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I've come across the following question, which I think must be easy for experts: is there a complex elliptic curve $E$ with an isogeny of degree 2 to itself?

Of course one can ask the same question for isogenies whose degree is not a square, or for higher dimensional abelian varieties etc.

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up vote 13 down vote accepted

Expanding on Francois's answer, $E$ has an endomorphism of degree 2 if and only if its endomorphism ring $R=\operatorname{End}(E)$, which is an order in an imaginary quadratic field, has an element of norm 2. There are exactly three such orders, namely $\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{-2}]$, and $\mathbb{Z}[(1+\sqrt{-7})/2]$. So up to isomorphism over $\overline{\mathbb{Q}}$, there are exactly three elliptic curves with endomorphisms of degree 2. Equations for these curves and their degree 2 endomorphism are given in Advanced Topics in the Arithmetic of Elliptic Curves, Proposition II.2.3.1.

There are similarly only finitely many curves with a higher degree cyclic isogeny of fixed degree $d$. Using Velu's formulas, one could probably write them all down for small values of $d$.

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Thank you very much for the additional explanation (I actually needed it)! – rita May 3 '13 at 15:59

Yes, but the elliptic curve needs to have complex multiplication, since the multiplication-by-$n$ map has degree $n^2$. For an explicit example, you can take $E=\mathbf{C}/(\mathbf{Z}+i\mathbf{Z})$ with the isogeny being multiplication by $1+i$.

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Thank you for the answer. – rita May 3 '13 at 15:58

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