Isomorphic elliptic curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:23:08Z http://mathoverflow.net/feeds/question/15733 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15733/isomorphic-elliptic-curves Isomorphic elliptic curves Kayamba 2010-02-18T17:49:28Z 2010-02-20T20:47:57Z <p>If we have an elliptic curve E over a field k and we pick a non-square d in k-{0}. Suppose E is isomorphic to E^(d). (E^(d) is the quadratic twist) Why must j(E) = 1728 and why is k(sqrt(d)) = k(sqrt(-1))?</p> http://mathoverflow.net/questions/15733/isomorphic-elliptic-curves/15738#15738 Answer by Victor Ostrik for Isomorphic elliptic curves Victor Ostrik 2010-02-18T18:26:31Z 2010-02-18T18:26:31Z <p>Various forms of $E$ over $k$ are classified by $H^1(k,Aut(E))$ and the quadratic twist corresponds to the image of the class $d\in k^\times/(k^\times)^2=H^1(k,{\mathbb Z}/2)$ under the map $H^1(k,{\mathbb Z}/2)\to H^1(k,Aut(E))$ induced by the map ${\mathbb Z}/2\to Aut(E)$ which sends nontrivial element of ${\mathbb Z}/2$ to the automorphism $x\mapsto -x$ of $E$. You are interested in the case when the image of $d$ under this map is zero. At the very least this means that $Aut(E)$ is strictly greater than ${\mathbb Z}/2$; but there are very few elliptic curves with this property..</p> http://mathoverflow.net/questions/15733/isomorphic-elliptic-curves/15742#15742 Answer by Victor Miller for Isomorphic elliptic curves Victor Miller 2010-02-18T19:04:40Z 2010-02-18T19:27:18Z <p>To make things concrete, suppose that we're working in characteristic $\ne 2,3$ (we can do something similar in those cases, though it gets messier), and the equation of the curve is</p> <p>$E : y^2 = x^3 + a x + b$</p> <p>and the equation of the twist is:</p> <p>$E^{(d)} : d y^2 = x^3 + a x + b$.</p> <p>Multiplying this by $d^3$:</p> <p>$(d^2 y)^2 = (d x)^3 + a d^2 (d x) + d^3 b$</p> <p>for this to be isomorphic to $E$ over $k$ it is necessary and sufficient that there is a $\lambda \in k^*$ such that (see Silverman, or any other book on elliptic curves) such that $a d^2 = a \lambda^4$, $b d^3 = b \lambda^6$. If $a,b \ne 0$ then we must have $\lambda^6 = d^3$ and $\lambda^4 = d^2$, which shows that $\lambda^2 = d$ which you've ruled out. If $b=0$ you have, by taking square roots, $\lambda^2 = \pm d$, and only the minus sign is possible. If $a=0$ (and thus $b \ne 0$) then $\lambda^6 = d^3$. But then $(\lambda^3/d)^2 = d^3/d^2 = d$, which you've also ruled out. Thus, only $b=0$ is possible, which is $j=1728$.</p> http://mathoverflow.net/questions/15733/isomorphic-elliptic-curves/15930#15930 Answer by kayamba for Isomorphic elliptic curves kayamba 2010-02-20T20:47:57Z 2010-02-20T20:47:57Z <p>That's correct Mr. Choi. It's a paper I am reading did not quite understand everything so I posted the questions. Thanks for the responses they were very helpful.</p>