If $E$ is an elliptic curve over $\mathbb{Q}$, and $K$ is an imaginary quadratic field. If $rank E(K)\leq 1$ and both $E$ and the quadratic twist of $E$ by $K$ satisfy the full BSD conjecture, does the base change of $E$ to $K$ satisfy the full BSD conjecture?
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1 Answer
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Yes.
This follows from the fact that BSD is invariant under Weil restriction and isogeny (the Weil restriction of $E/K$ to $\mathbb{Q}$ is isogenous to the product of $E$ with its quadratic twist).
Note that you don't need to assume here that $K$ is imaginary nor anything about the rank of $E/K$.
answered Aug 31, 2014 at 10:27
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$\begingroup$ Thanks. In a lot of cases, we know that the $p$-part of full BSD of $E$ and the quadratic twist of $E$ holds, by Skinner et. all. So what is the answer if we focus the question on the $p$-part of full BSD conjecture? $\endgroup$ Commented Aug 31, 2014 at 10:43
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$\begingroup$ Im guessing the answer is probably yes here, though I don't know references. I would recommend examining the proofs of invariance of BSD under Weil restriction and isogeny, and see if they can be adapted to this setting. $\endgroup$ Commented Aug 31, 2014 at 11:04
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$\begingroup$ You are right. I chase the proofs of Milne in jmilne.org/math/articles/1972a.pdf and "Arithmetic duality theorem". The two invariances are proved by comparing each ingredient one by one. So the equivalence on $p$-part should holds. $\endgroup$ Commented Aug 31, 2014 at 11:18