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This is an algebraic result corresponding to etale morphism which I want to prove:

Let $k \to R$, $k$ is a field and $R$ is a local ring which is a finitely generated $k$-algebra, suppose the module of differential $\Omega_{R/k}=0$, show $R$ is an integral domain.

This is a special case of a result in algebraic geometry [Hartshorne Chapt 3. Ex10.3]:

$f$ is flat and $\Omega_{X/Y}=0\quad \Rightarrow \quad f$ is unramified.

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Maybe I'm missing something, but what if $L$ and $K$ are finite separable field extensions of $k$ and you set $R=L\oplus K$? –  Matt Apr 5 '12 at 17:04
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Right, if we set $R=k\times k$ will give a contradiction. –  wxu Apr 5 '12 at 17:05
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Absolutely true Matt and wxu : +1 to both . Indeed a finite algebra $R$ over a field $k$ is étale iff it is separable iff $\Omega_{R/k}=0$.There is no implication in either direction between these equivalent conditions and being an integral domain. –  Georges Elencwajg Apr 5 '12 at 17:51
    
@Matt ,wxu and Elencwajg : Thank you so much for pointing out this mistake, I should require $R$ to be a local ring. –  Li Zhan Apr 6 '12 at 0:35

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Write $R=k[x_{1},\ldots,x_n]/(f_{1},\ldots,f_{c})$, $I=(f_{1},\ldots,f_{c})$, if $\Omega_{R/k}=0$, then $R$ must has dimension zero hence an Artinian ring which has finite many maximal ideals. For every maximal ideal $\mathfrak{m}$ we have a exact sequence $$ \mathfrak{m}/\mathfrak{m}^{2}\to \Omega_{R/k}\otimes_{R} \kappa(\mathfrak{m})\to \Omega_{\kappa(\mathfrak{m})/k}\to 0. $$ By assumption $\Omega_{R/k}=0$, it follows that $\kappa(\mathfrak{m})/k$ is a finite separble algebraic extension. So our exact sequence can become $$ 0\to \mathfrak{m}/\mathfrak{m}^{2}\to \Omega_{R/k}\otimes_{R} \kappa(\mathfrak{m})\to \Omega_{\kappa(\mathfrak{m})/k}\to 0. $$ It follows that $I_{\mathfrak{m}}=\mathfrak{m}_{\mathfrak{m}}$ because $\mathfrak{m}/\mathfrak{m}^2=0 $ and NAK, so $R=\prod R_{\mathfrak{m}}=\prod \kappa(\mathfrak{m})$.

If we assume that $R$ is a local ring, we can also directly deduce that $R$ is an Artinian local ring without using the hypothesis $\Omega_{R/k}=0$, anyway we prove the result.

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Thank you very much! I can understand all your argument except you claim: $\kappa(m)/k$ is a finite separable extension, then $m/{m^2} \to \Omega_{R/k} \otimes_R \kappa(m)$ is injective, could you please tell me why this is true? Or you can just refer to some commonly used reference and I can check it myself. –  Li Zhan Apr 6 '12 at 23:40
    
@Li Zhan, Matsumura, commutative ring theory, theorem 25.2(second fundamental exact sequence) and theorem 25.3. –  wxu Apr 7 '12 at 0:27
    
GTM150,David Eisenbud, commutatative algebra with a view toward algebraic geometry, corollary 16.13. lemma 16.15. Corollary 16.16. etc. –  wxu Apr 7 '12 at 0:39

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