Sufficient conditions for vanishing module of Kähler differentials

It is a standard fact that the module of Kähler differentials $$\Omega_L$$ of a finite separable extension of a field $$k$$ is equal to 0. I also know that for a ring B and a finite extension $$k \rightarrow B$$ a sufficient condition for the module of differentials to vanish is that $$B$$ is a finite product of separable extensions of $$k$$.

Question. Is it possible to lift this result to more general extensions $$A\rightarrow B$$ where $$A$$ and $$B$$ are both rings? Of course with some additional hypothesis on the rings and on the extension.

• The phrase you are looking for is 'unramified' ring map. – Alex Youcis Apr 29 at 17:24
• I know about unramified ring maps, but my problem is having to prove that a certain map is unramified! So I'm asking what are sufficient conditions for the module of differentials to be zero in various situations. – user364213 Apr 29 at 17:48
• What kind of answer are you expecting? – Alex Youcis Apr 29 at 17:49
• Yeah, I'm not really sure how that will manifest in practice. If you can show one of the other equivalent definitions of unramifiedness then you're OK, but that might be no easier (unless, for example, you really only know the functors your rings represent etc.). – Alex Youcis Apr 29 at 18:00
• – Alex Youcis Apr 30 at 4:32

Let me answer the literal questions the OP asked, and then give a solution to the question he was really getting at (elucidated in the chat linked in the above comments).

The literal question

Let's suppose that $$X\to Y$$ is a finite flat map (let's assume that $$Y$$ is Noetherian to avoid headaches). Then, to check that $$X\to Y$$ has vanishing differentials is equivalent to checking that $$X\to Y$$ is etale. To do this without actually computing the differentials is, in general, not easy. There are a few ways one can do it in practice:

• Show that its unramified in the sense that for every $$x\in X$$ and $$y=f(x)\in Y$$ we have that the map $$\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$$ has the property that $$\mathfrak{m}_y\mathcal{O}_{X,x}=\mathfrak{m}_x$$ and $$k(x)/k(y)$$ is finite separable.
• Show that the map $$X\to Y$$ satisfies Grothendieck's infinitesimal lifting criterion (a priori you actually have to show unicity of the lifts to get etaleness, but since our map is finite showing smoothness is fine since smooth plus relative dimension $$0$$ implies etale).
• You can show that the discriminant of the trace form $$(x,y)\mapsto \mathrm{tr}(xy)$$ (defined affine locally on $$Y$$) is a perfect pairing.

Of course, these might all be harder than just computing the differentials given what you know about $$X\to Y$$.

Another useful technique (which I implement below) is that to check that $$X\to Y$$ is etale it suffices to check this over some fpqc cover of $$Y$$. More concretely, since you like commutative algebra, if we have a ring map $$A\to B$$ and we have some faithfully flat map $$A\to R$$ (e.g. like $$A\to\widehat{A}$$ if $$A$$ is local) then $$A\to B$$ is etale if and only if $$R\to B\otimes_A R$$ is. This falls into the general notion of 'fpqc descent'. But, one concrete way of seeing it is that the non-degeneratness of the trace pairing (mentioned in the third bullet above) can be checked after base changing to a faithfully flat extension.

As for the question in the comments (paraphrasing):

"Is it enough that $$Y=\mathrm{Spec}(A)$$, $$A$$ a Dedekind domain, and $$X=\mathrm{Spec}(B)$$ with $$B$$ a product of normal rings?"

The answer is no as $$\mathrm{Spec}(\mathbb{Z}[i])\to\mathrm{Spec}(\mathbb{Z})$$ shows. If you want them to be (even excellent!) DVRs you can consider $$\mathrm{Spec}(\mathbb{Z}[i]_{(1+i)})\to\mathrm{Spec}(\mathbb{Z}_{(2)})$$.

The question the OP was actually interested in

I will write here the actual question the OP was interested. If the OP decides to transfer it from our chat to a question on the main forum I will transfer this answer there.

