# From abelianized inertia to divisors

Posted on December 1, 2017 by Dima

tags: curves, Jacobians, finite fields

The goal of this note is to clarify some points from a paper of Bogomolov, Korotiaev and Tschinkel.

Bogomolov’s approach to birational anabelian geometry consists in looking at abelianisations of absolute Galois groups of function fields of varieties over a fixed algebraically closed field $$k$$ and discerning some extra structures that can be computed from the full absolute Galois group in order to reconstruct the function field.

If $$X$$ is a curve over $$k$$, then by results of Florian Pop $${\operatorname{Gal}}(k(X)^{alg}/k(X))$$ is profinite free on $$|X(k)|$$ generators (result independently established by David Harbater: MR1352282 (97b:14035) Harbater, David. Fundamental groups and embedding problems in characteristic p.), and so one cannot distinguish between the curves by the absolute Galois group of their function fields alone.

The anabelian problrem for function fields of curves is reformulated in the BKT paper as follows: let $$(G_1, \mathcal{I}_1)$$ and $$(G_2, \mathcal{I})$$ be a two pairs of abelianizations of absolute Galois groups of functions fields $$k(X_1)$$ and $$k(X_2)$$ of curves of genuse $$\geq 2$$ together with collections of abelianized inertia groups of points.

Let us unpack the terminology.

Let $$L \supset K$$ be a finite extension of fields and $$v: K^\times \to {\mathbb{Z}}$$ a discrete valuation. Assume $$v': L^\times \to {\mathbb{Z}}$$ is a valuation such that $$v'|_K = v|_K$$. The Galois group $${\operatorname{Gal}}(L/K)$$ acts transitively on all such extensions, and the decomposition group $$D_{v'}$$ of $$v'$$ is the subgroup of elements of $${\operatorname{Gal}}(L/K)$$ that fixes $$v'$$: $$v(\sigma\ x) = v(x)$$ for all $$x \in L^\times$$. Decomposition groups of different extensions of $$v$$ to $$L$$ are conjugated. Therefore if the extension $$L/K$$ is Abelian, all decomposition groups coincide. Decomposition group naturally maps to the Galois group of the extension of residue fields: residue field $${\mathcal{O}}_{v'}/{\mathfrak{m}}_{v'}$$ of $$v'$$ over the residue field of $$v$$, the kernel of this homomorphism is called the inertia group. In fact, everything said above applies to non-finitely generated extensions as well. We will denote the abelianized inertia group of any extension of $$v$$ as $$I_{v,a}$$ or simply $$I_v$$.

Here’s how to interpet the decomposition and inertia group in terms of Kummer theory. A cyclic Galois extension $$L/K$$ gives rise to the $$\mu_n$$-torsor $${\operatorname{Spec}}L \otimes_K L$$ defined over $$K$$, and such things are classified by $$H^1$$ of the absolute Galois group: $$H^1(K, \mu_n)$$. Since $$\mu_n$$ carries trivial Galois action, cocycle condition degenerates into the requirement of being a homomorphism from the absolute Galois group of $$K$$ to $$\mu_n$$, and each such homomorphism factors through the maximal abelian factor, let us denote it $${\mathcal{G}}_a$$ after BKT. Thus the $$H^1$$ in question is isomorphic to $${\operatorname{Hom}}({\mathcal{G}}_a, \mu_n)$$. On the other hand, applying group cohomology functor to the Kummer exact sequence gives $$K^\times/(K^\times)^n \cong H^1(K, \mu_n)$$. Now restricting attention to powers of a prime $$l$$, and taking limit, we finally get $\widehat{K^\times} \cong {\operatorname{Hom}}({\mathcal{G}}_{a,l}, {\mathbb{Z}}_l)$ where $$\widehat{K^\times}$$ is the pro-$$l$$ completion, and $${\mathcal{G}}_{a,l}$$ is the maximal Abelian pro-$$l$$ quotient of the absolute Galois group. This identification, when dualized, tells us that $${\operatorname{Hom}}(K^\times, {\mathbb{Z}}_l) \cong {\mathcal{G}}_{a,l}$$.

In terms of this correspondence $D_v = {\operatorname{Hom}}(K^\times/1+m_v, {\mathbb{Z}}_l) \qquad I_v = {\operatorname{Hom}}(K^\times/{\mathcal{O}}_v^\times, {\mathbb{Z}}_l)$ Indeed, there is a natural mapping ${\operatorname{Hom}}(K^\times/1+m_v, {\mathbb{Z}}_l) \to {\operatorname{Hom}}({\mathcal{O}}^\times/1+{\mathfrak{m}}_v, {\mathbb{Z}}_l)$ where $${\mathcal{O}}^\times/1+{\mathfrak{m}}_v$$ is the multiplicative group of the residue field, so the homomorphisms to $${\mathbb{Z}}_l$$ that lie in the inertia must vanish on $${\mathcal{O}}_v^\times$$.

