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Let $X$ be a variety over a field $k$ (we always assume $k=\bar k$, but I think this doesn't matter here), and let $k_p$ denote the skyscraper sheaf on $X$ w.r.t. the point $p\in X$. I want to find all the morphisms from the structure sheaf $\mathcal{O}_X$ to $k_p$.

I know that the answer is that there are only the evaluation morphisms, i.e., morphisms of the form $\phi\mapsto\phi(p)$ on open subsets that contain $p$, but I don't really see why.

First of all, wouldn't I need to know which kind of morphisms I am looking for? That is, morphisms of sheaves of abelian groups, or $\mathcal{O}_X$-modules, or... But from the very short solution I have here, I think that the $\mathcal{O}_X$-module one is being searched for.

So let $U$ be an open subset which contains $p$ (otherwise $k_p(U)=0$ and there is only the zero map). Then we took $\frac{g}{h}\in\mathcal{O}_X(U)$, and if $f$ is the wanted morphism we must have $f_U(\frac{g}{h})=\frac{g}{h}(p)\cdot f_U(1_U)=\frac{g}{h}(p)\cdot f_X(1_X)$. I really don't quite understand this calculation.

My guess: the $\mathcal{O}_X$-module structure on $k_p$ should be given by $\mathcal{O}_X(U)\to k_p(U)=k$, $\phi\mapsto\phi(p)$, and then we'd have to pull out $\frac{g}{h}$ as above. But why is this the 'canonical' structure (I searched my notes for something regarding this, but there is nothing which mentions this structure)? And why do we have to use that $f_U(1)=f_X(1_X)$ at all?

Thanks in advance!

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What other $\mathscr{O}_X$-module structure would you want to put on $k_p$? –  Zhen Lin Nov 13 '12 at 8:18
    
Hmm, are you familiar with the general nonsense of adjoint functors? If you are, your question follows from the adjointness of the stalk functor and the skyscraper functor. –  only Nov 13 '12 at 11:56

1 Answer 1

up vote 3 down vote accepted

Prelude : a silly remark.
Suppose you have an algebra $B$ over a ring $A$ given by the morphism of rings $u:A\to B$, so that $B$ is an $A$-module with scalar multiplication given by $a*b=u(a)b$ .
Then a morphism of $A$-algebras $v:A\to B$ satisfies $$v(a)=v(a*1_A)=a*v(1_A)=a*1_B=u(a)1_B=u(a) \quad (\bigstar)$$ so that $v=u$
Application to the actual question
The sky-scraper sheaf $k_P$ is an $\mathcal O_X$-module exactly the way you describe it: by taking the evaluation morphisms of rings $u_U:\mathcal O_X(U)\to k_P(U): f\mapsto f(P)$ (with $u_U=0$ if $P\notin U$).
The sheafification of the above silly remark then immediately proves that the only morphism of $\mathcal O_X$-algebras $v:\mathcal O_X\to k_P$ is $v=u$, the evaluation morphism.
The calculation you ask about is then a particular case of the (sheafified) displayed calculation $(\bigstar)$ in the silly remark.

Warning
The result is of course no longer true if you allow morphisms of groups $\mathcal O_X\to k_P$ instead of morphisms of $\mathcal O_X$-modules: for example $v=2 u\neq u$ is another morphism of sheaves of groups $v:\mathcal O_X\to k_P$

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