J-invariant and isomorphism of elliptic curves over $\mathbb{Q}$

Question

If two elliptic curves share the same j-invariant then they may not be isomorphic to each other over $\mathbb{Q}$.

Example:

$E_1: y^2 = x^3 + x$

j-inavriat: $1728$

Torsion points: $[(0 : 0 : 1), (0 : 1 : 0)]$

Rank $0$.

$ $

$E_2: y^2 = x^3 + 3 x$

j-inavriat: $1728$

Torsion points: $[(0 : 0 : 1), (0 : 1 : 0)]$

Rank $1$ - generator point $[(1 : 2 : 1)]$

Is there some other invariant or can we define a new type of invariant that if two elliptic curves share the same such invariant then they are isomorphic over $\mathbb{Q}$?

(they can be birationally transformed to each other over $\mathbb{Q}$)

Answer 1

Since Silverman answered consider this as a comment, showing that it is a good thing to experiment with quadratic twists

• Given an elliptic curve $E/\Bbb{Q}$ we get an homomorphism $$\rho_E: Gal(\overline{\Bbb{Q}}/\Bbb{Q})\to Aut(E_{tors})$$

  If $E$ is isomorphic to $E'$ over $\Bbb{Q}$ then $\rho_{E'}= f \circ \rho_E \circ f^{-1}$

• With $E:y^2=x^3+ax+b$, $E_d : dy^2=x^3+ax+b$, $f(x,y)=(x,y/\sqrt{d})$ we get $$\rho_{E_d}(\sigma)= [\chi_d(\sigma)]\circ f \circ \rho_E(\sigma) \circ f^{-1}$$ where $\chi_d(\sigma) = \frac{\sigma(\sqrt{d})}{\sqrt{d}}=\pm 1$ and $[-1](x,y)=(x,-y)$ commutes with $f,\rho_E$.

Thus $E\cong E_d$ over $\Bbb{Q}$ iff $d\in (\Bbb{Q}^*)^2$

And the so called quadratic twists $E_d,d\in \Bbb{Q}^*/(\Bbb{Q}^*)^2$ are infinitely many pairwise distinct $\Bbb{Q}$-isomorphism classes of elliptic curves, they become isomorphic only over $\Bbb{Q}( \{ \sqrt{p}\})$.

This suggests that (in most cases...) a sufficient data determining the $\Bbb{Q}$-isomorphism class is the $j$-invariant plus the Galois module or the L-function.

Answer 2

You really don't need to use $L$-functions or representations. Let's work over a field $K$ of characteristic not equal to 2 or 3, so for example $K=\mathbb Q$. Then an elliptic curve $E/K$ always has a Weierstrass model $$ E:y^2=x^3+Ax+B, $$ but the model is not unique. The $j$-invariant $$ j(E) = 1728\cdot\frac{4A^3}{4A^3+27B^2} $$ classifies $E$ up to $\overline K$ isomorphism. You're interested in $K$-isomorphism. Assuming that $j(E)\ne0$ and $j(E)\ne1728$ (i.e., assume that $AB\ne0$), define a new invariant $$ \gamma(E) = B/A \bmod{{K^*}^2} \in K^*/{K^*}^2. $$ One can check that $\gamma(E)$ is well-defined modulo squares in $K$. Then $$ \text{$E\cong E'$ over $K$} \quad\Longleftrightarrow\quad \text{$j(E)=j(E')$ and $\gamma(E)=\gamma(E')$.} $$ If $j(E)=0$, then $A=0$ and there's a similar criterion in terms of $B$ modulo ${K^*}^6$, and if $j(E)=1728$, then $B=0$ and there's a criterion in terms of $A$ modulo ${K^*}^4$.

However, probably the right way to understand this is to use the fact that for a given $E/K$, the collection of $E'/K$ that are $\overline{K}$-isomorphic to $E$ are classified by the cohomology group $$ H^1\bigl(\operatorname{Gal}(\overline K/K),\operatorname{Aut}(E)\bigr). $$ The three cases correspond to $\operatorname{Aut}(E)$ being $\mu_2$, $\mu_6$, and $\mu_4$, respectively, and one knows (Hilbert Theorem 90) that $$ H^1\bigl(\operatorname{Gal}(\overline K/K),\mu_n\bigr)\cong K^*/{K^*}^n. $$ This unifies the three cases, and gives a quite general way to describe the $\overline{K}/K$-twists of an algebraic variety.