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\title{A drastically oversimplified partial account of Hrushovski's approach 
to Manin-Mumford}
\author{Martin Bays}
\date{2023-05-17
\\\ Blockseminar ``Model Theory meets Algebraic Geometry''
\\\ Landhaus Rothenberge, Universität Münster}

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!newop End
!newop Aut
!newop ACF
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!newop alg
!newop SU

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\begin{document}
\maketitle

\tableofcontents

# Torsion
Definition:
  Let $A$ be a commutative algebraic group.
  ***) $A[n] := \{ x : nx = 0 \} <= A$, the $n$-torsion subgroup
  ** $A[\infty] := \bigcup_{n (- \N} A[n]$, the torsion subgroup.
  ** For $p$ prime, $$A[p]' := \bigcup_{\{n : (n,p) = 1\}} A[n],$$
  the prime-to-$p$-torsion subgroup.
  ** Also $A[p^\infty] := \bigcup_n A_p^n$.
  *e)*
  .

Fact:
  If $A$ is an abelian variety and $n$ is prime to the characteristic,
  $A[n] ~= (\Z/n\Z)^{2\dim A}$.
  .

# Manin-Mumford
Theorem*[``Manin-Mumford Conjecture''; Raynaud 1983 ($A$ abelian), Hindry 1987]:--

    Let $A$ be a commutative algebraic group over a number field $K$.
    Let $\Gamma := A[\infty]$ be the torsion subgroup.
    Let $X (= A$ be a subvariety.
    Then $X \cap \Gamma$ is a finite union of cosets of subgroups of $A$.
    .

Hrushovski gave (1996) a new proof of this theorem.

Very briefly, the proof proceeds as follows:
***) Consider $A$ as a definable group in a model of $\ACFA_0$.
** Let $p$ be a suitable prime (almost all primes will be suitable).
** We find a finite rank definable subgroup $B <= A$ containing $A[p]'$.
** Using the trichotomy for $\ACFA_0$, we show that $B$ is a 1-based group.
** Then $X \cap B$ is a boolean combination of cosets.
** Hence $X \cap A[p]'$ is a finite union of cosets.
** Considering two different primes, conclude the same for $X \cap A[\infty]$.
*e)*

Here, we sketch the broad outlines of the proof of this latter result,
with a number of drastic simplifying \underline{assumptions}.

Definition:
  Let $A$ be an abelian variety.
  ***) $A$ is \defn{simple} if it has no infinite proper algebraic (i.e. 
  Zariski closed, equiv.\ $\ACF$-definable) subgroup.
  ** $\End(A)$ is the ring of algebraic endomorphisms (equiv.\ 
  $\ACF$-definable endomorphisms).
  For $n (- \Z$, the multiplication-by-$n$ map $[n] : x -> nx$ is an 
  endomorphism,
  so we consider $\Z <= \End(A)$ embedded this way.
  So $\End(A) = \Z$ means there are no other algebraic endomorphisms.
  *e)*
  .

We sketch Hrushovski's strategy in the following special case:

Theorem:--
  Let $A$ be a \underline{simple} abelian variety \underline{over $\Q$} with 
  \underline{$\End(A) = \Z$}.
  %(This is ``generically'' the case.)
  Then for any proper subvariety $X (!= A$,
  $X \cap A[\infty]$ is finite.
  .


# Definable endomorphisms and subgroups
Let $(L,\sigma) |= \ACFA_0$ be a monster model. Identify $A$ with $A(L)$.

Fact{subgroups}:
  Let $G$ be an algebraic group over $L$.
  Any definable subgroup $H <= G$ is of finite index in a subgroup of the form
  $$\{ x :(x,\sigma x,..., \sigma^{n-1}x) (- S \}$$
  for some algebraic subgroup
  $$S <= G\times G^\sigma \times ... \times G^{\sigma^{n-1}}.$$
  .

Using this and the structure of algebraic subgroups of abelian varieties,
we can understand definable subgroups of $A$ as follows.

