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Fix $(R,m)$ a complete DVR of mixed characteristic $(0,p)$ with perfect residue field, and consider finite flat commutative group schemes $G = Spec(A)$ over $R$. One can associate a differential invariant to $G$, an integer $d \geq 0$, in two ways:

  1. the "absolute different", such that the dual $A^*$ of $A$ under the $R$-trace pairing is equal to $m^{-d}A$, or

  2. the "cotangent length", equal to the $R$-length of the cotangent space $e^*\Omega_{G/R}$ (where $e$ is the unit section).

My question is: why is the cotangent length $d$ additive in short exact sequences of $G$?

A direct argument for the cotangent length would be great, though I'm open to an argument that compares the two definitions and proves additivity for the absolute different. For what it's worth, the comparison is clear in some cases by the "differential characterization of the different", and the additivity for the absolute different is clear in some cases by the "transitivity of the different", but I don't know where either of these classical facts is stated in the full generality needed here.

(Context: In Raynaud's famous "... type $(p,\ldots,p)$" paper, the theorem in Section 4.1 computes the action of inertia on the determinant of the generic fiber of $G$ in terms of the absolute different. The argument given proceeds in two steps. First, in the case where $G$ is simple, the result is explicitly verified using the classification worked out earlier in the paper. I have no issues with this argument, and "differential characterization of the different" even applies to compare the two definitions of $d$ in this case. Second, there is a reduction to the simple case by dévissage. But the required additivity of $d$ is never explicitly addressed, and moreover my needs would prefer the cotangent definition. Hence the question.)

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By Proposition 5.1(i) in Mazur and Roberts, Local Euler characteristics, Invent. Math. 9 (1970), 201-234, $G$ fits in an exact sequence $0 \to G \to A \to B \to 0$, where $A$ and $B$ are smooth commutative affine group schemes over $R$ of the same relative dimension, say $n$. Then the cotangent space $e^* \Omega_{G/R}$ is isomorphic to the cokernel of the induced homomorphism $\psi \colon e^* \Omega_{B/R} \to e^* \Omega_{A/R}$ of free $R$-modules of rank $n$. Its length $d(G)$ equals the valuation of $\det \psi$.

Given a short exact sequence $0 \to G' \to G \to G'' \to 0$, one can choose compatible resolutions of $G'$, $G$, $G''$ by using the same pushout constructions one would use for modules, to get a $3 \times 3$ commutative diagram with short exact rows and columns. The homomorphisms $\psi'$, $\psi$, $\psi''$ define a morphism of short exact sequences of cotangent modules, so $(\det \psi) = (\det \psi')(\det \psi'')$ as ideals of $R$. Hence $d(G)=d(G')+d(G'')$.

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  • $\begingroup$ Thanks! To summarize what's behind that Mazur–Roberts reference, for the sake of anyone reading this post later on: Writing G* for the Cartier dual of G, one embeds G = Hom(G*,G_m) into A = Map(G*,G_m) (maps of S-schemes), and takes B to be the quotient. (The Map in question is representable and that A,B have the stated properties.) $\endgroup$
    – Jay
    Commented Jul 27, 2021 at 21:56
  • $\begingroup$ The rule sending G to this particular s.e.s. [0 to G to A to B to 0] is concrete enough that it should be possible to verify directly that it is exact in G, no pushouts neded? The remaining argument on differentials then makes sense (to me). $\endgroup$
    – Jay
    Commented Jul 27, 2021 at 21:56
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    $\begingroup$ I think that that rule is not exact. It sends $1 \to \mu_2 \to \mu_6 \to \mu_3 \to 1$ to $1 \to \mathbf{G}_m^2 \to \mathbf{G}_m^6 \to \mathbf{G}_m^3 \to 1$, and $2 + 3 \ne 6$. Anyway, the pushout argument is not hard: Choose $G' \hookrightarrow A'$ and $G \hookrightarrow \mathcal{A}$; then $0 \to G' \to G \to G/G' \to 0$ embeds in $0 \to A' \to (A' \times \mathcal{A})/G' \to \mathcal{A}/G' \to 0$ (with $G'$ in the quotient in the middle being the antidiagonal copy), and then one can take cokernels to get $0 \to B' \to B \to B'' \to 0$. $\endgroup$
    – Bjorn Poonen
    Commented Jul 28, 2021 at 1:37
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    $\begingroup$ When I wrote "that rule is not exact", I meant specifically that the functor sending $G$ to $\operatorname{Map}(G^*,\mathbf{G}_m)$ is not exact, so certainly the functor sending $G$ to the Mazur-Roberts sequence $0 \to G \to A \to B \to 0$ is not exact. $\endgroup$
    – Bjorn Poonen
    Commented Jul 28, 2021 at 1:48

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