# Kahler differentials and quotient rings.

I am dealing with some nice rings that are always isomorphic to some fairly nice quotient ring of a polynomial ring. A typical example is:

$$\mathbb{C}[X,XY,XY^2] \cong \frac{\mathbb{C}[U,V,W]}{\langle V^2 - UW \rangle}.$$

I would like a nice way to write the Kahler differentials of such rings. For example when we have the following ring:

$$\mathbb{C}[X^{ \pm 1}] \cong A := \frac{\mathbb{C}[U,V]}{\langle UV \rangle}$$ There is already a nice way of writing the differential - $d(f(X)) = \frac{\partial f}{\partial X} dX$

but also all $\ f(U,V) + \langle UV \rangle \ \in A \$ can be written uniquely as $h(U) + g(V) + \langle UV \rangle$ for polynomials $h$ and $g$.

Then we can write something like: $d( f(U,V) + \langle UV \rangle ) = (\frac{\partial h}{\partial U} +\langle UV \rangle)dU + (\frac{\partial g}{\partial V} + \langle UV \rangle)dV + \langle d(UV)\rangle$

Note this is equivalent to the standard way to write the differential.

Can this be generalized? For example can I do this with the first example I gave?

really this is me trying to just get a nice way to write these maps, as they behave a lot like the standard way of writing the Kahler differential but the notation means i can't write $\frac{\partial f}{\partial X}$ for example.

I think the general answer to your question is given by the second fundamental sequence $$\mathfrak{m}/\mathfrak{m}^2 \rightarrow \Omega_{A/k}\otimes_k B \rightarrow \Omega_{B/k} \rightarrow 0$$ where $B=A/\mathfrak{m}$.

In practice this means that for a ring like $B=\mathbb{C}[u,v,w]/(v^2-uw)$,

$\Omega_B$ is generated by $du,dv,dw$ subject to the relation,

$$2vdv=udw+wdu$$

• Does this give me a nice way to write the differential map $A \rightarrow \Omega^1$ (hopefully in terms of the partial derivatives)? I feel like the quotient will mess things up. – JC574 Aug 12 '14 at 17:37
• Isnt the differential map just $v\mapsto dv$ etc. – Rene Schipperus Aug 12 '14 at 17:41
• Sorry, I guess I mean in the case $C[x,y]$ I can write $d: f \mapsto \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$, and then it's easy to work out closed and exact forms and kernels etc. I'd like something like that here – JC574 Aug 12 '14 at 17:44
• yeah doesnt that formula follow automatically, since $d$ is a derivation. I havent thought about how you define the isomorphism between your two expression, do you mean what happens to $d$ under this iso ? – Rene Schipperus Aug 12 '14 at 17:48
• I think the point is that the differentials $dx, dy$ etc are no longer indedependent. – Rene Schipperus Aug 12 '14 at 17:52

In differential geometry, if you have a differential form on a manifold $X$ and a smooth submanifold $Y \subseteq X$, then you can always "pull back" any differential form on $X$ to a differential form on $Y$ -- all of the algebra of differential forms, scalar fields, exterior derivatives, and such remains the same: all that pulling back does is add new relations between things. (tangent vectors go in the opposite direction, from $Y$ to $X$, so things become complicated if you are also looking at partial derivatives and such)

That's how I think of it here. I understand differentials on $\mathbb{C}[U,V,W]$. The process of passing to the quotient ring by the relation $V^2 - UW = 0$ means that on the module of differentials, you take the quotient by the relation

$$0 = d(V^2 - UW) = 2V dV - U dW - W dU$$

And in general, I just mod out the module of differentials by the submodule generated by the differentials of the new relations I'm putting to my ring.

Disclaimer: I tend to work with relatively nice rings, so this line of thought may have some problems in full generality.

• this is awesome, thanks. My rings are all very nice, so it should be fine. – JC574 Aug 12 '14 at 17:31
• good answer but i gave it to rene. nothing personal, thanks for all the help – JC574 Aug 12 '14 at 17:54