$A$ is over $\Q$, so $A^\sigma = A$,
and $\sigma$ induces a group automorphism of $A$.
Hence any Laurent polynomial $f(\sigma) (- \Z[\sigma,\sigma^{-1}]$ induces a 
group endomorphism of $A$;
e.g. $(\sigma^{-4} + 3 + 2\sigma^2)(x) = \sigma^{-4}(x) + 3x + 2\sigma^2(x)$.

Definition*:--
    ***)
    Let $\End^*(A)$ be the ring of definable group endomorphisms of $A$.
    ** Let $E^*(A) := \Q \otimes \End^*(A)$.
    *e)*
    .
Fact{End}:--
    ***(I) $E^*(A)$ is the Laurent polynomial ring $\Q[\sigma,\sigma^{-1}]$.
    % Probably also \End^*(A) = \Z[\sigma,\sigma^{-1}]?
    ** Any definable subgroup $H <= A$ is a finite index subgroup of the kernel 
    of a definable endomorphism,
    i.e. $H <= \ker(f)$ and $[\ker(f) : H] < \infty$ for some $f (- \End^*(A)$.
    ** Any proper definable subgroup $H < A$ has finite rank.
    ** Let $f,g (- \End^*(A)$.
    Then $\ker(f) \cap \ker(g)$ has finite index in $\ker(f)$ iff $f|g$ in 
    $E^*(A)$
    \\\ (and $\ker(f) <= \ker(g)$ iff $f|g$ in $\End^*(A)$).
    *ee*
    .

# Good reduction
Definition*:
    $f(T) (- \Z[T]$ has \defn{no cyclotomic factors} if no cyclotomic 
    polynomial divides $f$;
    equivalently, no root of unity $\zeta (- \C$ satisfies $f(\zeta)=0$.
    .
Proposition{fExists}:
    For all but finitely many primes $p$,
    there is $f(T) (- \Z[T]$ with no cyclotomic factors and $\sigma (- 
    \Aut(\Qbar)$
    \\\such that $A[p]' <= \ker(f(\sigma))$.
    .
Proof:
    Let $p (- \N$ be prime, let $\psi : \Z --> \F_p$ be the reduction 
    map.
    Reducing the coefficients of the defining polynomials over $\Z$,
    we obtain a reduced variety $A_p$ over $\F_p$,
    and a reduced ``addition'' $+_p : (A_p)^2 --> A_p$ (possibly ill-defined).

    Say $p$ is \defn{of good reduction} for $A$ if $(A_p,+_p)$ is an 
    abelian variety of the same dimension as $A$.

    Fact*:
        All but finitely many $p$ are of good reduction.
        .
    Proof:
      Write above conditions as an $\L_{\"ring}$-sentence which holds in 
      $\"ACF_0$,
      and hence by compactness in $\"ACF_p$ for all large enough $p$.
      %\qedhere$_{\mathit{Fact}}$
      .

    Fact*[Weil]:--
        Let $p$ be of good reduction.
        Let $\phi_p (- \Aut(\F_p^\alg)$ be Frobenius $x |-> x^p$.
        Then there is $f(T) (- \Z[T]$ with no cyclotomic factors such that 
        $f(\phi_p)$ vanishes on $A_p$.
        .
    Proof[Sketch proof]:
        Let $l/=p$ be prime.

        Let $T_l := \varprojlim_n A_p[l^n]$ as a $\Z_l$-module (the $l$-adic 
        Tate module).

        Let $V_l := \Q_l (x)_{\Z_l} T_l$, a finite-dimensional $\Q_l$-vector 
        space.

        Then $\phi_p$ acts on $T_l$ and hence on $V_l$ as a linear 
        endomorphism.
        Let $f(T) := \det(TI - \phi_p|_{V_l})$ be the characteristic 
        polynomial of this action.
        \emph{A priori} $f(T) (- \Q_l[T]$, but it turns out that $f(T) (- 
        \Z[T]$.
        Now if $\zeta$ is an $n$th root of unity and $f(\zeta)=0$,
        then $\zeta$ is an eigenvalue of $\phi_p|_{V_l}$,
        and hence $1$ is an eigenvalue of $\phi_p^n|_{V_l}$.
        Then $\phi_p^n$ has infinitely many fixed points in $A_p[\infty]$,
        contradicting the fact that the fixed field of $\phi_p^n$ is the finite field $\F_{p^n}$.

