MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

4 edited body

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 $d = GCD(d_v)$ LCM(d_v)$where the GCD LCM ranges over all possible$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$v_p(q)$. 3 added 1788 characters in body 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$d = GCD(d_v)$where the GCD 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$v_p(q)$. 2 added 1050 characters in body 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.

1