How to prove that $A$ is supersingular iff the Picard number $\rho(A)$ is equal to the second $l$-adic Betti number $b_2(A) = 6$?

Question

Let $A$ be an abelian surface over algebraically closed field $k$ of characteristic $p > 2$. How to prove that $A$ is supersingular (in other words, there is an isogeny between $A$ and $E^2$, where $E$ is a supersingular elliptic curve) iff the Picard number $\rho(A)$ is equal to the second $l$-adic Betti number $b_2(A) = 6$? The proof is contained in Shioda's article "Algebraic cycles on certain K3 surfaces in characteristic $p$", but I can't find it.

Is this theorem right if $k$ is a finite field, $\rho(A)$ is redefined by $\rho(A) := \mathrm{rank}(NS_k(A))$ (divisors over $k$), and the isogeny is defined over $k$?

Answer 1

Now let $k$ be a finite field. If $E$ is an elliptic curve over $k$ then $End(E)\otimes\mathbf{Q}$ is either an imaginary quadratic field or a definite quaternion algebra over $\mathbf{Q}$.

Warning: $E$ may be supersingular even if $End(E)\otimes\mathbf{Q}$ is an imaginary quadratic field; this means that not all endomorphisms of $E$ are defined over $k$.

Again, if $A$ is not $k$-simple then it is isogenous either to a product $E_1\times E_2$ of two mutually non-isogenous elliptic curves $E_1$ and $E_2$ over $k$ (and $\rho(A)=2$) or to a square $E^2$ of an elliptic curve $E$ over $K$ (and $\rho(A)=4$ or $6$). More precisely, $\rho(A)=6$ if and only if $A$ is isogenous to $E^2$ where $E$ is a supersingular elliptic curve, all whose endomorphisms are defined over $k$. (Indeed, it may happen that both $E_1$ and $E_2$ are supersingular but not isogenous over $k$ and then $\rho(A)=2$)

If $A$ is simple then one of the following conditions holds.

I) $End(A)\otimes \mathbf{Q}$ is a CM-field of degree $4$ and $\rho(A)=2$.

2) $End(A)\otimes \mathbf{Q}$ is a quaternion algebra over an imaginary quadratic field. Then $A$ is supersingular but not all its endomorphisms are defined over $k$ and $\rho(A)=4$.

3) $End(A)\otimes \mathbf{Q}$ is a totally definite quaternion algebra over a real quadratic field. Then $A$ is supersingular but not all its endomorphisms are defined over $k$ and $\rho(A)=2$. (See Tate's Bourbaki talk http://archive.numdam.org/ARCHIVE/SB/SB_1968-1969__11_/SB_1968-1969__11__95_0/SB_1968-1969__11__95_0.pdf for a classification of simple abelian varieties over finite fields up to an isogeny.)

In the cases 2) and 3) $A$ becomes isogenous to a square of a supersingular elliptic curve over an algebraic closure of $k$.

Answer 2

If $A$ is not simple then it is isogenous to a product $E_1 \times E_2$ of elliptic curves $E_1$ and $E_2$, and the Picard numbers of $A$ and $E_1\times E_2$ do coincide. If $E_1$ is not isogenous to $E_2$ then $\rho(E_1 \times E_2)=2,$ which is not the case. Therefore, $A$ is isogenous to a square $E^2$ of an elliptic curve $E=E_1$. It is known that $End(E)\otimes\mathbf{Q}$ is either the field $\mathbf{Q}$ of rational numbers or an imaginary quadratic field or a definite quaternion algebra over $\mathbf{Q}$ (in the latter case $E$ is a supersingular elliptic curve and $\rho(E^2)=6$.) Clearly, in the first case $\rho(E^2)=3$, in the second case $\rho(E^2)=4$. So, if $A$ is non-simple then it is isogenous to a square of a supersingular elliptic curve.

Suppose $A$ is simple. Here is the list of all possible division algebras $End(A)\otimes\mathbf{Q}$ over $\mathbf{Q}$ (see Section 6 of F. Oort, Endomorphism algebras of abelian varieties, MR0977774 (90j:11049)).

1. $\mathbf{Q}$, a real quadratic field, a CM field of degree 4 over $\mathbf{Q}$. In this case $\rho(A)=1$ (in the first case) or $2$ (in the remaining cases).
2. An indefinite quaternion algebra over $\mathbf{Q}$. In this case $\rho(A)=3$. (You may find the description of the corresponding $NS(A)\otimes\mathbf{Q}$ in Sect. 21 of Mumford's Abelian varieties). This implies that $A$ is not simple, which finishes the proof when $k$ is algebraically closed.