Exercise: Let $$Y=\mathrm{Spec}(A)$$ with $$A$$ a DVR with uniformizer $$\pi$$. Let $$G$$ be a finite flat group scheme over $$Y$$. Show that $$G$$ is etale if and only if $$G=\mathrm{Spec}(B)$$ with $$\displaystyle B\cong \prod_{i=1}^n B_i$$ with each $$B_i$$ a normal domain.

My proof might be overly complicated. I'm sure there is some simpler proof just messing around with augmentation ideals. I also tried to write the below as commutative algebra-y as possible at the request of the OP--but I don't like commutative algebra, so it didn't end up optimally algebra-y.

Solution: Let us suppose first that $$G$$ is finite etale over $$Y$$. We can decompose $$G$$ into connected components $$\displaystyle G=\bigsqcup_i \mathrm{Spec}(B_i)$$. We claim that each $$B_i$$ is a domain. The point is that since $$B_i$$ is Noetherian we know that it suffices to show that all the local rings of $$B_i$$ at its maximal ideals are domains. Note though that since $$A\to B_i$$ is unramified we have for any maximal $$\mathfrak{q}$$ of $$B_i$$ the map $$A\to B_\mathfrak{q}$$ has the property that $$\pi B_\mathfrak{q}$$ is equal to $$\mathfrak{q}$$. This means that $$B_\mathfrak{q}$$ is a regular local ring (since $$\dim B_\mathfrak{q}=1$$ since it's finite over $$A$$) and thus a domain.

Remark: The above is a half-way proof to the general fact that the connected finite etale covers of an integral normal scheme are integral normal.

Suppose, conversely, that $$\displaystyle G=\bigsqcup_i \mathrm{Spec}(B_i)$$ with $$B_i$$ normal domains. Note that since $$\dim B_i=1$$ we see that normality implies that for each closed point $$\mathfrak{q}$$ of $$B_i$$ we have that $$(B_i)_\mathfrak{q}$$ is a DVR and so is its completion. Note then that we have that

$$H:=G_{\widehat{A}}\cong \bigsqcup_i \mathrm{Spec}(B_i\otimes_A \widehat{A})=\bigsqcup_i \bigsqcup_{\mathfrak{q}\in\mathrm{MaxSpec}(B_i)}\mathrm{Spec}(\widehat{(B_i)_{\mathfrak{q}}})$$

We deduce that $$H$$ has connected components which are normal domains.

But, by the usual connected-etale sequence we have that $$H$$ is etale over $$\widehat{A}$$ iff its connected component of the identity $$H^\circ$$ is trivial (e.g. see section 8.1 of this). Note though that by our above decomposition, all the connected components of $$H$$ are normal and integral. In particular, $$H^\circ=\mathrm{Spec}(B_0)$$ for some normal domain $$B_0$$. This then implies that its generic fiber $$H_\eta^\circ:=\mathrm{Spec}(B_0\otimes_{\widehat{A}}K)$$ (with $$K:=\mathrm{Frac}(\widehat{A})$$) is normal and integral.

But, since $$H^\circ_\eta$$ is finite over $$\mathrm{Spec}(K)$$ and integral we know that $$H^\circ_\eta=\mathrm{Spec}(C)$$ where $$C$$ is a field. Note though that we have an identity section $$\mathrm{Spec}(K)\to H^\circ_\eta$$ since $$H^\circ_\eta$$ is a group over $$\mathrm{Spec}(K)$$. This implies that we have a map $$C\to K$$ of $$K$$-algebras. Since $$C$$ is a field this must be an isomorphism. Thus, $$C=K$$. So we see that $$B_0\otimes_{\widehat{A}} K\cong K$$. Since $$B_0$$ is finite free over $$\widehat{A}$$ this implies that $$B_0$$ is rank $$1$$. The conclusion follows. $$\blacksquare$$