        Finally, $f(\phi_p|_{V_l}) = 0$ (Cayley-Hamilton),
        so $f(\phi_p)$ is zero on $A_p[l^\infty]$,
        and hence on $A_p$, since (Fact) $A_p[l^\infty]$ is Zariski dense in 
        $A_p$.

        %Namely, $f(T) = \det(T - \tau)$ where $\tau$ is the map on the $l$-adic Tate 
        %module ($l$ prime to $p$) induced by $\phi_p$.
        %\qedhere$_{\mathit{Fact}}$
        .

        %\emph{(For $\G_m$: can take any $p$ and $f(T) := T-p$.)}

    Now by some valuation theory we'll omit:
    ***) $\phi_p$ lifts to an automorphism $\sigma$ of $\Qbar$.
    ** If $m$ is coprime to $p$, any $m$-torsion point of $A_p$ lifts to an 
    $m$-torsion point of $A$.
    *e)*
    Now $$|A[m]| = m^{2\dim A} = |A_p[m]|$$ for $m$ coprime to $p$,
    so the homomorphism $$\psi : A[p]' --> A_p[p]'$$ is an isomorphism.

    Claim:
      $A[p]' <= \ker(f(\sigma))$.
      .
    Proof:
      Let $\xi (- A[p]'$. Then
      $$\psi(f(\sigma)(\xi)) = f(\phi_p)(\psi(\xi)) = 0,$$
      so
      $$f(\sigma)(\xi) = 0.$$
      .
    \qedhere$_{\mathit{Proposition~\ref{p:fExists}}}$
    .

Now embed $(\Qbar,\sigma)$ into a monster model $(L,\sigma) |= \ACFA$.
From now on, we work in $(L,\sigma)$, and identify $A$ with $A(L)$.

Let $B := \ker(f(\sigma)) <= A$, a definable subgroup of $A$.
We have
$$ A[p]' <= B <= A .$$

# Modularity

Proposition{BModular}:--
    $B <= A(L)$ is modular:
    if $a,b (- B^{<\omega}$ are tuples, then $\Cb(a/b) (= \acl(a)$.
    .
Proof:
    Assume for now that $f$ is irreducible in $E^*(A)$.

    $B$ is infinite since $A[p]'$ is,
    so there exists an SU-rank 1 type $p(x)$ containing $x (- B$.

    Suppose $p$ is not modular.
    By trichotomy, for some $C$,
    replacing $p$ with a non-forking extension to $C$,
    $p(L) (= \acl(C,k)$ where $k = \Fix(\sigma)$.

    Fact*[Zilber indecomposability theorem]:
      For some $n$, $B' := (p(L)p(L)^{-1})^n$ is a subgroup.
      .

    Hence $B' (= \acl(C,k)$.

    Fact*:
      $B'$ is the intersection of a countable chain of definable subgroups of 
      $B$.
      .

    Lemma*:--
        Any definable infinite subgroup $H$ of $B$ has finite index in $B$.
        .
    Proof:
        By Fact~{End}(II,IV):
        $H$ has finite index in $\ker(g)$ say,
        and so $\ker(g) \cap \ker(f)$ has finite index in $\ker(g)$,
        so $g|f$ in $E^*(A)$, so by irreducibility of $f$ also $f|g$,
        so $\ker(g)\cap\ker(f)$ has finite index in $\ker(f)=B$,
        hence so does $H$.
        .

    So $B/B'$ is profinite, of cardinality $<=2^{\aleph_0}$.
    Hence, enlarging $C$ by representatives of the cosets of $B'$, we have $B (= 
    \acl(C,k)$.

    Fact*:
    It follows that for some finite normal $K <= B$,
    $$B/K (= \dcl(C,k).$$
    .

    So, since $k$ eliminate imaginaries and is stably embedded,
    $B/K$ is definably isomorphic over $C$ to a definable group in the pseudofinite field $k$.

