Let $R=\mathbb C[x,y]/(y(x-a)(x-b)-1)$ where $a,b$ are distinct complex numbers. Show that the cohomology of the de Rham complex $$0\to R\to \Omega_{R/\mathbb C}\to 0$$ is $\mathbb C$ in degree zero and $\mathbb C^2$ in degree one.

I have no experience in computing de Rham cohomology whatsoever (maybe except the case $R=\mathbb C[x]$, not with quotients), but first I need to at least identify $\Omega _{R/\mathbb C}$ and to write down to write the map $R\to \Omega_{R/\mathbb C}$. How do I do that?

  • $\begingroup$ Can you check your source if your equation is $y(x-a)(x-b)-1$ instead? Otherwise, I do not think you have $C^2$ in degree one if $b=0$. $\endgroup$ – Youngsu Apr 28 at 5:36
  • $\begingroup$ @Youngsu Thanks, it was a typo. $\endgroup$ – user437309 Apr 28 at 16:13
  • 1
    $\begingroup$ $R$ is the ring of rational functions $f(x)$ of the form $\tfrac{g(x)}{(x-a)^m (x-b)^n}$, for $g$ a polynomial. The differential takes $f(x)$ to $f'(x) dx$. What have you learned in calculus about derivatives and integrals of rational functions? $\endgroup$ – David E Speyer May 1 at 17:17

You have $R=\Bbb C[x,y]/(F(x,y))$ for your given polynomial $F(x,y)$. The differentials $\Omega_{R/\Bbb C}$ form an $R$-module, generated by symbols $dx$ and $dy$ subject to the single relation $F_x(x,y)\,dx+F_y(x,y)\,dy=0$ where $F_x$ and $F_y$ are partial derivatives. The map $d:R\to\Omega_{R/\Bbb C}$ takes $g(x,y)$ to $g_x(x,y)\,dx+g_y(x,y)\,dy$.

The kernel of $d$ certainly contains the elements of $\Bbb C$. What you now need to do next is to show that if $g$ is a polynomial with $(g_x,g_y)=(hF_x,hF_y)$ for some polynomial $h$, then $g\in\Bbb C$. Over to you!

  • $\begingroup$ So do I not need to know explicitly what $\Omega_{R/\mathbb C}$ is isomophic to? And just to make sure I understood what you wrote in the second paragraph: is this argument supposed to show that $\ker d\subset \mathbb C$, and since $\ker d$ is the cohomology in degree zero, that would finish the first part? So we take $g$ from the kernel, meaning that $g_xdx+g_ydy$ is an $R$-multiple of $F_xdx+F_ydy$, that's how you got the equality of pairs? And also should we write $\overline g$, $\overline h$, $\overline F_x$ etc. everywhere since $R$ is a quotient ring? $\endgroup$ – user437309 Apr 27 at 3:47
  • $\begingroup$ I was trying to avoid quotients and express everything in terms of polynomials. Indeed the $0$-cohomology is the kernel of $d$ and the $1$-cohomology is the cokernel. In the last line I should have said that $g\in C+(F(x,y)$. @user437309 $\endgroup$ – Lord Shark the Unknown Apr 27 at 3:51
  • $\begingroup$ But if $g\in \mathbb C+(F(x,y))$, then how does this show (if it does) that $\ker d\subseteq \mathbb C$? Or is the point of showing $g\in \mathbb C+(F(x,y))$ different? $\endgroup$ – user437309 Apr 27 at 3:55
  • $\begingroup$ $g\in\Bbb C+(F(x,y))$ is the same as saying that $\overline g\in\Bbb C$ within the quotient ring $R$. $\endgroup$ – Lord Shark the Unknown Apr 27 at 3:56
  • $\begingroup$ I see, thank you. That's why the abuse of notation (talking about residues as the actual elements of the ring) is confusing for me. I will try to prove that inclusion, but meanwhile do you have any hints on how to find the cokernel? I guess that's where we would need to identify $\Omega_{R/\mathbb C}$... $\endgroup$ – user437309 Apr 27 at 4:01

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