ is a surjective map of Noetherian rings, with kernel $I$, and that $M$ is a finitely generated $A$-module.  Then the coherent sheaf associated to $M$ is supported on Spec $B$ if and only if some power of $I$ 
is contained in the annihilator of $M$. 
Now performing $i^{-1}$ in this context (i.e. to a sheaf which is supported on $Y$) does nothing: it is just a "change of perspective functor", which says that a sheaf on $X$ ($=$ Spec $A$ in our situation), which is supported on the closed subset $Y$ ($=$ Spec $B$ in our situation), is the same as a sheaf on $Y$.
On the other hand, performing $i^*$ corresponds, on the level of modules, to passing form $M$ to $B\otimes_A M = M/IM$.  
Now there is a natural map $M \to M/IM$, but it will only be an isomorphism if 
$IM = 0$, i.e. if $M$ is annihilated by $I$.
Thus we have two related, but different, conditions:


*

*The sheaf attached to $M$ is supported on $Y$ if some power of $I$ is contained in $Ann(M)$.

*$i^*$ and $i^{-1}$ give the same result when applied to the sheaf $M$ supported on $Y$ if $I \subset Ann(M)$.
This is borne out in Georges example: The ideal $I = T \mathbb C[T]$, and
the module $M$ is annihilated by $I^2$, but not by $I$.
