Two Weierstrass models $E$, $E'$ over a field $K$ with Weierstrass coefficients $[a_1,a_2,a_3,a_4,a_6]$ and $[a'_1,a'_2,a'_3,a'_4,a'_6]$ are **isomorphic over $K$** if there exist $u\in K^*$ and $r,s,t\in K$ such that
$$
\begin{aligned} u a_1' &= a_1+2s, \\
u^2a_2' &= a_2-sa_1+3r-s^2, \\
u^3a_3' &= a_3+ra_1+2t, \\
u^4a_4' &= a_4-sa_3+2ra_2-(t+rs)a_1+3r^2-2st, \\
u^6a_6' &= a_6+ra_4+r^2a_2+r^3-ta_3-t^2-rta_1. \\
\end{aligned}
$$
The set of transformations with parameters $[u,r,s,t]\in K^*\times K^3$ form the group of **Weierstrass isomorphisms**, which acts on both the set of all Weierstrass models over $K$ and also on the subset of smooth models, preserving the point at infinity. The discriminants $\Delta$, $\Delta'$ of the two models are related by
$$
u^{12}\Delta' = \Delta.
$$

In the smooth case such a Weierstrass isomorphism $[u,r,s,t]$ induces an isomorphism between the two elliptic curves $E$, $E'$ they define. In terms of affine coordinates this is given by $$ (x,y) \mapsto (x',y') $$ where $$ \begin{aligned} x &= u^2x' + r \\ y &= u^3y' + su^2x' + t \\ \end{aligned} $$

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- Review status: reviewed
- Last edited by David Roe on 2018-12-13 14:17:44

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- 2018-12-13 14:17:44 by David Roe (Reviewed)