Let $E \subset \mathbb{P}_\mathbb{C}^2$ be an elliptic curve. If $E$ has complex multiplication (by anything) then the theory of complex multiplication in particular tells us that if $\sigma \in \textrm{Aut}(\mathbb{C})$ then $E^\sigma$ will be isogenous to $E$. Suppose $\textrm{End}(E) \cong \mathbb{Z}$ do we still have that $E^\sigma \cong E$ or are there some obvious counterexamples? What if we insist that the $j$-invariant, $j(E)$ is an algebraic number?
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$\begingroup$ Isogenous is not the same as isomorphic (ie $\simeq$). Are you asking whether in general an elliptic curve over $\bf C$ is isogenous to any of its conjugates by an automorphism of $\bf C$ ? This is very likely not true although I don't have a counterexample at hand. $\endgroup$– Damian RösslerCommented Sep 16, 2012 at 5:56
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2$\begingroup$ I mean it's clear that if $j$ is transcendental then an automorphism of $\mathbb C$ takes it to any other transcendental $j$ invariant, which is an uncountable set, but there are uncountably many isogenous curves. But there should also be counterexamples for $j$ algebraic. $\endgroup$– Will SawinCommented Sep 16, 2012 at 7:08
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1$\begingroup$ A curve over a galois extension of $\mathbb{Q}$ that isomorphic to all of its conjugates is called a $\mathbb{Q}$-curve, and by work of Ribet and the recent proof of Serre's conjecture, these are known to be modular (classic $GL_2(\mathbb{A}_\mathbb{Q})$). I think this forces the $j$-invariant, being some integration of a modular form on the upper half plane, to be in either a CM, or a totally real, extension of $\mathbb{Q}$. So any $j$-invariant with field of definition not CM or totally real, should provide a counter example. An expert should verify this, or, more probable, refute this. $\endgroup$– Dror SpeiserCommented Sep 16, 2012 at 8:50
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1$\begingroup$ Dror: you probably mean isogenous, not isomorphic. $\endgroup$– R.P.Commented Sep 16, 2012 at 10:00
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1$\begingroup$ @Will Savin: you mean "countably many isogenous curves"... $\endgroup$– Damian RösslerCommented Sep 16, 2012 at 13:29
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2 Answers
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We say that an elliptic curve $E$ over a number field $K$ is an elliptic $\mathbf{Q}$-curve if it is is isogenous to its Galois conjugates $E^\sigma$. These were first studied by Benedict Gross, but were later studied by Elkies, Ellenberg, Ribet, and many others. It's possible to show that $\mathbf{Q}$ curves are modular, independently of the BCDTW modularity theorem, and that even though they are defined over $K$, their $p$-torsion Galois representations (at least away from the degrees of said isogenies) descend down to $\mathbf{Q}$. Elkies also famously showed if $E_{/K}$ is a $\mathbf{Q}$-curve without CM, there must be a geometrically isogenous curve $E'_{/L}$ such that $Gal(L/\mathbf{Q}) \cong (\mathbf{Z}/2\mathbf{Z})^r$ for some integer $r$. This comes down to showing that the moduli space of elliptic $\mathbf{Q}$-curves of degree $N$ form a certain very special quotient of $X_0(N)$. For details on most of this, see Jordan Ellenberg's "$\mathbf{Q}$-curves and Galois representations" ( http://www.math.wisc.edu/~ellenber/MCAV.pdf )
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One way to get a counterexample is to take any elliptic curve $E$ over a quadratic field $K$ whose conductor is a prime ideal $\mathfrak{p}$ lying over a split prime $p$ in $\mathbb{Q}$. This means that $E$ has multiplicative reduction at $\mathfrak{p}$. If $\sigma$ is the non-trivial automorphism of $K/\mathbb{Q}$, then $E^\sigma$ has conductor $\mathfrak{p}^\sigma \neq \mathfrak{p}$, so $E^\sigma$ has good reduction at $\mathfrak{p}$. Any isogeny $E \to E^\sigma$ would be defined over a number field, and multiplicative/good reduction are both stable under finite field extension. Since isogenous curves have the same reduction type, $E$ and $E^\sigma$ cannot be isogenous.
An example: the two curves of minimal norm conductor over $\mathbb{Q}(\sqrt{5})$ have norm conductor 31, but their conductors are the two different ideals of norm 31. Check out William Stein's table: http://modular.math.washington.edu/Tables/hmf/sqrt5/ellcurve_aplists.txt
As a double check, you can even tell from the Hecke eigenvalues in the table that the curves are conjugate to one another.
