Do rational points on $X(1)$ correspond to elliptic curves up to rational isomorphism?

Question

I'm not sure whether this is obvious or not. The curve $X(1)$ parametrices all elliptic curves up to isomorphism class over $\mathbb{ C}$, and a $K$-rational point corresponds to an elliptic curve whose $j$-invariant lies in $K$, but if I remember correctly the $K$-rational points actually classify elliptic curves up to isomorphism over $K$. Is there an obvious reason for this?

Answer 1

The non-obvious fact you may have heard is that points of $X(1)$ parameterize elliptic curves over $K$ up to $\overline{K}$-isomorphism.

It's clear from the fact that points of $X(1)$ over an algebraically closed field parameterize elliptic curves up to isomorphism that two elliptic curves over $K$ correspond to the same point if and only if they are isomorphic over $\overline{K}$, but a priori there could be $K$-points that do not correspond to any elliptic curves over $K$.

However, one can check that every $K$-point corresponds to an elliptic curve. I think the best way to do this is to simply write down an explicit formula for an elliptic curve with a given $j$-invariant.

Answer 2

That's not actually true in general. If $K$ is the field of rational numbers then any elliptic curve $E/K$ has infinitely many "quadratic twists" that have the same $j$-invariant but are not isomorphic with $E$ over $K$. Namely, if $E$ is the curve $y^2 = P(x)$ for some cubic polynomial $P$ then the quadratic twist $E_D : D y^2 = P(x)$ has the same $j$-invariant as $E$ but $E_D$ and $E_{D'}$ are not isomorphic unless $D'/D$ is a square in $K$. It works the same way for any field not of characteristic $2$ (though some fields like ${\bf C}$ have the property that every element is a square and in that case all $E_D$ are isomorphic).

Answer 3

In general suppose I have an object $E$ living over a field $K$ (this could be a $K$-algebra of some kind or a scheme over $K$) such that I can talk about its extension of scalars $E_L$ along a field extension $K \to L$. If $E, E'$ are two objects over $K$ that become isomorphic after an extension, they are said to be forms of each other. When $L/K$ is a Galois extension with Galois group $G$, and under some mild conditions on what is meant by "object," forms of $E$ are classified by a Galois cohomology set (not group, sadly)

$$H^1(G, \text{Aut}(E_L))$$

and in particular, objects generally have multiple forms.

Specializing now to the case of elliptic curves, the $j$-invariant is invariant under extension of scalars, in the sense that if an elliptic curve $E$ over a field $K$ has $j$-invariant $j(E) \in K$ and we then extend scalars to $E_L$ via a map $f : K \to L$, then $j(E_L) = f(j(E))$. So two elliptic curves over $K$ with the same $j$-invariant continue to have the same $j$-invariant after arbitrary extension of scalars; and general theory says this implies, hence is equivalent to, the statement that they are isomorphic over $\overline{K}$.

However, they need not be isomorphic over $K$, so could be nontrivial forms of each other. Generically (let me stick to characteristic $0$) an elliptic curve has automorphism group $\text{Aut}(E_L) \cong C_2$ given in terms of the group law by $z \mapsto -z$, and the Galois group acts trivially on this. So in this nice special case, forms are classified by

$$\text{Hom}(G, C_2)$$

and by the Galois correspondence, nontrivial homomorphisms $G \to C_2$ correspond to quadratic subextensions of $L/K$; the corresponding forms of $E$ are the quadratic twists that Noam Elkies described.

In two exceptional cases $j = 0, 1728$ there are extra automorphisms: these elliptic curves have larger automorphism groups $C_6, C_4$ corresponding to sixth resp. fourth roots of unity, and now the Galois action is nontrivial and given by the action of $G$ on roots of unity. So the Galois cohomology in this case is

$$H^1(G, \mu_6), H^1(G, \mu_4)$$

which, if we now take $L$ to be the algebraic closure, can be identified with $K^{\times}/(K^{\times})^6, K^{\times}/(K^{\times})^4$ by Kummer theory (similarly in the previous case we have $H^1(G, \mu_2) \cong K^{\times}/(K^{\times})^2$). So we get sextic and quartic twists, as Will Sawin mentioned. This can all be made more explicit in coordinates.