I'm reading about Minimal Weierstrass Equations in VII.1 of Silverman's Arithmetic of Elliptic Curves.
Any elliptic curve $E/K$ can be represented by an affine cubic
$$ Y^2+a_1XY+a_3Y=X^3+a_2X^2+a_4X+a_6 \tag{$\dagger$} $$
If moreover $\text{char}(K)\neq 2,3$ then we can simplify
$$ Y^2 = X^3-27c_4X-54c_6 \tag{$\ddagger$} $$
By Proposition 3.1, any two Weierstrass equations for $E/K$ are related by the following change of variables with $u\in K^*$ and $r,s,t\in K$
$$ X = u^2X'+r $$
$$ Y = u^3Y' + su^2X' + t $$
Suppose $\text{char}(K)\neq 2,3$. By Remark 1.1, if $v(c_4) < 4$, then the Weierstrass equation is minimal, since any change of variables that preserves the form of $(\ddagger)$ satisfies $c_4' = u^{-4}c_4$. In particular, any such change of variables satisfies for some $u\in K^*$
$$ X = u^2X' $$
$$ Y = u^3Y' $$
Requiring that the form $(\ddagger)$ be maintained is obviously a priori more restrictive than the general change of variables outlined in the previous paragraph. My question is: Why is this enough? Can there be an elliptic curve $E/K$ with $\text{char}(K)\neq 2,3$ such that there are two Weierstrass equations: one with form $(\dagger)$ and discriminant $\Delta$, another one with form $(\ddagger)$ and discriminant $\Delta'$, both with coefficients in $R$, the ring of integers of $K$, and satisfying the strict inequality
$$ 0 \leq v(\Delta) < v(\Delta') ?$$
References are included below.
Best Answer
The point is that there must exist a short minimal Weierstrass equation for $E$ - note that you can change any short Weierstrass equation to one with $a_1, a_2,$ or $a_3 $ nonzero, and the same discriminant, simply by making a change of coordinates $$x = x' + r \qquad y = y + sx' + t$$ with suitably chosen $r,s, t$.
Conversely suppose that the residue field $k$ of the local field $K$ has characteristic not $2$ or $3$ (equivalently $2$ and $3$ are units in $\mathcal{O}_K$) and that $E$ admits a (general, integral) Weierstrass equation with discriminant $\Delta$, say $$y^2 + a_1xy + a_3y = x^3 + ...$$ completing the square to eliminate the extra $y$ terms is a change of coordinates of the form $y = 1/2(y' - a_1x - a_3)$ and $x = x$- thus does not change the valuation of the discriminant (since $2$ is a unit) - there is a further change to be done to make sure the cubic is monic, but again, this only uses multiplication by powers of $2$.
Similarly when one "completes the cube" to get rid of the $x^2$ term, this is where we use that $3$ is a unit.
Thus we do not change the valuation of the discriminant by putting the equation in short Weierstrass form. Hence we may assume a minimal Weierstrass equation has $a_1, a_2, a_3$ all zero.