How to change coordinates so that $\pi|a_3,a_4,a_6$ in the second step of Tate's algorithm.

Question

I am trying to apply Tate's algorithm for the elliptic curve $$y^2-2(7T+3)xy-36(7T+3)y-x^3-2(16T^2-7)x^2+324x+648(16T^2-7)$$ over the rational function field $\mathbb{Q}(T)$ at the primes $\pi=T+1,T-1,9T-7,27T-23,3T-1$. I expect that they should be of Kodaira type $I_2,I_2,I_2,I_1,I_1,I_1$ respectively but when I reach the second step where it should give me these types I do not know how to change coordinates so that $\pi$ divides $a_3,a_4,a_6$. Upon doing this, we should expect that $\pi$ does not divide $b_2$ and then the Kodaira type should be $I_n$ where $n$ is the valuation of the selected prime. Is there a systematic way to find a change of coordinates that does this but does not simply scale all the coefficients by nonzero powers of $\pi$ as this does not yield any result from Tate's algorithm.

Answer 1

I'll apply Tate's algorithm at the prime $\pi = T-1$ and leave the rest to you. The relevant formulas can be found in $\S9$ of Silverman's Advanced Topics in the Arithmetic of Elliptic Curves.

We compute that the discriminant of $E$ is $$ \Delta = 2^{18} 3^{4} (T-1)^2 (T+1)^2 (27T-23) (3T+1) (9T+7)^2 $$ so $\pi \mid \Delta$. We reduce mod $T-1$ by setting $T=1$, obtaining the curve $$ \widetilde{E}: y^2 - 20xy - 360 y = x^3 + 18 x^2 + 324 x + 5832 \, . $$ Computing the vanishing of partial derivatives, we find that this curve has a singular point at $(-18, 0)$. To translate this point to the origin, we replace $x \leftarrow x-18$ and substitute this into the equation for $E$, which yields the curve given by $$ y^2 - (14T + 6) xy = x^3 + (32 T^2 - 68) x^2 - 1152 (T^2 - 1)x \, . $$ (In more detail, we let $x' = x + 18$ so that $(x,y) = (-18,0)$ corresponds to $(x', y) = (0,0)$. Then we substitute $x = x' - 18$ into the equation for $E$ and expand in terms of $x'$.) To ensure we did the substitution correctly, we verify that $T-1$ divides $a_3 = a_6 = 0$ and $a_4 = - 1152 (T^2 - 1)$. Since $$ b_2 = a_1^2 + 4 a_2 = (- (14T + 6))^2 + 4 (32 T^2 - 68) $$ which doesn't vanish when $T = 1$, then $T-1 = \pi \nmid b_2$. Since $v_{\pi}(\Delta) = 2$, we conclude that the Kodaira type is $I_2$.