Now let $$X$$ be a smooth projective curve over an algebraically closed field $$k$$, and pick a point $$x \in X(k)$$. In view of the identification above, $$I_x$$ corresponds to the homomorphism obtained as follows. Let $$v_x: K^\times \to {\mathbb{Z}}$$ be the valuation associated with the point $$x$$. Then $$I_{x,a}$$ is the subgroup of $${\operatorname{Hom}}(K^\times, {\mathbb{Z}}_l)$$ topologically generated by $$v_x$$, or, equivalently, the subgroup of maps $$\varphi: K^\times \to {\mathbb{Z}}_l$$ such that $$\varphi({\mathcal{O}}_{v_x}^\times) = 0$$.

Now let $$k=\mathbb{F}_p^{alg}$$, and let $$X$$ be a curve of genuse $$\geq 2$$, $$K=k(X)$$. Then $0 \to K^\times/k^\times \to {\operatorname{Div}}(X) \to {\operatorname{Pic}}(X) \to 0$ Dualize it: $0 \to {\operatorname{Hom}}({\operatorname{Pic}}, {\mathbb{Z}}_l) \to {\operatorname{Hom}}({\operatorname{Div}}, {\mathbb{Z}}_l) \to {\operatorname{Hom}}(K^\times, {\mathbb{Z}}_l) \to {\operatorname{Ext}}^1({\operatorname{Pic}}, {\mathbb{Z}}_l) \to 0$ (and $${\operatorname{Ext}}^1({\operatorname{Div}}, {\mathbb{Z}}_l)$$ vanishes, since $${\operatorname{Div}}$$ is projective as a direct sum of $$X(k)$$ copies of $${\mathbb{Z}}$$). Some identifications:

• $${\operatorname{Hom}}(K^\times, {\mathbb{Z}}_l) \cong {\mathcal{G}}_{a,l}$$ by Kummer theory (above), note that all morphisms here vanish on $$k^\times$$ since it is divisible;
• $${\operatorname{Hom}}({\operatorname{Pic}}(X), {\mathbb{Z}}_l) \cong {\mathbb{Z}}_l$$, since $${\operatorname{Pic}}^0$$ is torsion;
• $${\operatorname{Hom}}({\operatorname{Div}}(X), {\mathbb{Z}}_l)$$ is the $${\mathbb{Z}}_l$$-linear space of maps $$X(k) \to {\mathbb{Z}}_l$$, and $${\operatorname{Hom}}({\operatorname{Pic}}, {\mathbb{Z}}_l)$$ are the constant maps;

Consider the pairing ${\operatorname{Maps}}(X(k), {\mathbb{Q}}_l) \times K^\times/k^\times \to {\mathbb{Q}}_l, \qquad [\mu, f] = \sum_x v_x(f) \mu(x)$ Then $I_x := \{ [a \cdot \delta_x, -]\}_{a \in {\mathbb{Z}}_l}$ where $$\delta_x(x)=1$$, $$\delta_x(y)=0$$ for all $$y \neq x$$.

And elements of the abelianization of the Galois group are of the form $$[\mu, -]$$ for a map $$\mu$$ such that $$[\mu, f] \in {\mathbb{Z}}_l$$ for all $$f \in K^\times$$.

By identifying $${\operatorname{Hom}}({\operatorname{Div}}, {\mathbb{Z}}_l)$$ with $${\operatorname{Maps}}(X(k), {\mathbb{Z}}_l)$$ one observes that ${\operatorname{Hom}}({\operatorname{Div}}(X), {\mathbb{Z}}_l) = \prod_{x \in X(k)} I_x$

Further, since the kernel of the map from the product of inertia groups into $${\mathcal{G}}_a$$ is $${\mathbb{Z}}$$, and it can be expressed as $$\prod_x a\cdot \delta_{x} = 0$$ in $${\mathcal{G}}_a$$. As $$I_x \cong Z_l$$, a generator can be chosen up to $${\mathbb{Z}}_l^\times$$. But because of the relation above, as soon as one chooses a generator $$a \delta_{x_0}$$, the only generators in $$I_y$$, $$y\neq x$$ that fall in the relation above are $$a \cdot \delta_y$$.

Fact. $${\operatorname{Ext}}^1({\operatorname{Pic}}, {\mathbb{Z}}_l)$$ is isomorphic to $$\pi_{1,l} \cong {\mathbb{Z}}_l^{2g}$$. Its Pontryagin dual is isomorphic to $$J_X\{l\} \cong {\mathbb{Q}}_l^{2g}/{\mathbb{Z}}_l^{2g}$$.

