Let $E/K$ be an elliptic curve with complex multiplication over an imaginary quadratic field $K$. Then, I heard that it is well-known that the Tate module $V_{p}(E)$ over $\mathbb{Q}_{p}$ decomposes as direct sum of two $Gal(\overline{K}/K)$-modules.

I cannot find any literature and so let me know the elementary (and concrete) proof.

  • $\begingroup$ Proposition 5.4 in Rubin's part of the Cetraro Notes "Arithmetic theory of elliptic curves" states that $E[\mathfrak{a}]$ is a free rank $1$ module over $\mathcal{O}_K/\mathfrak{a}$ for all integral ideals $\mathfrak{a}$. So depending on what $p$ does in $K$, something happens to $\mathcal{O}_K/p^n$. $\endgroup$ – Chris Wuthrich Jun 6 '16 at 8:40
  • $\begingroup$ In fact, $T_{p}(E)$ is a free module over $K_{p} := K \otimes \mathbb{Z}_{p}$. See Theorem 5 of Serre and Tate's \textit{Good reduction of abelian varieties}. $\endgroup$ – Jeff Yelton Jun 6 '16 at 12:10
  • $\begingroup$ Correction: I meant $V_{p}(E) := T_{p}(E) \otimes \mathbb{Q}_{p}$. $\endgroup$ – Jeff Yelton Jun 6 '16 at 12:40

This is covered at the end of Serre's book "Abelian l-adic Representations and Elliptic Curves".

Basically, since $\mathrm{Gal}(\bar{K}/K)$ acts semi-simply on $V_p(E)$ (you might have to go to Lang's book "Abelian Varieties" for details on this) the sequence of $\mathrm{Gal}(\bar{K}/K)$-modules

$$0\to V_p(E(p)^\circ) \to V_p(E) \to Y \to 0$$

splits, where $Y$ are the points of order $p^n$ of $E$, and you get the decomposition $V_p(E)=X_1 \oplus X_2$.

Serre also proves the converse of this, which is the hard part!


Your Answer

