Twist of Elliptic curve $E_D: Dy^2=x^3+ax+b$ is not isomorphic to $E:y^2=x^3+ax+b$?

Question

Let $E/K$ be an elliptic curve over a field $K$.

Let $E/K: y^2=x^3+ax+b$ be an elliptic curve.

Let $E_D: Dy^2=x^3+ax+b$ is called a quadratic twist of $E/K$ by a square free integer $D$.

This curve is isomorphic over $K(\sqrt{D})$ because there is an isomorphism $E(K(\sqrt{D})\cong E_D(K(\sqrt{D}))$ given by $(x,y)\to (x,y/\sqrt{D})$.

My question is, are $E$ and $E_D$ not isomorphic over $K$? How can I prove that?

Thank you in advance.

Answer 1

In general they are not isomorphic. Here is an argument I think should work.

Suppose there were an isomorphism $\phi:E\to E_D$ defined over $K$. Now take your isomorphism over $K(\sqrt{D})$ and compose to get an automorphism $E\to E$ defined over $K(\sqrt{D})$. By general theory, you can pick $E$ such that its only automorphisms are the identity and the inverse homomorphism $[-1]$ (say by making sure the $j$-invariant is not equal to $0$ or $1728$ and using Theorem III.10.1 of Silverman’s AEC). Then by precomposing the isomorphism $E\to E_D$ by $[-1]$ if you have to, you know that $E\to E_D$ is an isomorphism over $K$ which is the identity. This is impossible unless $D$ were $1$.