# Field of Definition of Quotient of Elliptic Curve

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?

• 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 $K$. – reuns Jan 10 at 21:44
• 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 $O+\Phi=\Phi$. – Chris Wuthrich Jan 10 at 23:12
• 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. – Felipe Voloch Jan 11 at 1:05
• @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 $\mathbb{Q}(\mu_8)$. – Chris Wuthrich Jan 11 at 10:00
• @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 – reuns Jan 12 at 0:45