    A \defn{virtual isogeny} between groups $G$ and $H$ is a homomorphism $h : 
    G' --> H'$ where
    ***) $G'$ is a finite index subgroup of $G$,
    ** $H'$ is a finite index subgroup of $H$,
    ** $\ker(h)$ is finite.
    *e)*

    Fact*[Hrushovski-Pillay]:--
        Any definable group in $k$ is definably virtually isogenous with a 
        group $H(k)$ where $H$ is an algebraic group over $k$.
        % Note: something stronger is claimed in Hrushovski's paper, without 
        % the virtuality on the left, but this isn't proven in 
        % Hrushovski-Pillay. Luckily, we don't need it.
        .

    Let $H(k_1)$ be minimal such that:
    ***) $k_1 >= k$ is a finite extension of $k$,
    ** $H$ is an algebraic group over $k_1$, and
    ** there is a definable virtual isogeny $h$ of $B$ with a subgroup of 
    $H(k_1)$.
    *e)*

    Fact*:
    $H$ is an abelian variety.
    .

    Now the graph of $h$ is a definable subgroup of $A \times H$.

    Lemma*:
      An abelian variety over $L$ has only countably many definable subgroups.
      .
    Proof:
      An abelian variety has only countably many algebraic subgroups.
      So by Fact~{subgroups}, every definable subgroup has finite index in one 
      of countably many definable subgroups.
      So STS: a definable subgroup $H$ has only finitely many subgroups $H'$ 
      of index $n$.
      But indeed, $[n] : H --> H$ has finite kernel, so $\SU(nH) = \SU(H)$,
      so $[H : nH] < \infty$.
      But $nH <= H'$ since $n(H/H') = 0$,
      so $H'$ is a union of cosets of $nH$.
      .

    %Hence $h$ is defined over $\acl_\sigma(k_1) = k^\alg = \bigcup_n 
    %\Fix(\sigma^n)$.
    Hence $h$ has only finitely many conjugates over $k_1$.
    So say $\tau := \sigma^t$ is a power of $\sigma$ such that $h^\tau = h$ and $k_1 (= \Fix(\tau)$.
    Then for $x (- B$, $$h(\tau(x)) = \tau(h(x)) = h(x),$$ so $$\zeta_x := 
    \tau(x) - x (- \ker h.$$

    Now $\ker h$ is finite and preserved by $\tau$,
    so say $N$ is such that $\tau^N |_{\ker h} = id$,
    and say $M\ker h = 0$.

    Then for $x (- \dom(h)$,
    $$\tau^{NM}(x) - x = \sum_{i <NM} \tau^{i+1}(x) - \tau(x) = \sum_{i < NM} 
    \tau^i(\zeta_x) = M\sum_{i < N} \tau^i(\zeta_x) = 0.$$

    Let $n := tNM$.
    Then $\dom(h) <= \ker(f(\sigma)) \cap \ker(\sigma^n - 1)$,
    so $\ker(f(\sigma)) \cap \ker(\sigma^n - 1)$ has finite index in 
    $\ker(f(\sigma)) = B$,
    so by Fact~{End}(IV), $f(\sigma)|(\sigma^n - 1)$ in $E^*(A)$,
    contradicting $f$ having no cyclotomic factors.

    %Now $f(T)$ and $T^n-1$ are coprime,
    %so say $g(T),h(T) (- \Z[T]$ are s.t.
    %$g(T)f(T) + h(T)(T^n-1) = m (- \N_{>0}$
    %(to recall the algebra here:
    %first get $g,h$ in the Euclidean domain $\Q[T]$, then clear denominators).
    %Then for $b (- B$, $mb = g(\sigma)f(\sigma)b + h(\sigma)(\sigma^n-1)b = 0$,
    %So $B <= G[m]$, contradicting infinitude of $B$.

    So $X$ is modular.
    Hence $B' = (XX^{-1})^n$ is modular.

