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Section 1.12 Abelian Varieties over finite fields (Ricky)

Set \(q = p^m\text{,}\) \(p\) prime. Given \(X/\FF_q\) have geometric Frobenius \(\pi_X\colon X \to X\) which acts as \(\id\) on \(|X|\) and sends \(f\to f^q\) for \(f\in \sheaf O_X(U)\text{.}\)

Example 1.12.1.

\(X \hookrightarrow \PP^n\) then \(\pi_X(a_0:\cdots :a_n) = (a_0^q : \cdots :a_n^q)\text{.}\)

We also have absolute Frobenius

\begin{equation*} F\colon X\to X^{(p)}\text{.} \end{equation*}
Example 1.12.2.
\begin{equation*} X \colon y^2 = x^3 + i / \FF_q \end{equation*}
\begin{equation*} X^{(p)} \colon y^2 = x^3 + i^3 = x^3 - i / \FF_q \end{equation*}

We see that \(X^{(p^m)} = X\) and \(F^m = \pi_X\text{.}\)

\begin{equation*} \xymatrix{ X \ar[drr] \ar[dr]\ar[ddr] & &\\ & X^{(p)} \ar[d]\ar[r] & X\ar[d]\\ & \FF_q \ar[r]_p &\FF_q } \end{equation*}

If \(f\colon X \to Y\) of \(\FF_q\)-schemes then \(\pi_Y \circ f = f\circ \pi_X\text{.}\) Now let \(X\) be an abelian variety over \(\FF_q\text{.}\) From above, we have \(\pi_X\) commutes with all elements of \(\End^0(X) = \End^0(X)\otimes \QQ\text{.}\) Let \(f_X\) be the characteristic polynomial of \(T_l(\pi_X) \colon V_l(X) \to V_l(X)\) for \(l \ne p\text{.}\)

An alternative definition is to take \(f_X\in \ZZ\lb X\rb\) monic of degree \(2g \text{,}\) \(g = \dim X\) s.t.

\begin{equation*} f_X(n) = \deg([n] -\pi_X)\text{,} \end{equation*}

see 12.8.

Since \(X\) is elementary \(Z(\End^0(X))\) is a field containing \(\QQ\lb \pi_X\rb\text{.}\) Let \(g\) be the minimal polynomial of \(\pi_X\) over \(\QQ\text{.}\) Let \(\alpha\) be a root of \(f\text{.}\) Then \(g(\alpha)\) is an eigenvalue of \(g(V_l(\pi_X)) = V_l(g(\pi_X)) = V_l(0) = 0\text{.}\) Hence \(g(\alpha) = 0\text{.}\)

We need some facts before proving this: Ref 5.20, 5.21

  • There exists
    \begin{equation*} V\colon X^{(p)} \to X \end{equation*}
    such that
    \begin{equation*} V\circ F = [p]_X \end{equation*}
    and
    \begin{equation*} F \circ V = [p]_{X^{(p)}}\text{.} \end{equation*}
    Using \(\deg F = p^g \) get \(\deg V = p^g\)
  • By induction \(\lb p^m\rb = V^m \circ F^m\text{.}\)

We also need some facts about \(F\) and \(V\) relative to \(X^\vee\text{.}\)

\begin{equation*} F_X^\vee = V_{X^\vee} \colon (X^\vee)^{(p)} \to X^\vee \end{equation*}

identifying \((X^\vee)^{(p)} = (X^{(p)})^\vee\text{,}\) Ref 7.33, 7.34.

Reduce to the case where \(X\) is simple, we have

\begin{equation*} h\colon X\to X_1 \times X_2 \times \cdots \times X_s \end{equation*}

an isogeny with \(X_i\) simple, then \(h\) induces an isomorphism

\begin{equation*} h\colon V_l(X)\xrightarrow{\sim} \bigoplus_i V_l(X_i) \end{equation*}

so \(f_X = f_{X_1} \cdots f_{X_s}\text{.}\) Hence we can assume \(X\) is simple.

