Let $K$ be a complete extension of $\mathbb{Q}_{p}$ with valuation $v$ over $p$, valuation ring $R$, maximal ideal $\mathfrak{m}$ and residue field $k$. It is well known that if $E/K$ is an elliptic curve, then the valuation of the $j$-invariant of $E$ determines it's (potential) reduction type; namely, if $v(j(E))\geq 0$ then $E$ has potentially good reduction, and potentially bad multiplicative reduction otherwise.

In the former case, if $\tilde{E}$ the reduction mod $v$ of $E$ is an ordinary elliptic curve, then $E$ is distinguished from other liftings of $\tilde{E}$ by it's Serre-Tate coordinate $\lambda \in \widehat{\mathbb{G}_{m}}(R)=1+\mathfrak{m}R$. Serre-Tate theory tells us that lifting $\tilde{E}$ is equivalent to lifting its $p$-divisible group, which is equivalent to specifying an extension of $G_{K}$-modules:

$$0\longrightarrow \mu_{p^{\infty}} \longrightarrow E_{p^{\infty}} \longrightarrow \mathbb{Z}_{p} \longrightarrow 0$$

This is specified by the Serre-Tate coordinate in the sense that the Galois action on $p^{n}$-torsion is determined by the action on $p^{n}$-th roots of $\lambda$. In particular:

$$K(E[p^{n}]) = K(\mu_{p^{n}},\lambda^{1/p^{n}})$$

On the other hand, if $v(j(E)) < 0$ then $E$ has (potentially) bad multiplicative reduction and is isomorphic to a Tate curve: $$E_{q}: y^{2}+xy=x^{3}+a_{4}(q)x+a_{6}(q)$$ where $q\in K$ has $v(q)>0$ and $a_{4}(q)$ and $a_{6}(q)$ are convergent power series in $q$ with $v(a_{4}(q)),v(a_{6}(q))>0$. Thus, $E_{q}$ reduces mod $v$ to the nodal cubic: $$\tilde{E_{q}}:y^{2}+xy=x^{3}$$ Similarly, the uniformising parameter $q$ determines the Galois action on torsion of $E_{q}$ and we have: $$K(E[m])=K(\mu_{m},q^{1/m})$$

It seems to me that the Serre-Tate coordinate and uniformising parameter play the same role in terms of describing the Galois action on torsion (in precisely the same way), however, Tate's uniformisation is different in how one can write equations for a lifting of $\tilde{E}_{q}$ in terms of $q$.

I was wondering if the Serre-Tate coordinate $\lambda$ and uniformising parameter $q$ are really playing the same role that I am suggesting, and if so is there a similar sense in which one can 'write equations' for a lifting of an ordinary elliptic curve $\tilde{E}$ in terms of a Serre-Tate coordinate $\lambda\in 1+\mathfrak{m}R$. Serre-Tate theory suggests that $\lambda$ completely determines the lift $E$ so in this sense, an equation for $E$ depends only on the data ($\tilde{E},\lambda$) but I am not aware of how one, in principle, 'constructs' $E$ from this data.


1 Answer 1


There is a similar story: One can write $E$ as a $y^2 +a_1(\lambda-1) xy + a_3(\lambda-1) y = x^3+ a_2(\lambda-1)x^2 + a_4(\lambda-1) x + a_6 (\lambda-1)$ where $a_1,a_2,a_3,a_4,a_6$ are power series with coefficients in $R$.

For $\lambda \in 1 + \mathfrak m R$, these power series will of course be convergent. One can of course simplify this expression in the usual ways depending on the characteristic, e.g. set $a_1,a_2,a_3$ to zero in characteristic $>3$

This follows from the classical Serre-Tate theory, which among other things states that there is an isomorphism between the formal moduli space of deformations of $E$ and the formal multiplicative group $\widehat{\mathbb G_m}$, which is the formal spectrum of the ring of power series in the variable $\lambda-1$. The coefficients $a_1,a_2,a_3,a_4,a_6$ can certainly be written as functions on the formal moduli space of deformations of $E$ by choosing a Weierstrass model for the universal deformation, and thus can be expressed as power series in the variable $\lambda-1$, with integral $p$-adic coefficients.

I don't know if or how one can make these power series explicit.

  • $\begingroup$ Thanks for the answer Will! It would be very interesting if one could give explicit power series for the coefficients as in the Tate curve case. $\endgroup$ Jan 24 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.