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

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 $$\mathbb{Q}$$)

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.

• Does all of this work over fields that are not perfect? Jun 6, 2021 at 7:31
• @Guenterino For characteristic not equal to 2 or 3, the first part of my answer should work verbatim. For non-perfect fields, the second part of my answer clearly would need to be modified. It should work to give a description of "separable twists". But I'm not sure offhand what happens for "inseparable twists" (if that's even the right term for it) in characteristics 2 and 3 if $K$ is not perfect. I don't work with that situation myself, but I expect there are people who know the answer. You could try asking as a new question here, and if you don't get an answer, then try MathOverflow. Jun 6, 2021 at 10:49
• Thank you very much for your answer. I have created a new thread (math.stackexchange.com/questions/4165089/…) as you suggested. Jun 6, 2021 at 16:14

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.

• So nothing simpler than L-function is needed for general cases. But for some special cases like yours $E$ and $E_d$ there may be a simpler method of determining when they are $\Bbb{Q}$-isomorphic. Jan 12, 2020 at 12:37