TITLE: Isogeny classes of elliptic curves
QUESTION [8 upvotes]: 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?

REPLY [11 votes]: 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.