# Characterizing a module of Kahler differentials

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 $$f,g\in R)$$.

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

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 $$2.$$