(many thanks to Adam Topaz for pointing me to this fact)

Therefore, given two isomorphic pairs $$({\mathcal{G}}_1, I_1)$$ and $$({\mathcal{G}}_2, I_2)$$ we have an isomorphism $$J_{X_1}\{l\} \cong J_{X_2}\{l\}$$.

Further, we are going to take degree zero part of the short exact sequence above and apply two functors: tensor it with $${\mathbb{Z}}_l$$, and take a pro-$$l$$ completion (tensor with $${\mathbb{Z}}/l^n{\mathbb{Z}}$$ and take a limit). Clearly, the result of the application of the first functor has a natural map to the result of the application of the second functor.

$\begin{array}{ccccccccccc} & & 0 & \to & K^\times/k^\times \otimes {\mathbb{Z}}_l & \to & {\operatorname{Div}}^0(X) \otimes {\mathbb{Z}}_l & \to & {\operatorname{Pic}}^0(X)_l & \to & 0\\ & & & & \downarrow & & & & & & \\ 0 & \to & \mathcal{T}_1(X) & \to & \widehat{K^\times} & \to & \widehat{Div}^0(X) & \to & 0 & & \\ \end{array}$

We know that $${\operatorname{Pic}}(X)$$ is a torsion group, and in fact each $$l^n$$-torsion factor is finitely generated. Therefore $${\operatorname{Pic}}\otimes Z_l$$ is the $$l$$-primary part of $${\operatorname{Pic}}$$. Indeed $${\mathbb{Z}}/l^n{\mathbb{Z}}\otimes {\mathbb{Z}}_l \cong {\mathbb{Z}}/l^n{\mathbb{Z}}$$, since any $$x \in {\mathbb{Z}}^n$$ is of the form $$x = l^n\cdot y + z$$ where $$z \in {\mathbb{Z}}$$, but on the other hand $$Z_l$$ is $$l'$$-divisible for any prime $$l'\neq l$$.

Then, notice that pro-$$l$$ completion of $${\operatorname{Pic}}$$ is in fact trivial, since $${\operatorname{Pic}}$$ is divisible — because the ground field is algebraically closed. Denote $$\mathcal{T}_1(X) = \mathrm{lim}_n \mathrm{Tor}_1({\operatorname{Pic}}^0(X), {\mathbb{Z}}/l^n{\mathbb{Z}})$$.

Pick a point $$x_0 \in X(k)$$, a generator of $$I_{x_0}$$, which will be of the form $$a \cdot \delta_{x_0} \in I_{x_0}$$, then the generators $$a \cdot \delta_{x} \in I_x$$ for all $$x \neq x_0$$ are chosen automatically.

Define a subgroup $$FS$$ of $$\hat{K^\times}$$ as the one generated by elements $$\alpha_y$$ (for all $$y \in X(k)$$) such that

• $$\alpha(\delta_{x_0}) = -1$$,
• $$\alpha(\delta_{y}) = 1$$,
• $$\alpha(\delta_{x}) = 0$$ for $$x \neq x_0, y$$.

By definition, this subgroup is mapped to the image of $${\operatorname{Div}}(X) \otimes Z_l$$ in $$\widehat{{\operatorname{Div}}(X)}$$ by the morphism $$\widehat{K^\times} \to \widehat{{\operatorname{Div}}(X)^\times}$$ in the lower sequence for the above diagram. By composing it with the map $${\operatorname{Div}}^0(X) \otimes {\mathbb{Z}}_l \to {\operatorname{Pic}}^0(X)_l$$ we get a map $$FS \to {\operatorname{Pic}}^0(X)_l$$ with kernel $$\mathcal{T}_l(X)\cong {\mathbb{Z}}_l^{2g}$$.

The latter map gives rise to the Abel-Jacobi map, by associating a point $$x \in X(k)$$ to $$\alpha_x \in FS$$.

Note that we have defined $$FS$$ purely group-theoreticaly. The upper sequence in the digaram above, althogh group-theoretic, comes from our knowledge that the consdired pair $$({\mathcal{G}}, I)$$ comes from an algebraic curve.

Therefore, if $$J\{p\} = 0$$, we have reduced our anabelian problem to the following.

Let $$(J_{X_1}, X_1)$$ and $$(J_{X_2}, X_2)$$ be two pairs of Jacobians of curves and Abel-Jacobi images of these curves, isomorphic as abstract groups with the isomorphism mapping $$X_1$$ to a shift of $$X_2$$. Then $$X_1$$ should be isomorphic to $$X_2$$ (possibly to some base change $$X_2 \otimes_k k$$).