Consider the $\mathbb C$-algebra $R=\mathbb C[x,y,z]/(z(y^2-x^3)-1)$. How to prove that the module of Kahler differentials $\Omega_{R/\mathbb C}$ of $R$ over $\mathbb C$ is a free $R$-module of rank 2?

This $R$-module is generated by $\{d(f):f\in R\}$ modulo the relations $$d(bb')=bd(b')-d(b)b'\\d(ab+a'b')=ad(b)+a'd(b')$$ for all $a,a'\in \mathbb C,b,b'\in R$ where $$d:R\to \Omega_{R/\mathbb C}$$ is a derivation (a group homomorphism such that $d(fg)=fd(g)+d(f)g$ for all $f,g\in R)$.

  • 1
    $\begingroup$ I think you're probably going to be well-served to look at Key Fact 21.2.3 of Vakil. $\endgroup$ – Alex Youcis Apr 25 at 3:40
  • $\begingroup$ You may calculate the module directly using the fundamental sequences or use the fact that $R$ is a regular ring of dim $2$. $\endgroup$ – Youngsu Apr 26 at 15:31

There are two observations which make this calculation go much more smoothly.

  1. First, note that $$R\cong\Bbb{C}\left[x,y,\frac{1}{y^2 - x^3}\right]\cong S^{-1}\Bbb{C}[x,y],$$ where $S = \{1, y^2 - x^3, (y^2 - x^3)^2,\dots\}.$
  2. Secondly, K\"ahler differentials are compatible with localization: namely, if $B$ is an $A$-algebra and $S\subseteq B$ is a multiplicative subset, then $$ S^{-1}\Omega^1_{B/A}\cong\Omega^1_{S^{-1}B/A}. $$ (For a proof, see here.)

Now we can show the claim. Recall that $\Omega^1_{\Bbb{C}[x,y]/\Bbb{C}}\cong\Bbb{C}[x,y]dx\oplus\Bbb{C}[x,y]dy.$ Then if $S = \{1, y^2 - x^3, (y^2 - x^3)^2,\dots\}$ as above, \begin{align*} \Omega^1_{R/\Bbb{C}}&\cong \Omega^1_{S^{-1}\Bbb{C}[x,y]/\Bbb{C}}\\ &\cong S^{-1}\Omega^1_{\Bbb{C}[x,y]/\Bbb{C}}\\ &\cong S^{-1}\left(\Bbb{C}[x,y]dx\oplus\Bbb{C}[x,y]dy\right)\\ &\cong \left(S^{-1}\Bbb{C}[x,y]dx\right)\oplus\left(S^{-1}\Bbb{C}[x,y]dy\right)\\ &\cong Rdx\oplus Rdy\\ &\cong R^2. \end{align*}

Thus, $\Omega^1_{R/\Bbb{C}}$ is a free $R$-module of rank $2.$

  • $\begingroup$ Wow. Very nice. $\endgroup$ – Alex Youcis Apr 28 at 5:24
  • $\begingroup$ I think the current answer contains enough detail. $\endgroup$ – Youngsu Apr 28 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.