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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}

\newcommand{\Z}{{\mathbb{Z}}}
\newcommand{\Q}{{\mathbb{Q}}}
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\newcommand{\End}{{\operatorname{End}}}
\newcommand{\Aut}{{\operatorname{Aut}}}
\newcommand{\ACF}{{\operatorname{ACF}}}
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\newcommand{\SU}{{\operatorname{SU}}}

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\newcommand{\Qbar}{\bar\Q}
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\newtheorem*{exercise*}{Exercise}

% Rahim's nonforking symbol (I don't know where he got it from)
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0pt{\hss$#1\mid$\hss}\lower.9\ht0 \hbox to
0pt{\hss$#1\smile$\hss}\kern\wd0}
\def\nind{\mathop{\mathpalette\notind\emptyset }}

\begin{document}
\maketitle

\tableofcontents

\section{Torsion}
\begin{definition}
  Let $A$ be a commutative algebraic group.
  \begin{itemize}\item $A[n] := \{ x : nx = 0 \} \leq  A$, the $n$-torsion subgroup
  \item $A[\infty] := \bigcup_{n \in \N} A[n]$, the torsion subgroup.
  \item For $p$ prime, $$A[p]' := \bigcup_{\{n : (n,p) = 1\}} A[n],$$
  the prime-to-$p$-torsion subgroup.
  \item Also $A[p^\infty] := \bigcup_n A_p^n$.
  \end{itemize}
\end{definition}

\begin{fact}
  If $A$ is an abelian variety and $n$ is prime to the characteristic,
\\  $A[n] \cong  (\Z/n\Z)^{2\dim A}$.
\end{fact}

\section{Manin-Mumford}
\begin{theorem*}[``Manin-Mumford Conjecture''; Raynaud 1983 ($A$ abelian), Hindry 1987] \mbox{}

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

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

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

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

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

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

\begin{theorem} \mbox{}
  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 $X \subsetneq  A$,
\\  $X \cap A[\infty]$ is finite.
\end{theorem}


\section{Definable endomorphisms and subgroups}
Let $(L,\sigma) \vDash  \ACFA_0$ be a monster model. Identify $A$ with $A(L)$.

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

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 $f(\sigma) \in \Z[\sigma,\sigma^{-1}]$ induces a 
group endomorphism of $A$;
\\e.g.\ $(\sigma^{-4} + 3 + 2\sigma^2)(x) = \sigma^{-4}(x) + 3x + 2\sigma^2(x)$.

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

\section{Good reduction}
\begin{definition*}
    $f(T) \in \Z[T]$ has \defn{no cyclotomic factors} if no cyclotomic 
    polynomial divides $f$;
\\    equivalently, no root of unity $\zeta \in \C$ satisfies $f(\zeta)=0$.
\end{definition*}
\begin{proposition} \label{p:fExists}
    For all but finitely many primes $p$,
\\    there is $f(T) \in \Z[T]$ with no cyclotomic factors and $\sigma \in 
    \Aut(\Qbar)$
    \\such that $A[p]' \leq  \ker(f(\sigma))$.
\end{proposition}
\begin{proof}
    Let $p \in \N$ be prime, let $\psi : \Z \rightarrow  \F_p$ 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'' $+_p : (A_p)^2 \rightarrow  A_p$ (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 $A$.

    \begin{fact*}
        All but finitely many $p$ are of good reduction.
    \end{fact*}
    \begin{proof}
      Write above conditions as an $\L_{\operatorname{ring}}$-sentence which holds in 
      $\operatorname{ACF}_0$,
\\      and hence by compactness in $\operatorname{ACF}_p$ for all large enough $p$.
      %\qedhere$_{\mathit{Fact}}$
    \end{proof}

    \begin{fact*}[Weil] \mbox{}
        Let $p$ be of good reduction.
\\        Let $\phi_p \in \Aut(\F_p^\alg)$ be Frobenius $x \mapsto  x^p$.
\\        Then there is $f(T) \in \Z[T]$ with no cyclotomic factors such that 
        $f(\phi_p)$ vanishes on $A_p$.
    \end{fact*}
    \begin{proof}[Sketch proof]
        Let $l\neq p$ be prime.

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

        Let $V_l := \Q_l \otimes_{\Z_l} T_l$, 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} $f(T) \in \Q_l[T]$, but it turns out that $f(T) \in 
        \Z[T]$.
\\        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$_{\mathit{Fact}}$
    \end{proof}

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

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

    \begin{claim}
      $A[p]' \leq  \ker(f(\sigma))$.
    \end{claim}
    \begin{proof}
      Let $\xi \in A[p]'$. Then
      $$\psi(f(\sigma)(\xi)) = f(\phi_p)(\psi(\xi)) = 0,$$
      so
      $$f(\sigma)(\xi) = 0.$$
    \end{proof}
    \qedhere$_{\mathit{Proposition~\ref{p:fExists}}}$
\end{proof}

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

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

\section{Modularity}

\begin{proposition} \mbox{} \label{p:BModular}
    $B \leq  A(L)$ is modular:
\\    if $a,b \in B^{<\omega}$ are tuples, then $\Cb(a/b) \subseteq  \acl(a)$.
\end{proposition}
\begin{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 $x \in B$.

