Let $E/K$ be an elliptic curve with complex multiplication and let $E[p]$ be the group of $p$-torsion points of $E$ where $K$ is a number field and $p$ is a prime number.

My question is: as a $(\mathbb{Z}/p\mathbb{Z})[G_K]$-representation, does there exist a case such that $E[p]$ is irreducible? ($G_K$ denotes the absolute Galois group of $K$.)


  1. When $E/K$ has no CM, Serre's theorem implies $E[p]$ is irreducible for almost all $p$.

  2. When $E/K$ has CM over $K$, $E[p]$ can never be absolutely irreducible, at least for the case that $End_K(E)\cong$ full ring of integers of an imaginary quadratic field.


1 Answer 1


Certainly. It will (typically) be irreducible for primes where the curve has supersingular reduction. Indeed, for such primes, the p-torsion as a module over the endomorphism ring $O$ is isomorphic to $O/p \simeq \mathbf{F}_{p^2}$, and the image will (for all but finitely many such primes) will be all of $\mathbf{F}^*_{p^2}$ thought of as a subgroup of $\mathrm{GL}_2(\mathbf{F}_{p})$ via the isomorphism of groups $\mathbf{F}_{p^2} \simeq (\mathbf{F}_{p})^2$ (The image will be the so-called “non-split Cartan). This is certainly always irreducible - for example, reducible subgroups have order dividing the order $p(p-1)^2$ of the Borel.

An alternative way to think about it: reducibility for almost all $p$ implies the existence of $p$ isogenies for almost all $p$. Since the isogeny class of an elliptic curve is finite, this implies that for any infinite set of primes $S$ where p-isogenies exist there will be an endomoprhism of degree $pq$ with distinct primes $p$ and $q$ in $S$. (By pigeonhole principle the two isogenies are both from E to the same E’ then take one followed by the dual isogeny of the other.) But their only exist endomorphisms of E of orders which are norms in $O$ which forces $p$ and $q$ to split in the quadratic field.


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