# Maximal Ideals and Kahler Differentials

For an algebraic variety $V$, denote its ring of regular functions by ${\cal O}(V)$. The Kahler differentials of $V$ are the quotient of the kernel $M$ of the multiplication map $$m: {\cal O}(V) \otimes {\cal O}(V)\to {\cal O}(V)$$ by the ideal $M^2$.

What can be said about the maximal proper submodules of $M$?

Is there any sense/specific-case in which the submodule $M^2$ is maximal?

I am particularly interested in the homogeneous variety case, specifically the flag variety case. For example, is $M^2$ a maximal right $G$-invariant proper submodule when $V$ a a $G$-homogeneous variety.

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1) This definition of differentials is only correct when $V$ is affine. 2) $M$ is not a ring, therefore it does not make sense to talk about maximal ideals of $M$. Perhaps you mean maximal proper submodules? – Martin Brandenburg Oct 13 '11 at 17:47
All fixed now. Thanks for pointing out the mistakes. – John McCarthy Oct 13 '11 at 17:55
Maybe the map should be called $m$ or something... – Matt Oct 13 '11 at 20:08

If I understand your question correctly, you are asking how likely it is that $M/M^2$ would be a simple module. That is very rare.

Case I varieties

If $Z\subseteq V$ is a closed subvariety of $V$, then for any quasi-coherent $\mathscr O$-module $\Omega$, there exists a restriction morphism, $$\Omega \to \Omega\otimes_V \mathscr O_Z.$$ In other words, there is a corresponding submodule of $\Omega$: $$\mathscr I_Z\cdot \Omega \subsetneq \Omega.$$ As long as $\mathrm{supp} Z\neq V$, the support of the quotient is not the entire $V$, so this proper submodule is not zero.

In particular, if $\Omega$ is a simple module, then there cannot be such $Z$, so $V$ would have to be a point.

Case II homogenous spaces

[after inkspot's comment] If $f:V\to W$ is a non-trivial $G$-invariant morphism, for instance $G\to G/H$ for a proper non-trivial subgroup $H\subset G$, then $f^*\Omega_W\subset\Omega_V$ is a proper non-zero $G$-submodule.

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Are the flag manifolds simple homogeneous spaces? – John McCarthy Oct 13 '11 at 22:09
(1) I'm confused by the distinction between "homogeneous space" and "simple homogeneous space": how can a variety be homogeneous and at the same time have a proper $G$-invariant subspace? (2) If $H$ is a proper non-trivial subgroup of $G$ then put $Y=G/H$. The projection $\pi:G\to Y$ exhibits $\pi^*\Omega^1_Y$ as a proper $G$-invariant sub-bundle of $\Omega^1_G$. – inkspot Oct 14 '11 at 8:19
inkspot, you are completely right. I had something else in mind when I wrote that. Thanks for pointing that out. I will correct it. – Sándor Kovács Oct 14 '11 at 22:39