Additivity of Elliptic Curve Rank over Compositum of Fields Assume that BSD holds for number fields. Let $E/\mathbf{Q}$ be an elliptic curve. For simplicity, let's assume it has Mordell-Weil rank zero. Let $F_1/\mathbf{Q}$ and $F_2/\mathbf{Q}$ be finite, abelian, disjoint extensions, and let $F=F_1.F_2$ denote their compositum.

Question: Under what conditions can we conclude that
$$ \mathrm{rank}_{\mathbf{Z}}(E/F)=\mathrm{rank}_{\mathbf{Z}}(E/F_1) + \mathrm{rank}_{\mathbf{Z}}(E/F_2) ?$$
In other words, are there any sufficient conditions to ensure that any Mordell-Weil rank growth in $F$ actually in arises either $F_1$ or $F_2$?

On the one hand, this seems too good to be true. On the other hand, if we assume BSD for number fields, it seems that we should be able to make an argument using the factorization of $L$-functions as follows:
Let $\mathrm{Gal}(F_1/\mathbf{Q})=G_1$ and $\mathrm{Gal}(F_2/\mathbf{Q})=G_2$. Since $F_1 \cap F_2 = \mathbf{Q}$, we have $\mathrm{Gal}(F/\mathbf{Q})=G_1 \times G_2$. Using $^\hat{}$ to denote character groups, it seems we should have
\begin{align*}
L(E, F, s) &= \prod_{\chi\in \hat{G}} L(E, \mathbf{Q}, s, \chi)\\ &\simeq \prod_{\chi_1\in \hat{G_1}}\prod_{\chi_2\in \hat{G_2}} L(E, \mathbf{Q}, s, \chi_1 \chi_2)\\ &=L(E, F_1, s)\times L(E, F_2, s) \times\prod_{\substack{\chi_1\in \hat{G_1} \\ \chi_1\neq 1}}\prod_{\substack{\chi_2\in \hat{G_2} \\ \chi_2\neq 1}} L(E, \mathbf{Q}, s, \chi_1 \chi_2)
\end{align*}
where the products are taken over the irreducible characters. Thus, we have the desired additivity if we can control the terms of the form $L(E, \mathbf{Q}, s, \chi_1 \chi_2)$.
Are there any hypotheses which push this across the finish line?

Edit: Ariel Weiss has posted an answer which highlights the following situation:
$$E=X_0(11), F_1=\mathbf{Q}(\sqrt{-7}), F_1=\mathbf{Q}(\sqrt{-8}), F=\mathbf{Q}(\sqrt{-7},\sqrt{-8})$$ Then we have
$$\mathrm{rank}_{\mathbf{Z}}(E/\mathbf{Q})=0, \mathrm{rank}_{\mathbf{Z}}(E/F_1)=1, \mathrm{rank}_{\mathbf{Z}}(E/F_2)=1, \mathrm{rank}_{\mathbf{Z}}(E/F)=2$$ This is precisely the sort of situation I'm looking to capture in general.
 A: It's quite easy to find counterexamples.
For example, take an elliptic curve $E/\mathbb Q$ of rank 0, and choose two quadratic twists, $E_{d}, E_{d'}$ of rank 1. By root number considerations, $E_{dd'}$ has even rank, so it probably has rank $0$. For example, you could take $E = X_0(11)$, $d = -7$, $d' = -8$.
Then $E(\mathbb Q(\sqrt{dd'})$ has rank 0 and $E(\mathbb Q(\sqrt{d})$ has rank 1. But $E(\mathbb Q(\sqrt d, \sqrt{d'})$ has rank at least $2$, with the additional point of infinite order coming from $E(\mathbb Q(\sqrt{d'})$, which also has rank $1$.
In your BSD example, the L-function of $E$ over $\mathbb Q(\sqrt d, \sqrt{d'})$ factors as
$$L(E, s)L(E_d, s)L(E_{d'}, s)L(E_{dd'}, s),$$
Here, the extra $L$-factor $L(E, \mathbb Q, s, \chi_1\chi_2)$ is measuring the contribution to the rank from some other subextension of the compositum of $F_1$ and $F_2$.
This type of argument works more generally under your assumptions that $F_1, F_2$ are Galois. So somehow, your assumption needs to be that no other subextension of $F_1\cdot F_2$ contributes to the rank.
A: If you're willing to expand your question a bit, there are interesting relations to be found. More precisely, let $E/K$ be an elliptic curve defined over a number field, and let $L/K$ be a Galois extension such that $G(L/K)$ has a non-trivial idempotent relation (also sometimes called a Brauer relation). For example, this will be true if if $G(L/K)$ is a non-cyclic abelian group. Each idempotent relation yields a corresponding relation among the ranks of $E(F)$ as $F$ ranges over the  fields lying between $K$ and $L$. This, and much more, may be found in the paper
Kani, Ernst; Rosen, Michael, Idempotent relations among arithmetic invariants attached to number fields and algebraic varieties, J. Number Theory 46, No. 2, 230-254 (1994). ZBL0853.14011.
