Show that $j(E^{(d)}) = j(E)$

Question

Let $E$ be given by $y^2 = x^3 + Ax + B$ over a field $K$ and let $d \in K^\times$. The twist of $E$ by $d$ is the elliptic curve $E^{(d)}$ given by $y^2 = x^3 + Ad^2x + Bd^3$.

I know $j(E) = 1728 \frac{4A^3}{4A^3 + 27B^2}$ and that a twist is when two different elliptic curves of the same field have the same j-invariant.

The book is super vague, but it mentions some kind of change of base where $A_1 = \mu^4A$ and $B_1 = \mu^6B$, but I don't understand it or how to apply it.

I'm completely lost, any help would be amazing.

@MarianoSuárez-Álvarez yes, I end up with $A^3/(4A^3 + 27B^2) = A^3d^6 / 4A^3d^6 + 27B^3d^6)$ — XRBtoTheMOON, Mar 15 '18 at 05:35
@MarianoSuárez-Álvarez I'd say no. I know I'm wrong tho, so I don't understand how they are equal. — XRBtoTheMOON, Mar 15 '18 at 05:41
@MarianoSuárez-Álvarez ugh, I see it now. Im going to delete my account sigh — XRBtoTheMOON, Mar 15 '18 at 05:45

score 1 · Accepted Answer · answered Mar 15 '18 at 05:15

1

You will get the result by applying the formula for $j(E)$ for the new coefficients $A' := Ad^2$ and $B' := Bd^3$.

answered Mar 15 '18 at 05:15

Bernd

115

I guess this is what I dont understand, because then arent tho only equal in that one specific case? Wouldnt any two elliptic curves be twists then if all I have to do is set $A_1$ to whatever is in front of $x$? I realize these are stupid questions, but im struggling pretty bad with this – XRBtoTheMOON Mar 15 '18 at 05:32
Asyou have calculated in your comment, you can cancel the factor d^6. Just try a different example e.g. A' = d and B' = d, calculate the two j-invariants and compare the results. – Bernd Mar 15 '18 at 05:41
@Bend omfg, I have to be the dumbest person ever. So embarrassing. – XRBtoTheMOON Mar 15 '18 at 05:44

Show that $j(E^{(d)}) = j(E)$

1 Answers1