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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 $f,g\in R)$.

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    $\begingroup$ 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
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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 $2.$

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  • $\begingroup$ 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

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