Can it be explained in a concrete fashion what isogeny is? Apparently some of the proofs of Fermat's last theorem are essentially this but it is not clear to me what that actually is. (It would be good to see examples which aren't Fermat's theorem though)


2 Answers 2


An isogeny is a special kind of morphism between elliptic curves.

An elliptic curve is, first of all, a curve; we want to be able to map from one curve to another. We already have a good notion of maps between curves, namely rational maps.

But an elliptic curve is more than just a curve: it also has a distinguished point (the point at infinity) and a group structure (the group of $K$-rational points for whatever field $K$ you are working on). So we would like to consider maps between elliptic curves that do more than simply preserve the fact that they are curves (the way rational maps do), but that also preserve this extra structure on the curves.

A group morphism would of course map the identity element to the identity element and take the image of a sum to the sum of the images. For groups, the former is a consequence of the latter. For elliptic curves, the latter is a consequence of the former as well.

So: an isogeny between two elliptic curves $E_1$ and $E_2$ is a rational morphism $\phi\colon E_1\to E_2$ that maps the point at infinity $O$ of $E_1$ to the point at infinity $O$ of $E_2$.

Because rational maps between curves are either constant or surjective, an isogeny either maps all of $E_1$ to $O$, or is surjective onto $E_2$, and in that case must be a finite map.

Added. Silverman refers the reader to Hartshorne or to Shafarevich for the proof that a rational map between curves is either constant or onto. Hartshorne proves it late in the game (Chapter II, Section 6, after he has schemes and divisors on hand). The proposition in question says:

Proposition (II.6.8 in Hartshorne) Let $X$ be a complete nonsingular curve over $k$, $Y$ any curve over $k$, and $f\colon X\to Y$ a morphism. Then either $f(X)$ is a point, or $f(X)=Y$. In the second case, $K(X)$ is a finite extension field of $K(Y)$, $f$ is a finite morphism, and $Y$ is also complete.

The proof of the first part is that since $X$ is complete (proper over $k$; equivalently for curves, projective) it's image must be closed in $Y$ and proper over $\mathrm{Spec} k$. But since $f(X)$ must also be irreducible, it is either a point or all of $Y$.

Shafarevich proves it earlier (without schemes; in Chapter I, Section 5.3, Theorem 4), but in essence Silverman is assuming you know the map will be finite: Shafarevich defines a finite map $f$ to be a map $f\colon X\to Y$ such that the induced inclusion $k[Y]\to k[X]$ makes $k[X]$ integral over $k[Y]$. He then proves finite maps are surjective: if $y\in Y$ and $\mathfrak{m}_y$ is the ideal of $k[Y]$ of functions that vanish on $y$, and $t_1,\ldots,t_n$ are the coordinate functions on $Y$, and $y=(\alpha_1,\ldots,\alpha_n)$; then $\mathfrak{m}_Y = (t_1-\alpha_1,\ldots,t_n-\alpha_n)$. Viewing $k[Y]$ as a subring of $k[X]$, $f^{-1}(y)=\emptyset$ if and only if the $t_i-\alpha_i$ generate the trivial ideal of $k[X]$; this is equivalent to $\mathfrak{m}_yk[Y]=k[X]$, but since $k[X]$ is integral over $k[Y]$ by definition, it is a finite $k[Y]$-module, and by Nakayama's Lemma we cannot have $\mathfrak{m}_yk[Y]=k[X]$. Hence $f^{-1}(y)\neq\emptyset$, so $f$ is onto.

An example of an isogeny is the map from an elliptic curve to itself that maps a point $P$ to $nP$, for any $n\in\mathbb{Z}$. This is an isogeny, because the multiplication map can be expressed with rational functions on the coordinates of the point.

See for example Chapter 3, Section 4, of The Arithmetic of Elliptic Curves by Silverman (titled "Isogenies").

Isogeny comes from iso and genus, "equal origin."

Added. As Adrián Barquero rightly points out, this is then extended to any abelian variety (which is a variety equipped with a binary operation that makes it into an group, with the addition and the inverse maps being rational functions; elliptic curves are a special case of abelian varieties), so that isogeny is a rational map between two abelian varieties that is either constant or has finite kernel, and maps the point at infinity of the domain to the point at infinity of the codomain. (Matt E confirms that the trivial map is sometimes accepted as an isogeny, though sometimes not, so careful there...)

  • $\begingroup$ Thank you for the very simple explanation! You said "such maps between curves are either constant or surjective" I was just wondering if that is a very difficult theorem or is it obvious from the definition? $\endgroup$
    – quanta
    May 3, 2011 at 17:22
  • $\begingroup$ According to Wikipedia, an isogeny is more generally defined as a morphism between abelian varieties. $\endgroup$ May 3, 2011 at 17:24
  • 4
    $\begingroup$ @Jiangwei: Speaking as someone who works in number theory, where isogenies of elliptic curves and abelian varieties come up frequently, the trivial morphism is sometimes accepted as an isogeny, and sometimes not, depending on the context. I don't think there is any universal convention one way or the other. Regards, $\endgroup$
    – Matt E
    May 3, 2011 at 18:52
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    $\begingroup$ @quanta: Dear quanta, One can easily prove the constant/surjective dichotomy over a field of characteristic zero using complex analysis. Standard arguments (the Lefschetz principle) reduce to the case of a morphism of smooth projective curves over $\mathbb C$. Since projective curves are compact, the image is compact. Basic complex analysis proves that a non-constant holomorphic map on an open disk in $\mathbb C$ is open, and so a non-constant isogeny is open. Thus the image of a non-constant isogeny is both open and closed. Since (in this discussion, at least) curves are connected ... $\endgroup$
    – Matt E
    May 3, 2011 at 18:57
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    $\begingroup$ ... the non-constant isogeny is necessarily surjective. Regards, $\endgroup$
    – Matt E
    May 3, 2011 at 18:57

Regarding the motivation for your question, it might be that you've been reading statements such as "any elliptic curve is isogenous to an elliptic curve contained in the Jacobian of a modular curve", or statements related to Faltings's theorem/Tate's conjecture: that two elliptic curves with the same $L$-function are isogenous.

Firstly: over the complex numbers, an elliptic curve $E$ is of the form $\mathbb C/\Lambda$, for a lattice $\Lambda$ in $\mathbb C$. The curves isogenous to $E$ are precisely those of the form $\mathbb C/\Lambda'$, where $\Lambda'$ is another lattice which is contained in $\mathbb Q\otimes \Lambda$ (i.e. $\Lambda'$ is commensurable with $\Lambda$).

Why does this come up? Well, in some sense, isogenous elliptic curves over $\mathbb Q$ have almost the same number-theoretic properties. For example, they have good reduction at the same set of primes $p$, and if $p$ is a prime of good reduction, then the number of solutions mod $p$ of the two curves will be the same. This is why they have the same $L$-function.(The $L$-function is constructed using as input the number of solutions to the elliptic curve modulo primes $p$.)

Since the relationship between elliptic curves and modular forms (the one that Wiles proved) is mediated via their $L$-functions, isogenous elliptic curves come from the same modular form. This is why the concept of isogeny comes up all the time in discussion of FLT and Wiles's proof.

  • $\begingroup$ My guess is that the OP was reading about how Fermat's descent arguments can be interpreted using $2$-isogenies or $3$-isogenies. $\endgroup$ May 3, 2011 at 21:12
  • $\begingroup$ @Qiaochu: Dear Qiaochu, This hadn't occurred to me, but now that you say it, I'm sure you're that you're right! Regards, $\endgroup$
    – Matt E
    May 4, 2011 at 1:17

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