NOTE: scroll down to read my latest edit first if you're reading this for the first time :)
My aim is to calculate the de Rham cohomology of the variety $U = \text{Spec} \ A$, where: $$A = \frac{\mathbb{C}[U,V,W]}{\langle V^2 - UW\rangle} $$
So far I have defined:
$$\Omega^1_{A/\mathbb{C}} := \frac{\Omega^1_{\mathbb{C}[U,V,W]/ \mathbb{C}}}{\langle d(V^2-UW)\rangle } \ \ , \qquad \Omega^p_{A/\mathbb{C}} := \wedge^p \Omega^1_{A/\mathbb{C}} $$
and:
$$ d^0(f+I) := \frac{ \partial f}{ \partial U} \,dU + \frac{ \partial f}{ \partial V} dV + \frac{ \partial f}{ \partial W} \, dW + \langle d(V^2-UW)\rangle $$ (letting $ \ I = \ \langle V^2 - UV\rangle \ \ $)
Now, assuming what I have above is ok, we have $ \langle d(V^2-UW) \rangle \ = \langle 2V\, dV - W\,dU - U\,dW \rangle $
So ($U$ affine etc) : $$\small{ H^0_{dR}(U) \cong \ker d^0 = \lbrace f+I : \left(\frac{ \partial f}{ \partial U} + I \right) dU + \left(\frac{ \partial f}{ \partial V} + I \right) dV + \left(\frac{ \partial f}{ \partial W} + I \right) dW \in \langle 2V \,dV - W\,dU - U\,dW \rangle \rbrace }$$
Here is where I get stuck. I know we should end up with $H^0_{dR}(U) \cong \mathbb{C} $, but these maps are pretty confusing, especially now as our generators are no longer independent thanks to the new relation we've introduced.
My questions are:
1) is what i've done above correct so far?
2) if it is, then how would one go about showing $H^0_{dR}(U) \cong \mathbb{C} $ continuing what i've done above? (this can be just a hint in the right direction if you want)
EDIT:
I've looked through the suggested MO thread, and commented below on it. It provides a good proof of (2) using a more general method. I would still like to get at it through this calculation if that's possible.
EDIT 2:
Also I tried this earlier; is this be correct for the ring $\mathbb{C}[x,y]/\langle xy \rangle $ ?:
Suppose $\frac{ \partial f}{ \partial x} dx + \frac{ \partial f}{ \partial y} dy \in \langle ydx + xdy \rangle $ (working in $ \Omega^1_{\mathbb{C}[x,y]} $)
Now $ \exists g \in \mathbb{C[x,y]} \ \ \frac{ \partial f}{ \partial x} = yg(x,y) \ , \frac{ \partial f}{ \partial y} = xg(x,y)$
Choose $G_x(x,y) \ , \ G_y(x,y)$ such that $\frac{ \partial G_x}{ \partial x} = g \ , \ \frac{ \partial G_y}{ \partial y} = g$.
Now we see $f = yG_x + H_1(y) $ and $ f = xG_y + H_2(x) $ for some $H_1,H_2$
without loss of generality then $G_x$ is divisible by $x$ and similarly $y | G_y$. Then we see $H_1(y) = f = H_2(x) $ (since $xy=0$), so $H_1,H_2$ (and hence also $f$) constant.
I realize I've argued this fairly loosely and it could do with more rigour.
EDIT...4?:
David below has pointed out that this calculation will not give me what I want. I would still like to know a sketch of the method that should be used to find kernels and images of linear maps like this, because it's this which I find tough. I guess the situation I am thinking about is:
$ \phi : \ V \rightarrow W$, where $V,W$ are infinite dimensional $\mathbb{C}$-vector spaces and in general do not have a basis, (but the case where $V$ has a basis is also important) and where we know all the relations imposed on $V,W$.
Is there a standard method to find kernels and/or images in this case?
(this will get the bounty)