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Let $A$ be an abelian variety defined over a field $K$ of characteristic $p>0$. Let $A[\ell]$ be the group of $\ell$-torsion points, $\ell\neq p$ a prime. Are there positive constants $C, \eta$ depending on $A$ and $K$ only such that $[K(A[\ell]):K]> C \ell^{\eta}$? What about $[K(P):K]> C' \ell^{\eta'}$, $P$ an $\ell$-torsion point? In some cases I know the Lie algebra to be $sp(4,Q_{\ell}) \oplus Q_{\ell} Id$ and $sp(6,Q_{\ell}) \oplus Q_{\ell} Id$. Is this enough?

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You want C,eta to be independent of ell? Then No for trivial reasons because K could be algebraically closed. – Kevin Buzzard Mar 11 '10 at 19:37
Right. Here $K$ is finitely genereated. – Oscar Villareal Jun 22 '10 at 22:56
up vote 3 down vote accepted

Now let me address your last question. For the sake of simplicity, let us assume that $K$ is a global field of characteristic $p>2$ and the ring $End(A)$ of all endomorphisms of $A$ (over an algebraic closure of $K$) is the ring $Z$ of integers. Then my old results (Math. Notes: 21 (1978), 415--419 and 22 (1978), 493--498) imply that for all but finitely many primes $\ell$ the Galois module $A[\ell]$ is absolutely simple and the Galois group $Gal(K(A[\ell])/K)$ is noncommutative.

I claim that for all but finitely many primes $\ell$ the order of $Gal(K(A[\ell])/K)$ is divisible by $\ell$ and therefore $[K(A[\ell]):K] > \ell$.

Indeed, suppose that for a given $\ell$ the order of $Gal(K(A[\ell])/K)$ is not divisible by $\ell$. Let us call such $\ell$ exceptional. Then the natural representation of $Gal(K(A[\ell])/K)$ in $A[\ell]$ can be lifted to characteristic zero, i.e., $Gal(K(A[\ell])/K)$ is isomorphic to a (finite) subgroup of $GL(2\dim(A),C)$ where $C$ is the field of complex numbers. Now, by a theorem of Jordan, there exists a positive integer $N$ that depends only on $\dim(A)$ anf such that $Gal(K(A[\ell])/K)$ contains a normal commutative subgroup, whose index does not exceed $N$. This means that $K(A[\ell])/K$ contains a Galois subextension $K_{0,\ell}/K$ such that $[K_{0,\ell}:K] \le N$ and the field extension $K(A[\ell])/K_{0,\ell}$ is abelian.

Let $S$ be the (finite) set of places of bad reduction for $A$. The field extension $K(A[\ell])/K$ is unramified outside $S$; the same is true for the subextension $K_{0,\ell}/K$. (In the number field case one should add the divisors of $\ell$ but we live in characteristic $p$!). Recall that the Galois extensions of global $K$ with degree $\le N$ and ramification only at $S$ constitute a finite set. Let $L/K$ be the compositum of all such extensions. Clearly, $L$ is also a global field while $L/K$ is a finite Galois extension that contains $K_{0,\ell}$. In addition, the Galois group $Gal(L(A[\ell])/L)$ is commutative. Now considering $A$ as an abelian variety over $L$ and applying previously mentioned results, we obtain that the set of exceptional primes $\ell$ is finite.

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Let $E$ be a supersingular elliptic curve over a finite field $K$ of characteristic $p$. If $K$ is sufficiently large then the generator (Frobenius automorphism) of the absolute Galois group of $K$ acts on all $E[\ell]$ as multiplication by the square root $q'$ of $q$ where $q$ is the cardinality of $K$. Let $d=d(\ell)$ be the smallest positive integer such that $(q')^d-1$ is divisible by $\ell$. Then $[K(E[\ell]):K]=d$.

Now suppose that $p=2$, $q$ is, at least, $4$ and let $r$ be a prime such that $\ell=2^r-1$ is a (Mersenne) prime. Then $[K(E[\ell]):K]=d(\ell)<\log(\ell+1)$. So, if the set of Mersenne primes is infinite then we get a negative answer to your first question. Similarly, if the set of Fermat primes is infinite then we also get a negative answer to your first question. Of course, it would be good enough to know that infinitely many Mersenne (or Fermat) numbers have ``sufficiently large" prime factors.

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