# Example of an elliptic curve which is not defined over $\Bbb{Q}$

An elliptic curve $$E$$ is defined over $$\Bbb{Q}$$ if only if it is isomorphic over $$\overline{\Bbb{Q}}$$ to an elliptic curve whose coefficients are in $$\Bbb{Q}$$.

My question is, what are examples of an elliptic curve which is not defined over $$\Bbb{Q}$$ ?

Elliptic curve with coefficients not in $$\Bbb{Q}$$ is not enough, because it might be isomorphic to an elliptic curve with coefficients in $$\Bbb{Q}$$.

• A smart-ass example could be $y^2=x^3+\pi$ defined over $\Bbb{Q}(\pi)$. Its function field $\Bbb{Q}(\pi,x,y)$ has transcendence degree two over $\Bbb{Q}$, so it cannot be isomorphic to a curve defined over $\Bbb{Q}$. Presumably you want one defined over a finite extension of $\Bbb{Q}$. Commented Jan 28 at 9:36
• @Jyrki Lahtonen. Yes, for example, over $\Bbb{Q}(\sqrt{-1})$.　This question is related to a question. I posted today. math.stackexchange.com/questions/4852519/… Commented Jan 28 at 10:52
• Choose any elliptic curve whose $j$-invariant is not rational. For example, see this list from LMFDB of curves over $\mathbb Q(i)$. Commented Jan 28 at 16:27
• @Mathmo123 that looks like an answer to me - please record it as such below. Commented Jan 28 at 18:41
• Related Commented Mar 12 at 2:58

An elliptic curve $$E$$ over $$\overline{\mathbb{Q}}$$ can be defined over $$\mathbb{Q}$$ if and only if its j-invariant $$j(E)$$ is a rational number. More precisely, $$E$$ can be defined over a number field $$K\subset \overline{\mathbb{Q}}$$ if and only if $$j(E)\in K$$.

As Mathmo123 points out above, you can find many elliptic curves over $$\mathbb{Q}(i)$$ which can not be defined over $$\mathbb{Q}$$:

https://www.lmfdb.org/EllipticCurve/2.0.4.1/?field=2.0.4.1&Qcurves=non-Q-curve

But, it is actually quite easy to find examples yourself. Consider the elliptic curve $$y^2 = x^3 + Ax + 1$$. For which $$A$$ is the $$j$$-invariant a rational number?

Additional remark: If $$\lambda\in \overline{\mathbb{Q}}\setminus \{0,1]\}$$, then the Legendre elliptic curve $$E_{\lambda}$$ is defind by $$y^2 = x(x-1)(x-\lambda)$$. If $$\lambda$$ lies in $$K$$, then it is obviously defined over $$K$$.

The $$j$$-invariant of $$E_{\lambda}$$ is given by
$$j(E_{\lambda}) = \frac{256(1-\lambda(1-\lambda))^3}{(\lambda(1-\lambda))^2} = \frac{256(1-\lambda+\lambda^2)^3}{\lambda^2 (1-\lambda)^2}.$$ So, $$E_{\lambda}$$ is defined over $$\mathbb{Q}$$ if and only if there is a rational number $$q$$ such that $$q\lambda^2(1-\lambda)^2 - 256(1-\lambda+\lambda^2)^3 =0.$$ So, choosing $$\lambda$$ (for example) to be an eleventh root of $$2$$ obviously gives a non-rational $$j$$-invariant. (Because this $$\lambda$$ does not satisfy a polynomial relation over $$\mathbb{Q}$$ of degree smaller than $$11$$.) So, this is a relatively easy way of writing down elliptic curves which can not be defined over the rationals.

• Indeed, if $j(E) \notin \mathbb{Q}$, then $E/\mathbb{Q}$ is not isomorphic to any curve with a rational $j$-invariant. However, every elliptic curve defined over $\mathbb{Q}$ possesses a rational $j$-invariant because it is isomorphic over $\overline{\mathbb{Q}}$ to a curve whose coefficients are expressed as a rational function of $j$ (There is an explicit constructuction of an elliptic curve with the $j$-invariant $j$). Commented Mar 12 at 13:04
• Concerning $y^2 = x(x - 1)(x - \lambda)$, why does having $j = 2^8\frac{(\lambda^2 - \lambda + 1)^3}{\lambda^2(\lambda - 1)^2}$ as a rational number ensure that $\lambda$ is rational? $λ=\frac{1+\sqrt{-3}}{2}$ is not rational, but the j-invariant is a rational number $0$. Commented Mar 12 at 13:35
• @Pont thanks for the comments. I agree with your first comment. This is what I wrote in the first paragraph. (You meant to write $E/\overline{\mathbb{Q}}$ in your first sentence probably.) Will get back to you on $\lambda$-invariants as soon as I can. Commented Mar 13 at 9:05
• You're right about the first point you mentioned. Thank you. Also, the reason I added my initial comment was to address the rationale behind the first line of your answer. While it might be obvious to you, the explanation for that fact wasn't included. I believe the essential reason is that the coefficients of an elliptical curve with a given j-invariant can indeed be expressed as a rational function of the j-invariant. My comment was meant to mention this explanation, which you hadn't included. So, it's slightly different from merely reiterating what you were saying. Commented Mar 14 at 2:19
• @Pont "I believe the essential reason is that the coefficients of an elliptical curve with a given j-invariant can indeed be expressed as a rational function of the j-invariant. " That's correct. This is also explained in Silverman's book on elliptic curves (the first one). Will get back to you on lambda-invariants later hopefully. Commented Mar 14 at 6:02