How to show if Weierstrass equation is minimal then $v(\Delta) < 12$

Question

So if $E$ is an elliptic curve over a local field $K$ and it's residue field $k$ such that char$(k) \neq 2,3$, and $E$ is given by a minimal Weierstrass equation $$ y^2 + a_1xy + a_3y = x^3 + a_2 x^2 + a_4x + a_6$$ then I need to show that $v(\Delta) < 12$. Here $v$ is a valuation on $K$ with respect to which it complete.

Now we know by definition that the equation is minimal if it's coefficients $a_i$'s are in $R$, the valuation ring of $K$ and $v(\Delta)$ is minimized. And by the restriction on char, I guess we may have to assume that $\tilde{E}$ is given by $y^2 =x^3+ax+b$.

And maybe try to compute $v(\Delta) = v(4a^3+27b^2)$. But not sure where'll 12 come here?

I'd appreciate any kind of hint.

Thank you.

Answer 1

Here is a proof of the correct version, which says that a Weierstrass equation is minimal iff $v(\Delta) < 12$ or $v(c_4) < 4$ (defined in AEC chapter 3).

One direction is just some early remarks in AEC VII, the direction left to us is that if a Weierstrass equation is minimal that either $v(\Delta) < 12$ or $v(c_4) < 4$.

Since the characteristic of the residue field is not 2 or 3, both 2 and 3 are invertible in $K$ so when one traces through the reduction of a general Weierstrass equation to the nice form $$y^2 = x^3 + Ax + B$$ we see that they are all integral and preserve the valuation of the discriminant and $c_4$ (table 3.1 in AEC).

In that form, $$\Delta = -16(4A^3 + 27 B^2)\ \ \ \textrm{ and }\ \ \ c_4^3 = \Delta j = -1728 (4A)^3$$

Further, recall the only transformations preserving that form are $x = u^2 x'$ and $y = u^3 y'$ and the new equation satisfies $u^4 A' = A$ and $u^6 B' = B$ and $u^{12} \Delta ' = \Delta$ and $u^4 c_4' = c_4$.

We will prove the claim by contradiction; if we have a Weierstrass equation for which both $v(\Delta) \geq 12$ and $v(c_4) \geq 4$ then we can reduce the valuation of the discriminant. In particular, it is enough to show both $v(A) \geq 4$ and $v(B) \geq 6$ which since then if we set $u = 1/\pi$ for some uniformizer $\pi$, the new Weierstrass equation will still be integral because of the relations above, but it will have strictly smaller valuation $v(\Delta') = v(\Delta) - 12$.

First, $A$: from the equation above for $c_4$ and the fact that $v(2) = v(3) = 0$ we see that $$v(A) = v(c_4) \geq 4$$

Next, $B$: $$v(\Delta) = v(-16(4A^3 + 27B^2)) = v(4A^3 + 27B^2)$$

We would like to use the sharper ultrametric inequality, so there are two cases

1. the case we can't use it, so $v(4A^3) = v(27B^2)$ but that says that $3v(A) = 2v(B)$ and the LHS is at least 12 by the first part, so $2v(B) \geq 12$ i.e. $v(B) \geq 6$ are we're done anyways
2. the case we can use it, $v(4A^3) \neq v(27B^2)$. But actually, if we look again at (1), it is clear that it must be $v(4A^3) > v(27B^2)$ since otherwise the same argument will give $v(B) > 6$. In that case the sharp ultrametric inequality says that $$v(\Delta) = v(4A^3 + 27B^2) = v(27B^2) = 2v(B)$$ but once again the LHS is at least $12$ and so $v(B) \geq 12/2 = 6$.

Therefore we are in the position to make the change of variable outlined above, which reduces valuation of the discriminant, in contradiction to our choice of a minimal weierstrass equation.