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In Silverman's Arithmetic of Elliptic Curves, Chapter III, Proposition 4.12, we have the statement that if $E/F$ is an elliptic curve and $\Phi$ is a $\mathrm{Gal}(\bar{F}/F)$-invariant subgroup then the quotient $E/\Phi$ and the corresponding quotient map $\phi: E\to E/\Phi$ are defined over $F$.

Is the converse true? I.e., if $E/\Phi$ is defined over $F$, then is $\Phi$ necessarily $\mathrm{Gal}(\bar{F}/F)$-invariant?

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    $\begingroup$ If $f: E \to E'$ is an isogeny defined over $K\supset F$ then $\ker(f) $ is the roots of a polynomial equation in $K$, conversely if $\Phi$ is the roots of a polynomial equation in $K$, then $\phi : E \to E/\Phi$ is defined over $K$. $\endgroup$
    – reuns
    Commented Jan 10, 2019 at 21:44
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    $\begingroup$ If $E/\Phi$ is defined over $F$ then the identity $O$ on $E/\Phi$ is fixed by $\operatorname{Gal}(\bar{F}/F)$ and that is the coset $O+\Phi=\Phi$. $\endgroup$
    – Chris Wuthrich
    Commented Jan 10, 2019 at 23:12
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    $\begingroup$ Something wrong with the above comments. If only the curves are defined over $F$ , one can have an isogeny between them not defined over $F$ so the kernel is not Galois invariant. This happens, e.g., by taking both curves to be the same supersingular elliptic curve over $\mathbb{F}_p$. So, most endomorphisms are only defined over $\mathbb{F}_{p^2}$. But if both curves and the isogeny are defined over $F$, then clearly the kernel is also. $\endgroup$
    – Felipe Voloch
    Commented Jan 11, 2019 at 1:05
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    $\begingroup$ @Felipe is right. The question does not assume that the map $\phi$ is defined over $F$. For instance Cremona curves 32a3 and 64a4 are 2-isogenous over $\mathbb{Q}(\mu_8)$. $\endgroup$
    – Chris Wuthrich
    Commented Jan 11, 2019 at 10:00
  • $\begingroup$ @ChrisWuthrich Is it something like this : for two isogenies $f,f_2$ then $E/\ker(f) \cong E/\ker(f_2)$ iff $\ker(f \circ u) = \ker(f_2 \circ u_2)$ for some nonzero $u,u_2\in End(E)$ so $E/\ker(f) \cong E/\ker(f_2 \circ u_2 \circ u^*)$ and it may happen $u_2 \circ u^* \in End(E)$ is only defined over a larger field than $f,f_2$. And if the whole $End(E)$ are defined over $F$ then that subtlety won't happen $\endgroup$
    – reuns
    Commented Jan 12, 2019 at 0:45

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