Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a commutative ring, $RMod$ its category of modules and $CRing$ the category of commutative rings.

There's an embedding $RMod \rightarrow CRing/R$ that sends an $R$-module $M$ to the ring $$R \oplus M$$ (the direct sum taken as modules) with multiplication $(r_0,m_0)(r_1,m_1) = (r_0 r_1, r_0 m_1 + r_1 m_0)$. This functor restricts to an equivalence of categories between $RMod$ and $Ab(CRing/R)$, the category of abelian group objects in the slice category.

The projection $R \oplus M \rightarrow R$ makes this ring into a square-zero extension of $R$. My understanding is that in algebraic geometry, one thinks of a square zero extension of a ring as a kind of infinitesimal extension of $Spec (R)$. So the category of $R$-modules can be viewed geometrically as parameterizing a certain class of infinitesimal objects related to $R$.

On the other hand, of course, the category $RMod$ is equivalent to the category of quasicoherent sheaves on $Spec(R)$, which seems, to me at least, totally unrelated to my previous description.

So my question is: are these two views of the same category somehow related? When I think of the sheaf associated to a module $M$, does it somehow contain information about the corresponding infinitesimal extension? What about when I look at cohomology with coefficients in that sheaf?

share|improve this question
    
I think this is implicit in Emerton's answer below, but it might be helpful to notice that, if $R$ is a $k$-algebra, then sections of $R \oplus M \to R$ which are $k$-algebra maps are exactly the $k$-derivations of $A$ into $M$. –  Mike Skirvin Aug 19 '10 at 7:32
    
That should say derivations of $R$ into $M$ in my comment above. –  Mike Skirvin Aug 19 '10 at 7:33
add comment

1 Answer

If $X$ is a scheme and $\mathcal M$ is a quasi-coherent sheaf on $X$ then we can form a sheaf of rings $\mathcal A := \mathcal O_X \oplus \mathcal M$, on which multiplication of sections is given just by the same formula as for $R \oplus M$.

The pair $(X,\mathcal A)$ is then a scheme which is an infinitesimal thickening of $X$, and this is precisely how you pass from a quasi-coherent sheaf to the corresponding thickening; it is just a sheafified version of the construction in your posting.

(Regarding cohomology, in your question you seemed most interested in the case when $X =$ Spec $R$ is affine, in which case quasi-coherent sheaves have vanishing higher cohomology, so I'm not sure there is much to say about this.)

Added in response to comment below: To see how these come up geometrically, consider for example a $k$-scheme $X$ embedded diagonally into $X \times X$. (Here $k$ is a field, and everything is happening over Spec $k$.)

Let $\mathcal I_X$ be the ideal sheaf on $X \times X$ cutting out the diagonal, and consider the square-zero thickening $\mathcal O_{X\times X}/\mathcal I_X^2$ of $X$.

This sits in the short exact sequence $$0 \to \Omega^1_X = \mathcal I_X/\mathcal I_X^2 \to \mathcal O_{X\times X}/\mathcal I_X^2 \to \mathcal O_{X\times X}/I_X = \mathcal O_X \to 0.$$ The projection $p_1:X\times X \to X$ gives a spliting of this short exact sequence, and so we find that $\mathcal O_{X\times X}/\mathcal I_X^2 = \mathcal O_X \oplus \Omega^1_X$.

Recapitulating, we see that in the special case $\mathcal M = \Omega^1_X$, then $(X, \mathcal O_X \oplus \Omega^1_X)$ is equal to the first order infinitesimal neighbourhood of $X$ in $X\times X$.

Suppose for example that $X$ is a smooth curve, so that $\Omega^1_X$ is a line-bundle. Then $(X,\mathcal O_X \oplus \Omega^1_X)$ is locally like the dual numbers (as you observe in your comment) but is globally twisted (unless $X$ is an elliptic curve, i.e. the genus is 1, which is the one case when $\Omega^1_X$ is actually trivial).

This should give you some sense of how these kinds of objects arise geometrically (and why one would consider other examples rather than just the dual numbers).

share|improve this answer
    
Cool, so this construction works more generally. I guess my question stems from the fact that we usually pick out certain sheaves for specific jobs, like say, the sheaf of differentials, or the structure sheaf itself. So, after performing this construction, do these sheaves "loose their identity", or do they pick out special infinitesimal thickenings? When $X$ is affine, I think $\mathcal{O}(X) \oplus \mathcal{O}(X)$ will be the ring of dual numbers. Is there something more to be said geometrically about the result of this operation? Sorry if I'm not phrasing this well. –  Eric Finster Aug 19 '10 at 3:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.