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Let $A$ be an Abelian variety defined over the finite field with $q$ elements. Let $P_i(T)$ be the characteristic polynomial of the action of the Frobenius on the $i^{th}$ étale cohomology group.

Is the following assertion true:

"The Abelian variety $A$ is simple over the finite field with $q$ elements if and only if $P_1(T)$ is irreducible over $\mathbb Q$" ?

One implication is obvious, what about the other one ?

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This is true if the abelian variety is ordinary since in this case Frobenius generates the endomorphism algebra. – ulrich Feb 15 '12 at 14:43
To Ulrich : what means ordinary ? – Xavier49 Feb 15 '12 at 14:48
One way to define ordinary is as follows: if $A$ is $g$ dimensional then $A$ is ordinary if it has $p^g$ points of order $p$ over the algebraic closure of the base field $k$, where $p$ is $char(k)$. – ulrich Feb 15 '12 at 15:07
up vote 13 down vote accepted

The following result follows from Tate-Honda theory

Let $A$ be an abelian variety over a finite field $k$, and let $f_A$ be the characteristic polynomial of $A$. Then $A$ is isogenous to a power of a simple abelian variety if and only if $f_A$ is a power of an irreducible polynomial.

I can't find a set of online notes which contains this statement. Kirsten Eisenträger's notes are generally very good, but they get this result wrong on the first page -- Theorem 1.1 claims that, if $f_A$ is a power of an irreducible polynomial, then $A$ is simple, ignoring the possibility that $A$ is a power of a simple variety.

Let $A$ be isogenous to $\bigoplus A_i^{n_i}$, where the $A_i$ are simple and mutually non-isogenous. Every abelian variety has such a decomposition. Then $f_A = \prod f_{A_i}^{n_i}$.

Suppose that $f_A$ is a power of an irreducible polynomial. Then all of the $f_{A_i}$ must also be powers of that polynomial. In particular, for any $i$ and $j$, either $f_{A_i}$ divides $f_{A_j}$ or vice versa; without loss of generality, suppose $f_{A_i} | f_{A_j}$. By a result of Tate, this means that $A_i$ is isogenous to a subvariety of $A_j$. Since $A_i$ and $A_j$ are simple, this means that $A_i$ and $A_j$ are isogenous. Since we assumed that the $A_i$ were mutually nonisogenous, there must in fact be only one summand in our decomposition of $A$, and $A$ is isogenous to $A_1^{n_1}$ for some simple $A_1$ and some $n_1$.

Suppose now that $A$ is isogenous to $B^{n}$ for $B$ simple. Then $f_A = f_B^n$. So our goal is to show that $f_B$ is a power of an irreducible polynomial. If not, write $f_B = gh$ where $g$ and $h$ are relatively prime of positive degree. By a result of Honda, there exist abelian varieties $C$ and $D$ with characteristic polynomials $g$ and $h$. By the result of Tate cited above, $C$ and $D$ are isogenous to subvarieties of $B$, contradicting that $B$ is simple. $\square$

The answer to the question in your title is "yes", we can decide whether $A$ is irreducible by knowing its $\zeta$ function. Fix a prime power $q$. Let $k$ be the field with $q$ elements. Let $W(q)$ be the set of irreducible monic polynomials over $\mathbb{Q}$ all of whose roots have norm $q^{1/2}$. The main result of Honda-Tate theory (Theorem 4.1 in Kirsten's notes) is that there is a bijection between isogeny classes of $k$-simple abelian varieties over $k$ and $W(q)$. For each polynomial $g$ in $W(q)$, there is some positive integer $n(g,q)$ such that the characteristic polynomial of the corresponding simple abelian variety is $f^{n(g,q)}$. The tricky point is that $n(g,q)$ is not always $1$. For example, in Denis's answer, what is going on is that $n(x-p, p^2)=2$. So it is true that $A$ is $k$-simple if and only if $f_A$ is of the form $g^{n(g,q)}$; you just need to know how to compute that $n$ function. I think you should be able to extract this from sections 4 and 5 of Kirsten's notes, but I don't know the details.

UPDATE: Brian Conrad e-mails to spell out the recipe (hope I copied this correctly). Let $f$ be irreducible of the required form. Let $\pi$ be a root of $f$ and let $F$ be the field $\mathbb{Q}(\pi)$. For every $p$-adic place $v$ of $F$, let $d_v$ be the denominator of $v(\pi) [F_v:\mathbb{Q}_p]/v(q)$ when written in lowest terms. Let $d = LCM(d_v)$ where the LCM ranges over all possible $v$'s. Then $f^d$ is the characteristic polynomial of the simple abelian variety.

If I'm not mistaken, this condition can be stated in an elegant geometric way. For any polynomial $g$ over $\mathbb{Q}_p$, let $N(g)$ be the $p$-adic Newton polytope of $g$. We will subdivide the path $N$ as follows: Recall that, if $h$ is irreducible over $\mathbb{Q}_p$, then $N(h)$ is a line segment, and that, if $g$ factors as $\prod h_i^{r_i}$, then $N(g)$ is the concatenation of $r_i$ copies of each $N(h_i)$, ordered with increasing slope. We will decompose $N(g)$ into one piece for each distinct irreducible factor, with that piece being $r_i$ times $N(h_i)$. For example, $x^2-p^2$, $x^2+p^2$ and $x^2-2xp+p^2$ all have Newton polytope a line segment from $(2,0)$ to $(0,2)$. In the first case, we would subdivide this line segment into two line segments, touching at $(1,1)$, because the two factors $x+p$ and $x-p$ are distinct. In the second case, we would subdivide if $x^2+p^2$ factored in $\mathbb{Q}_p$ (i.e. if $p$ is $1 \mod 4$) but not if it remained irreducible (if $p$ is $3 \mod 4$). In the third case, we would not subdivide, because the factor $(x-p)$ is repeated.

Then I believe the condition is that $f$ is the characteristic polynomial of an abelian variety if and only if all the vertices of $N(f)$, subdivided as above, have heights that are integer multiples of $v_p(q)$.

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I will study that.Thanks a lot David. – Xavier49 Feb 15 '12 at 16:12

No, an elliptic curve over a field with $p^2$ elements may have zeta function $(X-p)^2=X^2-2pX+p^2$.

(See, e.g., section 4 of Waterhouse's paper "Abelian varieties over finite fields", though this probably goes back to Deuring.)

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Nice example, thank you. The implication I was thinking about was not so obvious. – Xavier49 Feb 15 '12 at 16:21

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