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In his famous Mordell paper, Faltings proved that two abelian varietes $A_1, A_2$ defined over a number field $K$ are isogenous if and only if the local $L$-factors of $A_1, A_2$ are equal at every finite place of $K$. Moreover, there is an analogue of the Birch/Swinnerton-Dyer conjecture for abelian varieties over number fields which predicts, among other things, that the $L$-funciton of $A/K$ ``knows" the Mordell-Weil rank of $A(K)$, since it is equal to the order of vanishing at $s = 1$ of the $L$-function. Further, in the same paper Faltings proved that it suffices to show that for all but finitely many places $v$ of $K$, the local factors agree.

Thus, the $K$-isogeny class of $A/K$ can be seen to 'know' important arithmetic information about $A$, and in the case when $A$ is the Jacobian of a curve $C$, also important information about $C$.

However, we know that there exist many examples (elliptic curves, say) of abelian varieties $A_1, A_2$ which are isomorphic over $\overline{\mathbb{Q}}$ (for example, quadratic twists of curves) but not isomorphic over their field of definition $K$, such that $A_1(K), A_2(K)$ have different ranks. For example, it is known that for any elliptic curve $E/\mathbb{Q}$, $E$ has many quadratic twists with rank at least two. Starting from a curve with rank 0 say, this produces many examples of elliptic curves that observe the aforementioned phenomenon. Therefore, the $\overline{\mathbb{Q}}$-isomorphism class of an abelian variety $A$ need not see much of the arithmetic over $K$. Indeed, we expect rank to behave as randomly in a twist family as we do over all curves (Goldfeld's conjecture).

Nevertheless, if $A_1, A_2$ are isomorphic over $\overline{\mathbb{Q}}$ then they will become isomorphic over some finite extension $M/K$. Thus, the $L$-series of $A_1, A_2$ over $M$ ought to become identical, since they will then be isogenous over $M$.

It thus seems that the $L$-function of an abelian variety $A/K$ depends subtly on the field $K$ itself. The question is, for a given abelian variety $A/K$, how does the $L$-function $L_M(A,s)$, given by viewing $A$ as an abelian variety over $M/K$ for a finite extension of $M$, behave as a function of $M$?

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  • $\begingroup$ Possibly related: mathoverflow.net/questions/149815/… $\endgroup$
    – Sylvain JULIEN
    Commented Jun 29, 2019 at 21:19
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    $\begingroup$ If the extension is Galois then the L-function of the base change is just the product of the original L-function twisted by the Artin representations, with multiplicities equal to the dimensions (as in the decomposition of the regular representation). $\endgroup$
    – François Brunault
    Commented Jun 29, 2019 at 21:32
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    $\begingroup$ @FrançoisBrunault So are you saying, for an elliptic curve $E/\Bbb{Q}$, omitting the finitely many bad reduction and ramified primes $L(E/K,s)=\prod_{p\text{ good}} \exp(\sum_{k \ge 1} \frac{p^{-sk}}{k}tr(\Pi(Frob_{p,L/\Bbb{Q}})^k) tr(\rho_K (Frob_{p,L/\Bbb{Q}})^k)$ with $Frob_{p,L/\Bbb{Q}}\in Gal(L/\Bbb{Q})$ the Frobenius of $O_L/P$, $L$ the normal closure of $K/\Bbb{Q}$, $\Pi(\sigma) \in GL_2(E[\ell^\infty])= GL_2(\Bbb{Z}_\ell)$ and $\rho_K$ the representation of $Gal(L/\Bbb{Q})$ permuting $Gal(L/\Bbb{Q})/Gal(L/K)$ ? $\endgroup$
    – reuns
    Commented Jun 29, 2019 at 22:59
  • $\begingroup$ @reuns This is correct. $\endgroup$
    – Will Sawin
    Commented Jun 30, 2019 at 1:55

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