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## Are Kahler differentials the same on the affine closure on a quasi-affine scheme?

Let $X$ be a quasi-affine scheme; that is, the natural map $$X\rightarrow \overline{X}:=Spec(\Gamma(X,\mathcal{O}_X))$$ is an inclusion. Each scheme has a quasi-coherent sheaf of Kahler differentials $\Omega$, and the above open inclusion induces a $\Gamma(X,\mathcal{O}_X)$-module map of global Kahler differentials

$$\Gamma(\Omega_{\overline{X}})\rightarrow \Gamma(\Omega_{X})$$

Is this map always an isomorphism?

Edit: Changed $\mathcal{O}_X$ to $\Gamma(X,\mathcal{O}_X)$.

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You mean $Spec(\Gamma(\mathcal O_X))$ ? Otherwise the spec of the sheaf is just $X$... – Qing Liu Mar 5 2011 at 0:59
Yes, corrected. – Greg Muller Mar 5 2011 at 3:19
What is the meaning of "inclusion" in this context? (I'm not claiming the usage is not standard, only that I am not familiar with it.) – Charles Staats Mar 5 2011 at 16:35
Its an open inclusion, so it induces an isomorphism of schemes onto an open subscheme. – Greg Muller Mar 6 2011 at 3:05

The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$.

In addition assume that $X$ is noetherian and $S_2$ (for instance normal).

In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.

More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.

Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$: First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$ and it also follows that $\mathrm{depth}_Z\mathscr F\geq 2$ even if $Z$ is not contained in an affine piece of $X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$.

EDIT 1 removed intro paragraph about the starting assumption.

EDIT 2 added "more generally" paragraph.

EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions.

 Yes, Sandor, you are correct in that I explicitly want $\overline{X}=Spec(\mathcal{O}_X)$. And its true that, if we know $\Omega_{\overline{X}}$ is reflexive, then we are done. – Greg Muller Mar 4 2011 at 23:23 On the other hand, it is almost equivalent to being reflexive. – Sándor Kovács Mar 4 2011 at 23:24 To work correctly with the depth, do you have to assume $\overline{X}$ is noetherian ? – Qing Liu Mar 5 2011 at 1:00 @Qing Liu: I suppose I secretly assumed it was noetherian. – Sándor Kovács Mar 5 2011 at 6:48 Sandor, on an S2 scheme, is it true that every reflexive sheaf is S2? I thought you needed S2 + G1 (do you know a reference?) – Karl Schwede Mar 5 2011 at 13:01