    Hence also $B$ is modular.
    Indeed: let $C$ be representatives of the cosets of $B'$ in $B$;
    adding parameters for $C$, any $a (- B^n$ is interalgebraic with some $a+c 
    (- (B')^n$, so $B$ is modular,
    hence $B$ is modular also without $C$,
    since (as in Addick's talk) modularity is preserved by ``deleting 
    parameters''.
    %then given $a (- B^n$ and $b$ from $L$,
    %WMA $a \|/_b C$,
    %but then say $c (- C^n$ is such that $a+c (- (B')^n$,
    %then $\acl(a+c) -) \aCb(a+c/bc) = \aCb(a/bc) = \aCb(a/b)$,
    %so $\aCb(a/b) (- \acl(a)$ (by $a \|/_b c$? Maybe not...).

    This concludes the case that $f$ is irreducible.
    The general case proceeds roughly as follows:
    If $f=gh$, then $h(\sigma)$ induces a short exact sequence
    $$ 0 --> \ker(h(\sigma)) --> \ker(f(\sigma)) --> \ker(g(\sigma)) --> 0 .$$
    Now the CBP can be read as saying:
    a finite-rank set $B$ is modular iff it is orthogonal to $k$,
    i.e. $b \|/_E c$ for any $b (- B$, $c (- k$, and small $E$.
    So if $\ker(f(\sigma))$ is non-modular,
    then $\ker(f(\sigma))$ is non-orthogonal to $k$,
    hence either $\ker(h(\sigma))$ or $\ker(g(\sigma))$ is non-orthogonal to 
    $k$ and hence non-modular.
    So we conclude inductively from the irreducible case.
    \qedhere$_{\mathit{Proposition~\ref{p:BModular}}}$
    .

# $V \cap A[p]'$
By an argument analagous to the case of stable 1-based groups
\cite[Proposition 4.7]{Ch},
qf-definable subsets of $B$ are boolean combinations of cosets of definable 
subgroups.

(In fact, by the trichotomy result of Chatzidakis-Hrushovski, $B$ is stable 
stably embedded,
so any definable subset is a boolean combination of cosets of 
$\acleq({})$-definable subgroups.
But I don't know how to obtain this from the jet spaces approach.)

Now let $V (!= A$ be a proper subvariety.

We want to show that $V \cap A[p]'$ is finite.

Fact*:
  The Zariski closure of a boolean combination of cosets of subgroups of $A$ 
  is a finite union of cosets.
  .
%FIXME: neat proof? Basic idea: can write boolean combination as $\bigcup_i 
%\alpha_i (G_i \\ C_i)$ where $G_i$ is a subgroup and $C_i$ is contained in a 
%union of cosets of infinite index subgroups of $G_i$. Then just need to see 
%that $\"Zcl(G_i \\ C_i) = \"Zcl(G_i)$.

Let $V' := \Zarcl{V\cap A[p]'}$.
Then $V' = \Zarcl{V' \cap B}$ is, by the above, a finite union of cosets of 
algebraic subgroups.
Since $A$ is simple, those algebraic subgroups are finite,
so $V'$ and hence $V\cap A[p']$ is finite.

# $V \cap A[\infty]$
Finally, we sketch how to deduce that $V\cap A[\infty]$ is finite,
following an approach of Pillay.

***) We may assume that $V\cap A[\infty]$ is dense in $V$,
hence $V$ is over a number field $K$.

** Now $A[\infty] = A[p]' + A[p^\infty]$.
** Let $D := \bigcup\{(V-t) \cap A[p]' : t (- A[p^\infty]\}$.

** Then $D$ is finite:
By compactness we obtain a uniform in $x$ bound on $|(V-x) \cap B|$,
hence on $|(V-t) \cap A[p]'|$,
hence on the degree over $K(A[p^\infty])$ of elements of $D$,
hence (Fact) on the order of elements of $D$,
hence on $|D|$.

** By considering another prime $p' /= p$ of good reduction,
$(V-x) \cap A[p^\infty] (= (V-x) \cap A[p']'$ is finite for any $x (- A$.

** If $t+t' (- V\cap A[\infty]$ where $t (- A[p^\infty]$ and $t' (- A[p]'$,
then $t' (- D$,
and so $t (- E := \bigcup_{s (- D} (V-s) \cap A[p^\infty]$,
which is finite.
** So $V \cap A[\infty] (= D + E$ is finite.
*e)*

\bibliography{HMM}
\end{document}
