# $k$-Module direct sum of Finite Separable Extension Fields iff Kahler Differentials vanish

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?

## 1 Answer

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.

• Concerning your argument that $R$ has no nilpotents: Essentially this follows from the inclusions $\sqrt{0}= \bigcap_{\mathfrak{p} \text{ prime}}\mathfrak{p} \subset \bigcap_{\mathfrak{p} \text{ prime}}\mathfrak{p}R_{\mathfrak{p}}=0$,right? Another point is how can I continue? Up to now we have an exact sequence (*) $$0 \to P \to R \to R/P=L\to 0$$ Taking into account the hint with dcc I guess that the strategy would be to deduce that $R$ splits to $R = L \oplus M$ with $M$ $k$-algebra with "same properties like $R$" such that Jul 14, 2019 at 23:04
• we can split up finite separable extensions successively and this procedure finishes after finite number of steps by dcc. But here I see following problems: What is the candidate for $M$? $P$ is a bad choice since it is simply not a ring with $1$. Futhermore does the sequence split? Or is there another way to obtain a splitting $R = L \oplus M$ and continue inductively? Jul 14, 2019 at 23:04
• It follows directly from the structure theory of artinian rings, that an artinian ring, with no nilpotents is the product of its localizations at its prime ideals. Jul 15, 2019 at 6:00
• The comment above is slightly misleadingly worded: In fact any artinian ring is the product of its localizations at its prime ideals, stacks.math.columbia.edu/tag/00J4, Lemma 10.52.5 Jul 15, 2019 at 10:09
• ...and every local artin ring without nilpotent elements is a field again by $\sqrt{0}= \bigcap_{\mathfrak{p} \text{ prime}}\mathfrak{p}$. So we are done. Thank you! Jul 15, 2019 at 22:17