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I'm reading "An introduction to the theory of $p$-adic representations" by Laurent Berger. In the page 14 it says:

Suppose $F$ is an unramified finite extension of $\mathbb Q_p$ and $E$ is an elliptic curve defined over $F$ with good reduction. Denote $V=\mathbb Q_p\otimes_{\mathbb Z_p}T_p(E)$, and consider $D_{cris}(V)$ (by comparison theorem it's equal to $H_{dR}^1(E)$), we have:

(denote the Frobenius on $D_{cris}(V)$ by $\varphi$)

  1. If $E$ has good ordinary reduction, then $\varphi$ is diagonalizable, and the valuation of two eigenvalues are $0,1$.

  2. If $E$ has good supersingular reduction, then $\phi$ is not diagonalizable.

I wonder how to get this result? How to get the matrix of $\varphi$ explicitly?

(Firstly I don't know how to pick a basis of $T_p(E)$ for calculation. Secondly, I'm trying to use de Rham cohomology, for example I know $\frac{d x}{y}$ and $\frac{xdx}{y}$ form a basis of $H_{dR}^1(E)$, to compute the matrix of $\varphi$ I need to lift $\varphi$ to a weak completion of $E$ and use $p$-adic cohomology，and I think this way is not computable since I need to deal with power series.)

Are there any references on this? Any related material would be nice, thanks.

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  • $\begingroup$ You should be able to extract an explicit matrix, in this basis, using arxiv.org/abs/math/0105031 . (But I'd love to see someone write up an answer with the details.) $\endgroup$
    – David E Speyer
    Commented Oct 19, 2023 at 17:06
  • $\begingroup$ @DavidESpeyer I think that paper only computes $p$-adic approximation of the matrix of $\varphi$, and it's difficult to get the matrix with entries in $F$(not mod $p^n$ approximation)? $\endgroup$
    – Richard
    Commented Oct 20, 2023 at 2:48
  • $\begingroup$ Yes, that's right. I don't think that the matrix entries are algebraic over $\mathbb{Q}$ though, so what would it mean to give them as elements of $\mathbb{Q}_p$ except to give high accuracy approximations? $\endgroup$
    – David E Speyer
    Commented Oct 20, 2023 at 2:56
  • $\begingroup$ @DavidESpeyer Actually in my research I need to calculate the eigenvectors of $\varphi$ when $E$ has ordinary reduction (write them as linear combinations of $\frac{dx}{y}$ and $\frac{xdx}{y}$), so such approximations may not be enough. $\endgroup$
    – Richard
    Commented Oct 20, 2023 at 3:15

1 Answer 1

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Both statements follow readily once you know that the Frobenius map on $D_{\mathrm{cris}}$ satisfies $\varphi^2 - a_p \varphi + p = 0$. This, in turn, can be deduced from the (much stronger) fact that $\varphi^2 - a_p \varphi + p = 0$ as endomorphisms of the special fibre $\overline{E}$ of $E$, which you can find proved in Silverman's book on elliptic curves (and many other places too).

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  • $\begingroup$ May I ask something further? By $p$-adic Hodge theory of admissible $\varphi$-modules I get: In good reduction case, $V$ is simple iff $E$ is supersingular ($V$ fails to be simple when $E$ has ordinary good reduction). I wonder am I right? $\endgroup$
    – Richard
    Commented Oct 20, 2023 at 16:01
  • $\begingroup$ If you have a follow-up question distinct from the original question, then ask it as a separate question. $\endgroup$
    – David Loeffler
    Commented Oct 21, 2023 at 14:15
  • $\begingroup$ I have post it at here, thank you for pointing out. $\endgroup$
    – Richard
    Commented Oct 22, 2023 at 2:48
  • $\begingroup$ Excuse me, I think $p$-power Frobenius is not an endomorphism of $\bar E$ since the residue field of $F$ (say, $\mathbb F_{p^r}$) is larger than $\mathbb F_p$. Hence $\varphi$ won't satisfy this equation, and instead we see $\varphi^{2r}-a_{p^r}\varphi^r+p^r=0$. Besides, supersingular is equivalent to $a_{p^r}\equiv 0 \pmod p$ (not for $a_p$). But with these I don't know how to deduce the statement then.. (for example I can't see if $x^2-a_{p^r}x+p^r$ is irreducible when $a_{p^r}\equiv 0\pmod p$) $\endgroup$
    – Richard
    Commented Oct 25, 2023 at 8:18

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