My question refers to an exercise from D. Mumford's "Red Book of Schemes" (page 144):
Problem Let $R$ be a finitely generated $k$-algebra. Show that $\Omega_{R/k} = (0)$ if and only if $R$ is a direct sum of finite separable extension fields over $k$.
The problem is that I don't fully understand the way to argue given in the hint (I know that there might be several ways to prove it but the main point of the interest is the way sketched in the hint).
The "$\Leftarrow$" is easy:
Let $R \cong L_1\times\dots\times L_d$ such that each extension $L_i/k$ is finite separable. Using primitive element Thm there exist $a_j\in L_j$ such that $L_j=k[a_j]$.
Let $f_j\in k[X]$ with $f_j=\sum_{i=1}^{n_i}r_i^jX^j$ be minimal polynomials of the $a_j$'s. Consider a derivation $d:L_1\times\dots\times L_d\rightarrow M$ to arbitrary $k$-module $M$.
In $R$ we have $f_j(a_j)=0$ and therefore
$$0= d(f_j(a_j))=d(\sum_{i=1}^{n_i}r_i^j a^i_j)=\sum_{i=1}^{n_i}r_i^jd(a^i_j)=\sum_{i=1}^{n_i}(r_i^jja^{i-1}_jd(a_j))=d(a_j)\sum_{i=1}^{n_i}(r_i^jja^{i-1}_j)$$.
Since all $L_j$ are separable we have $f'_j(a_j)\neq 0$ (=coefficients of $d(a_j)$) so we conclude $d(a_j)=0,$ for $i \in 1,...,d$ which forces any derivation to be the the zero map and therefore $\Omega_{R/k}=0$.
The hints for "$\Rightarrow$" I don't understand. Assume $\Omega_{R/k}=0$.
The first step is suggested to take a prime $P$ of $R$. Since $\Omega_{R/k} \to \Omega_{R/P/k}$ is surjective we obtain $ \Omega_{R/P/k}=0$ since $\Omega_{R/k}=0$ by assumption. Futhermore $Frac(R/P)=(R/P)_S$ is exactly the localization at multiplicative set $S:=R/P-\{0\}$ and therefore $\Omega_{Frac{R/P},k}= \Omega_{R/P/k} \otimes_{R/P} (R/P)_S= 0$ so $\Omega_{Frac{R/P},k}=0$.
This implies that $L:= Frac(R/P)$ is finite separable over $k$.
How can I proceed now? Naively I would like to split $R$ into something $R \cong L \oplus M$ with $M$ $k$-module (=vector space) and $\Omega_{M/k}$ and continue by induction. The problem is why I obtain such splitting $R \cong L \oplus M$?
Unclear stay: Why $R$ satisfy the dcc (descent chain condition)? Take into account that $R$ is finitely generated as $k$-algebra and not module! Here I see a problem.
And how can I show that nilpotent elements cannot occure?
We established, that $Q(R/P)$ is finite separable extension of $k$. So $R/P$ is an integral ring extension of $k$, therefore a field, and so every prime $P \subseteq R$ maximal. So $R$ is a noetherian ring of dimension $0$ therefore an Artin-ring, that fullfills a d.c.c.
As $Q(R/P)$ is separable over $k$, we have an isomorphism (see Corollary 16.13 in Eisenbud "Commutative Algebra with a View...")
$$(P/P^2) R_P = \Omega_{R|k} \otimes_R Q(R/P) = 0$$
So $P R_P = P^2 R_P$ and by Nakayama's lemma $P R_P = 0$ for all prime/maximal ideals $P \subseteq R$. This implies that there are no nilpotents.
Unfortunately I am sure, that Mumford must have had a simpler argument in mind, but I could not find it.