Let \(\lambda \colon X \to X^\vee\) be a polarization of \(X\) and \(\dagger\) be the corresponding Rosati involution on \(\End^0(X)\) we will show that \(\pi_X\pi_X^\dagger = q\text{.}\)

\begin{equation*} \pi_X \pi_X^\dagger = \pi_X \lambda^{-1} \pi_X^\vee \lambda = \lambda^{-1} \pi_{X^\vee} \pi_X^\vee \lambda = \lambda^{-1} \lb q \rb \lambda = \lb q \rb \end{equation*}

To see \(\pi_{X^\vee} = \pi_X^\vee = q\) we use \(\pi_X = F^m\) and \(\pi_X^\vee = V^m \text{.}\) So \(\pi_{X^\vee} \pi_X^\vee = F^MV^M = p^m = q\text{.}\) As \(X\) is simple \(\QQ\lb \pi_X\rb\) is a field. Thus \(f_X\) is a power of \(g\text{,}\) the minimal polynomial of \(\pi_X/\QQ\text{.}\) So the complex roots of \(f_X\) are \(\iota(\pi_X)\) for every embedding \(\QQ\lb \pi_X\rb\hookrightarrow \CC\text{.}\) since \(\pi_X^\dagger = q/ \pi_X\text{,}\) we see that

\begin{equation*} \QQ[\pi_X] \subseteq \End^0(X) \end{equation*}

is stable under \(\dagger\text{.}\) We have two cases for such a \(K = \QQ\lb \pi_X \rb \)

  1. \(K\) is totally real and \(\dagger = \id\text{.}\)
  2. \(K\) is a CM field and \(\dagger = \overline{\cdot}\text{.}\)

hence we get

\begin{equation*} \iota(\pi_X\pi_X^\dagger) = \iota(\pi_X) \overline{\iota(\pi_X)} = q \end{equation*}

for any \(\iota\colon K \to \CC\text{.}\)

If \(\pm \sqrt q\) is a root of \(f_X\) then we are in the case of \(K\) totally real. If \(\sqrt q\) has multiplicity \(n\text{.}\) Then \(-\sqrt q\) has multiplicity \(2g-n\text{.}\) Thus \(f_X(0) = (-1)^n q^g\text{.}\) But also \(f_X(0 ) = \deg(0 - \pi_X) = q^g\text{.}\) Hence \(n \) is even.

Honda-Tate.

The correspondence between isogeny classes of \(X/\FF_q\) and conjugacy classes of \(q\)-Weil numbers is a bijection. (i.e. algebraic integers \(\alpha\) s.t. \(|\iota \alpha| = \sqrt q\) for all \(\iota \colon \QQ(\alpha) \hookrightarrow \CC\)).

Using relations between a curve \(C/\FF_q\) and its Jacobian \(J(C) \text{,}\) one can show:

Hint: Use Lefschetz trace and \(H^1(C, \QQ_l) \simeq H^1(J(C) , \QQ_l)\text{.}\)

Application: Let \(J = J_0(103) = J(X_0(103))\text{.}\) \(J\sim J_+ \times J_-\text{.}\)

\begin{equation*} J_{\pm} = \im(w \pm \id) \end{equation*}

\(w\) Atkin-Lehner. \(\dim J = 8\) and \(\dim(J_-) = 6\text{.}\) In fact \(\exists f\in S_2( \Gamma_0(103))\) an eigenform s.t. if

\begin{equation*} f=\sum_{n\ge 1} a_n q^n \end{equation*}

then \(\lb \QQ(a_n)_{n \ge 1}: \QQ\rb =6\) and \(\tr( F_{J_-,p}; T_l(J_-)) = \tr_{K/\QQ}(a_p)\) for \(l \ne p, p\ne 103\) We can compute \(\tr_{K/\QQ} ( a_2) = 4\text{.}\) This implies that \(J_- \times \FF_2\) is not the Jacobian of a curve \(/\FF_2\text{,}\) if it were, then if \(J_- \times \FF_2 = J(C)\) then via Lefschetz trace formula

\begin{equation*} \#C(\FF_2) = 2+1 - 4 = -1 \end{equation*}

similar thing at 17.