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Suppose $X$ is an algebraic variety and $\delta : X \to X \times X$ is the diagonal map. I am defining the cotangent sheaf $\Omega^1_X$ as $\delta^{-1}(I/I^2)$ where $I$ is the ideal sheaf of functions in $\mathcal{O}_{X\times X}$ which vanishes on the diagonal. I'm then using the definition of the tangent sheaf as the dual sheaf $$ \Theta_X := \mathcal{H}om_{\mathcal{O}_X}(\Omega^1_X, \mathcal{O}_X). $$

I know that if we have an element $\alpha$ in $\Theta_X$ then precomposing with the map $d(f) = f\otimes 1 - 1 \otimes f \text{ mod } I^2$ gives us a derivation. But how can we go in the opposite direction and interpret a derivation of the structure sheaf as an element of the tangent sheaf? I'm not too worried about nitty gritty details, but an overall idea would be nice. Thanks for any help!

Mathmo1123
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1 Answers1

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Assume $X$ is an affine $k$-scheme, say $X = \operatorname{Spec} A$ where $A$ is a $k$-algebra. Your definition of the cotangent sheaf amounts to this: taking $I = \ker (A \otimes_k A \to A)$, $\Omega = I / I^2$ (regarded as an $A$-module). However, there is another definition: for every $A$-module $M$, there is a natural bijection between $A$-module homomorphisms $\Omega \to M$ and $k$-derivations $A \to M$.

Indeed, as you say, given an $A$-module homomorphism $\phi : \Omega \to M$, we can define a $k$-derivation $\psi : A \to M$ by $\psi (a) = \phi (a \otimes 1 - 1 \otimes a)$; and conversely, given a $k$-derivation $\psi : A \to M$, we can define an $A$-module homomorphism $\phi : \Omega \to M$ by $\phi (a \otimes b) = \psi (a) b$. It is straightforward to verify these are mutually inverse.

In particular, $A$-module homomorphisms $\Omega \to A$ correspond to $k$-derivations $A \to A$.

Zhen Lin
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  • Just out of interest, if $X$ is just an affine scheme not necessarily over $k$, then the tensor product should be taken over $\mathbb{Z}$, right? – Mathmo1123 Dec 18 '15 at 11:12
  • Sure. We could certainly take $k = \mathbb{Z}$, although I'm not sure one often considers tangent/cotangent sheaves in this setting. – Zhen Lin Dec 18 '15 at 11:36
  • Sorry to bug you, but if we're just looking at some affine scheme instead of a variety, when we take the kernel of $A\otimes A \to A$, don't we get slightly more than the ideal sheaf $I$, since if $f(x)g(x)$ vanishes everywhere on Spec$(A)$, this doesn't mean $fg$ is zero in the structure sheaf (in $A$)? – Mathmo1123 Dec 18 '15 at 21:02
  • What I mean is, if we have an element of $A \otimes A$ which vanishes on the diagonal, it need not lie in the kernel of the multiplication map, its image under the multiplication map just has to be nilpotent I think? But then the map $\varphi$ you define might not be a derivation? – Mathmo1123 Dec 18 '15 at 22:42
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    This is the correct definition for affine schemes. But one shouldn't really be thinking about differentials on non-reduced schemes anyway... – Zhen Lin Dec 19 '15 at 00:32