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

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

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

    \begin{fact*}
      $B'$ is the intersection of a countable chain of definable subgroups of 
      $B$.
    \end{fact*}

    \begin{lemma*} \mbox{}
        Any definable infinite subgroup $H$ of $B$ has finite index in $B$.
    \end{lemma*}
    \begin{proof}
        By Fact~\ref{f: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 $H$.
    \end{proof}

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

    \begin{fact*}
    It follows that for some finite normal $K \leq  B$,
\\    $$B/K \subseteq  \dcl(C,k).$$
    \end{fact*}

    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 $h : 
    G' \rightarrow  H'$ where
    \begin{itemize}\item $G'$ is a finite index subgroup of $G$,
    \item $H'$ is a finite index subgroup of $H$,
    \item $\ker(h)$ is finite.
    \end{itemize}

    \begin{fact*}[Hrushovski-Pillay] \mbox{}
        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.
    \end{fact*}

    Let $H(k_1)$ be minimal such that:
    \begin{itemize}\item $k_1 \geq  k$ is a finite extension of $k$,
    \item $H$ is an algebraic group over $k_1$, and
    \item there is a definable virtual isogeny $h$ of $B$ with a subgroup of 
    $H(k_1)$.
    \end{itemize}

    \begin{fact*}
    $H$ is an abelian variety.
    \end{fact*}

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

    \begin{lemma*}
      An abelian variety over $L$ has only countably many definable subgroups.
    \end{lemma*}
    \begin{proof}
      An abelian variety has only countably many algebraic subgroups.
\\      So by Fact~\ref{f: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, $[n] : H \rightarrow  H$ has finite kernel, so $\SU(nH) = \SU(H)$,
\\      so $[H : nH] < \infty$.
\\      But $nH \leq  H'$ since $n(H/H') = 0$,
\\      so $H'$ is a union of cosets of $nH$.
    \end{proof}

    %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 $k_1 \subseteq  \Fix(\tau)$.
\\    Then for $x \in B$, $$h(\tau(x)) = \tau(h(x)) = h(x),$$ so $$\zeta_x := 
    \tau(x) - x \in \ker h.$$

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

    Let $n := tNM$.
\\    Then $\dom(h) \leq  \ker(f(\sigma)) \cap \ker(\sigma^n - 1)$,
\\    so $\ker(f(\sigma)) \cap \ker(\sigma^n - 1)$ has finite index in 
    $\ker(f(\sigma)) = B$,
\\    so by Fact~\ref{f: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 $g(T),h(T) \in \Z[T]$ are s.t.
    %$g(T)f(T) + h(T)(T^n-1) = m \in \N_{>0}$
    %(to recall the algebra here:
    %first get $g,h$ in the Euclidean domain $\Q[T]$, then clear denominators).
    %Then for $b \in B$, $mb = g(\sigma)f(\sigma)b + h(\sigma)(\sigma^n-1)b = 0$,
    %So $B \leq  G[m]$, 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 $a \in B^n$ is interalgebraic with some $a+c 
    \in (B')^n$, 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 $a \in B^n$ and $b$ from $L$,
    %WMA $a \ind _b C$,
    %but then say $c \in C^n$ is such that $a+c \in (B')^n$,
    %then $\acl(a+c) \ni \aCb(a+c/bc) = \aCb(a/bc) = \aCb(a/b)$,
    %so $\aCb(a/b) \in \acl(a)$ (by $a \ind _b c$? Maybe not\ldots ).

    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
    $$ 0 \rightarrow  \ker(h(\sigma)) \rightarrow  \ker(f(\sigma)) \rightarrow  \ker(g(\sigma)) \rightarrow  0 .$$
    Now the CBP can be read as saying:
\\    a finite-rank set $B$ is modular iff it is orthogonal to $k$,
\\    i.e.\ $b \ind _E c$ for any $b \in B$, $c \in k$, 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$_{\mathit{Proposition~\ref{p:BModular}}}$
\end{proof}

\section{$V \cap A[p]'$}
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 
$\acleq(\emptyset )$-definable subgroups.
\\But I don't know how to obtain this from the jet spaces approach.)

Now let $V \subsetneq  A$ be a proper subvariety.

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

\begin{fact*}
  The Zariski closure of a boolean combination of cosets of subgroups of $A$ 
  is a finite union of cosets.
\end{fact*}
%FIXME: neat proof? Basic idea: can write boolean combination as $\bigcup_i 
%\alpha_i (G_i \setminus C_i)$ 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 $\operatorname{Zcl}(G_i \setminus C_i) = \operatorname{Zcl}(G_i)$.

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.

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

\begin{itemize}\item We may assume that $V\cap A[\infty]$ is dense in $V$,
\\hence $V$ is over a number field $K$.

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

\item 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 $|D|$.

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

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

\bibliography{HMM}
\end{document}
