**Epistemic status: **I have not checked this carefully for errors, it is entirely possible there are mistakes in this.

We will be mostly be concerned with Elliptic curves over number fields . However, let us start with a general lemma about Elliptic curves with endomorphism ring . In this post, I collect some properties about representations coming from Elliptic curves.

**Lemma 1: **Let be an elliptic curve over a field with endomorphism ring exactly . Then for elliptic curves with isogenies with non isomorphic cyclic kernels, then are not isomorphic.

**Proof: **Say the kernels have degree and suppose are isomorphic. Consider the isogeny where the central arrow is the supposed isomorphism and the last arrow is the dual of . This is an endomorphism of and therefore, by assumption is of the form (multiplication by ).

Moreover, the kernel of contains the kernel of which is a cyclic subgroup. Therefore, is contained in . That is, . Moreover, the quotient by the kernel of contains the cyclic subgroup . By a similar argument, this shows that .

That is to say, and and therefore contrary to assumption.

This might seem like quite a specialized lemma but it has surprising utility. For instance, this lemma features implicitly in showing that the various definitions of a supersingular elliptic curve are equivalent. We will use it together with Shafarevich’s theorem to prove the irreducibility of the Galois representation on the Tate module.

**Theorem 1[Shafarevich’s Theorem]: **If is an elliptic curve over a number field and is a finite set of finite places of , then there are only finitely many isomorphism classes of Elliptic curves with good reduction away from .

This has the important corollary:

**Corollary 1: **There are only finitely many isomorphism classes of Elliptic curves isogenous to .

I will not prove either theorem here but the proofs are not too hard. The idea is to bound the number of Elliptic curves that can occur by considering the Weierstrass equation and showing that there are only finitely many options that can occur for any given discriminant (by Siegel’s theorem). The number of discriminants that can occur is further bounded since they are units in determined upto a 12th power.

We can use the lemma and corollary to prove the irreducibility of Galois representations. Let be the rational Tate module of and the torsion of . Both of these are modules.

**Theorem 2: **For curves with no (rational) Complex multiplication, is irreducible for all and is irreducible for almost all .

**Proof: **If is reducible, then it contains a cyclic submodule defined over . Then, are elliptic curves defined over , isogenous to with cyclic kernels of order and therefore by the lemma are pairwise non isomorphic. However, by the corollary, there can only be finitely many such isomorphism classes and therefore, finitely many such that is reducible.

Similarly, suppose is reducible. Then it contains a cyclic submodule . Since the integral Tate module is also Galois-invariant, we can define defined over . Consider . These are cyclic subgroups of . As such, we can define to be curves over isogenous to with cyclic kernels.

As before, by the lemma they are pairwise distinct but by Shafarevich’s theorem this is impossible. In this case, we can also avoid Shafarevich’s theorem by the following argument:

By the lemma, we know that there for some . Note that in this case. Therefore, there is an isogeny with cyclic kernel but this is also an endomorphism since these curves were assumed to be isomorphic. Since does not have CM by assumption, this is an impossibility (endomorphisms cannot have cyclic kernels in this case).

This theorem is quite interesting but also quite far from the best known. A few remarks follow:

**Remark 1: **The above theorem is false for curves with complex multiplication. In that case, suppose is a field that contains the field of endomorphisms of . If is a prime that splits in , then is reducible. This is because there exists an isogeny defined over with kernel of size .

**Remark 2: **In fact, the assumption of not having rational CM is not hard to achieve at all. For instance, suppose our base field does not contain any quadratic imaginary number field (for instance, it could be or a totally real number field more generally), then by studying the action on the tangent space, it is easy to see that the endomorphism ring is always .

**Remark 3: **Serre proved a much stronger version of this where he showed that, under the hypothesis of the theorem, the image of in the is in fact isomorphic to the entire group for almost all . This is proved in his famous 1972 paper “Proprites galoisiennes des points d’ordre fini des courbes elliptiques”. This is of course much stronger than simple irreducibility.

In fact, we can beef up irreducibility to absolute irreducibility easily enough:

First, note that there is an element (called complex conjugation) in that has order and determinant by the Weil pairing. Hence, it has eigenvalues . In particular, we can find a rational basis of eigenvectors for complex conjugation. Since the action on is by the reduction of the action on , the same is true for the characteristic reduction.

This is to say, we have an element that has two eigenvectors with eigenvalues defined over the base field. This is sufficient to prove absolute irreducibility:

**Corollary 3: **Let be a two dimensional irreducible representation of a group over the field . Suppose that there is an element with two eigenvectors with eigenvalues respectively. Then, the representation is absolutely irreducible.

**Proof: **Let be a field extension such that is reducible. Then, there is an invariant subspace generated by a vector of the form . In particular, fixes it which implies that or is .

However, this immediately implies that the vector space spanned by or in is in fact invariant under and hence is irreducible.

Should the very last word of the post be “reducible” instead of “irreducible”?

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You are right!

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You want to be careful about what you mean by CM. Let’s say that ( a number field) has ‘rational CM’ if and that it has ‘geometric CM’ if has CM. For Theorem 2 and Remark 1 I think you want to refer to rational CM.

You might also point out that not only is the assumption that E has no rational CM not restrictive it applies, for examlpe, to *every* elliptic curve if . Indeed, one can check that if is an elliptic curve the map is a ring map and thus which implies that .

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Yes, you are right. Thanks for the suggestions!

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