Changing variables of elliptic curves

Question

$$y^2=x^3+Ax^2+Bx+C.$$ Change $x\longrightarrow x-A/3$, so that the new equation has the form $$y^2=x^3+ax+b.$$ Can you show step by step what operations we do in the first equation while obtaining the last equation?

Why did we use the $x\longrightarrow x-A/3$ exchange ?

In general, if you have a polynomial $x^n+ax^{n-1}+\dots$, the variable change $x\to x-a/n$ will eliminate the $x^{n-1}$ term. — Kenta S, May 27 '21 at 15:56
it is not really much of a theorem, you can simply prove it by expanding. — Kenta S, May 27 '21 at 16:04

Thomas Andrews · Accepted Answer · 2021-05-27T16:07:59.677

0

the sum of the three roots of $p(x)=x^3+Ax^2+Bx+C$ is $-A.$

If $r_1,r_2,r_3$ are the roots of $p(x)$ then $r_1+\frac A3,r_2+\frac A3,r_3+\frac A3$ are the roots of $q(x)=p\left(x-\frac A3\right),$ so the sum of the roots of $q(x)$ is $0.$

Since $q$ is also a monic cubic polynomial (why?,) this means that $q(x)=x^3+ax+b$ for some $a,b.$

edited May 27 '21 at 16:07

answered May 27 '21 at 16:02

Thomas Andrews

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Changing variables of elliptic curves

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