An algebraic result corresponding to etale morphism 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.
 A: 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.
