Finding an example for an elliptic curve over the p-adics with bad reduction but potential good reduction

abstract-algebraalgebraic-number-theoryelliptic-curvesp-adic-number-theory

Problem

I would like to find an elliptic curve $E$ over $\mathbb{Q}_p$ given by the equation
$$
E: \quad y^2 = x^3 – 27 c_4 x – 54 c_6.
$$
with the following properties:

  • $E$ does not have good reduction (i.e. $E$ has bad reduction),
  • $E$ does have potential good reduction,
  • $c_4$ and $c_6$ are both non-zero.

What I tried:

  • Proposition 5.1. on page 196 of Silverman's Arithmetic of Elliptic Curves states, assuming $E$ has minimal Weierstrass equation, that $E$ has good reduction if and only if the valuation of the discriminant is $0$, i.e. it is a unit in $\mathbb{Z}_p$.
  • Proposition 5.5. on page 197 of Silverman's Arithmetic of Elliptic Curves states that $E$ has potential good reduction if and only if its $j$-invariant is integral, i.e. $j(E) \in \mathbb{Z}_p$.
  • (cf. page 42 in Silverman) The discriminant of $E$ is $\Delta = \frac{c_4^3-c_6^2}{1728}$ and $j(E) = c_4^3/\Delta$.
  • (cf. Remark 1.1. of p. 186 in Silverman) For an elliptic curve $E: y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4x + a_6$, we have different criterions which are sufficient for the Weierstrass equation to be minimal. These are
    • $a_i \in \mathbb{Z}_p$ and $v_p(\Delta)<12$,
    • $a_i \in \mathbb{Z}_p$ and $v_p(c_4) < 4$,
    • $a_i \in \mathbb{Z}_p$ and $v_p(c_6) < 6$.

Here, I think one can describe $c_4$ and $c_6$ with the $a_i$'s.

  • I also tried to choose random values for $c_4$ and $c_6$ but it constantly happens that when $v_p(\Delta)>0$, then $v_p(j(E)) < 0$ at the same time, i.e. I cannot use the results above.

Now I do not know how to approach this problem more systematically. Could you please help me with this problem?

Best Answer

So like you said you want to find $p,c_4,c_6$ such that (assuming $p\ne 2,3$)

  1. $c_4\ne0$ and $c_6 \ne 0$
  2. $c_4^3 -c_6^2 \ne 0$
  3. $p|(c_4^3 -c_6^2)$
  4. $3v_p(c_4)=v_p(c_4^3)\ge v_p(c_4^3 -c_6^2)$ (integrality of $j$)

One way to get 3. to hold is for both $c_4,c_6$ to be divisible by $p$.

If $3v_p(c_4) =v_p(c_4^3) > v_p(c_6^2)=2v_p(c_6)$ then $v_p(c_4^3 -c_6^2) = v_p(c_6^2)$ so we get 4. for free.

So how to get $3v_p(c_4) > 2v_p(c_6)$ but $v_p(c_4)\ge 1$ and $v_p(c_6)\ge 1$, we can just take both to have valuation 1!

So why not try $c_4 = c_6 = p$, this satisfies all properties!

So for example $E : y^2 = x^3 -27\cdot5 x -54\cdot 5$ which is https://www.lmfdb.org/EllipticCurve/Q/10800dg1/

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