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Let $A$ be an abelian variety over $\mathbb{F}_p$. Then of course for every natural number $i$, we have that $\# A(\mathbb{F}_{p^i})$ divides $\# A(\mathbb{F}_{p^{i+1}})$.

But MAGMA says this is false: Here is my code:

P<x> := PolynomialRing((FiniteField(3)));
J := Jacobian(HyperellipticCurve(x^6 - 2 * x^5 + x^4 - 2 * x^3 + 6 * x^2 - 4 * x + 1));
for j in [1..10] do;
    Order(BaseChange(J, FiniteField(3, j)));
end for;

And the result is:

19 57 1444 5529 59299 467856 4976347 43264425 394975876 3458495577

What is wrong?

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  • $\begingroup$ Perhaps the code is buggy and does not check whether the curve is singular or not! The way you have written your hyperelliptic curve is confusing. Is the curve the same as the one associated with $x^6+x^5+x^4+2*x+1$? Perhaps that curve is singular over $\mathbb{F}_3$. $\endgroup$
    – Kapil
    Commented Apr 1, 2021 at 6:34
  • $\begingroup$ @Kapil Thank you for your comment. But Magma says this curve is smooth... $\endgroup$
    – zom
    Commented Apr 1, 2021 at 6:38
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    $\begingroup$ OK! I missed the point that you were looking at $\mathbb{F}_{p^i}$ and $\mathbb{F}_{p^{i+1}$. as pointed out by Alex J Best! Embarrassing for me too! $\endgroup$
    – Kapil
    Commented Apr 1, 2021 at 6:46

1 Answer 1

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$\mathbf F_{p^i}$ is only a subfield of $\mathbf F_{p^j}$ when $i |j$ so you only have the divisibility of group orders for $i|j$ not for $j = i+1$.

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    $\begingroup$ Thank you very much!!! It's my soooo simple mistake, I'm embarrassed... $\endgroup$
    – zom
    Commented Apr 1, 2021 at 6:40

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