confounding riddle about fine moduli schemes and twists of elliptic curves

Question

I've encountered a strange situation while thinking about modular curves... Consider the modular curve $Y(3)$ parametrizing elliptic curves with a symplectic basis for their 3-torsion. This curve has degree 12 over the $j$-line $Y(1)$. Let $y\in Y(1)$ be a $\mathbb{Q}$-rational point, then its fiber in $Y(3)$ has degree 12, and hence there exists a number field $K$ of degree dividing 12 over $\mathbb{Q}$ such that over $K$, the fiber above $y$ in $Y(3)_K$ is completely decomposed.

Now for any elliptic curve $E$ over $\mathbb{Q}$ with $j$-invariant $y$, $E[3]$ is defined over some number field $L$. Thus, over $L$, by picking a suitable basis for $E_L[3]$ we get an $L$-point of $Y(3)$ over $y$. Thus, by our construction of $K$ we may as well have taken $L = K$.

My question is this: Fix an elliptic curve $E$ over $K$ with $j$-invariant y, then there are infinitely many nonisomorphic twists $E^d$, parametrized by $d$ in $K^*/(K^*)^2$. Now fix a set of representatives $D_K$ for $K^*/(K^*)^2$, and consider the set:

$$\{E^d : d\in D_K\}$$

Now each of these twists $E^d$ is an elliptic curve over $K$ with $j$-invariant $y$, so by the above discussion, for each such twist $E^d$ with $d\in D_K$ we may choose a suitable basis $(P(d),Q(d))$ for $E^d[3]$ so that $(E^d/K,P(d),Q(d))$ corresponds to a $K$-point of $Y(3)$ above $y$. However, there are only 12 $K$-points of $Y(3)$ above $y$, but infinitely many nonisomorphic triples: $$\{(E^d/K,P(d),Q(d)) : d \in D_K\}$$

Where have I gone wrong?

$\newcommand{\QQ}{\mathbb{Q}}$

EDIT: The above question was mostly answered by Ari in his comment below, but I feel like it doesn't resolve my confusion. Here's another way of articulating my confusion:

Let $\mathcal{Y}(3)$ be the stacky version of $Y(3)$. Fix an elliptic curve $E$ over $\QQ$, corresponding to a $\QQ$-point of $\mathcal{M}_{1,1}$. The fiber of $\mathcal{Y}(3)$ over $E/\QQ$ is a representable stack, finite etale over $\QQ$, whose corresponding scheme is a degree 12 etale $\QQ$-algebra $F$. For any other nonisomorphic twist $E^d$ over $\QQ$, we may play the same game, and we find that the fiber of $\mathcal{Y}(3)$ over $E^d/\QQ$ is also a degree 12 etale $\QQ$-algebra which I will call $F^d$. Now, passing to coarse moduli schemes, the $F$'s and $F^d$'s are all somehow related to the etale algebra corresponding to the fiber $Y(3)_y$. In fact, by suitably picking a $y$, by Hilbert's irreducibility theorem we may assume that $Y(3)_y$ is connected (ie a field), so lets call it $K$. What is the relation between $F^d$ and $K$?

If there is no relation, then what is $K$? Surely it must have some moduli-theoretic meaning?

Okay, I'm beginning to suspect that there is a particular special twist of $E/\QQ$ such that $K$ is just $F^d$, where Spec $F^d$ is the scheme of $\Gamma(3)$-structures on $E^d$. This special $E^d$ might just be the twist representing the $K$-isomorphism class of the fiber of the universal elliptic curve $\mathcal{E}(3)_K/Y(3)_K$ at a $K$-point $x\in Y(3)$ lying above $y$. If this were true, it would seem to "imply" that all fibers of $\mathcal{E}(3)_K$ over points $x\in Y(3)$ lying above $y$ must be $K$-isomorphic?

Answer 1

I think there are two issues. One, as Ari noted, your triple $(E^d/K,P(d),Q(d))$ is not really defined over $K$, since the action of Galois is twisted by the character of $K(\sqrt d)/K$. But more importantly, you're not really getting different points on $Y(3)$, because the points of $Y(3)$ classify up to $\overline K$-isomorphism, not up to $K$-isomorphism. You can see this already on $Y(1)$, where it's clearer. A point $y\in Y(1)$ classifies the $\overline K$-isomorphism class of elliptic curves $E/\overline K$ with $j(E)=y$. All of your curves $E^d$ are in that same isomorphism class, so they give only one point of $Y(1)$. Now it happens that if $y\in Y(1)(K)$ for some field $K$, then there is an elliptic curve $E$ in the $y$-isomorphism class such that $E$ is defined over $K$. In fancier language, this is because for elliptic curves, the field of moduli is a field of definition. This is no longer true in higher dimension; there may be $K$-rational points $y\in\mathcal A_2(K)$, the moduli space of principally polarized abelian surfaces, such that the isomorphism class $y$ doesn't contain an abelian surface defined over $K$. What is true is that if $A\in y$, then every Galois conjugate $A^\sigma$ is $\overline K$-isomorphic to $A$. (As you can probably guess, there's some Galois cohomology underlying this phenomenon.)

Answer 2

$\newcommand{\QQ}{\mathbb{Q}}$ $\newcommand{\ZZ}{\mathbb{Z}}$

Okay, so the solution appears to be this (Thanks to Ari Shnidman, Joseph Silverman, nfdc23, and eric for their comments)

Fix an $N\ge 3$. Let $y\in Y(1)$ be a $\QQ$-point, then the fiber $Y(N)_y$ of $Y(N)$ above $y$ is a $\QQ$-algebra $A$ of degree $d_N := |PSL(2,\ZZ/N)|$. Since $Y(N)/Y(1)$ is galois, we find that $A = \prod_{i=1}^{k_y} K[x]/(x^e)$ is a product of $k_y$-many connected $\QQ$-algebras, where $e = 1$ if $y\ne 0, 1728$, otherwise $e = 3,2$ if $y = 0,1728$. Here we have $k_y\cdot e\cdot[K:\QQ] = d_N$. Note that by the Weil Pairing $K$ must contain a primitive $N$th root of unity $\zeta_N$.

Thus, we find that for any such $y$, there exists a field $K$ of degree over $\QQ$ dividing $d_N$ such that there exists a $K$-point on $Y(3)$ lying above $y$. This means, that there exists an elliptic curve $E/K$ with a pair of points $P,Q\in E(K)[N]$ which generate $E[N]$ and such that $e_N(P,Q) = \zeta_N$. In particular, $E[N]$ is defined over $K$. Note that given one such pair $(P,Q)$ there exist $SL_2(\ZZ/N)$-many such pairs with Weil pairing $\zeta_N$, whose equivalence classes modulo $Aut(E)$ occupy all of the points in the fiber $Y(3)_y$ (ie, over $y\ne 0,1728$ there are $d_N$ such equivalence classes. Otherwise there are $\frac{d_N}{3}$ or $\frac{d_N}{2}$ if $y = 0$ or $1728$).

The fact that all the points of $Y(3)_y$ is taken up by this single curve $E$ means that there is (up to $K$-isomorphism), precisely one elliptic curve $E$ with $E[N]$ defined over $K$ (this is the special twist I was referring to in the OP). This elliptic curve is precisely the fiber of the universal elliptic curve $\mathcal{E}(N)$ over $Y(N)$ at the $K$-point, which indeed implies that all fibers of $\mathcal{E}(N)$ with a particular $j$-invariant are isomorphic.

Lastly, in the OP I had said that the algebras $F^d$ are degree $d_N$ over $\QQ$. This is incorrect. They are actually degree $2d_N$ (the map from $\mathcal{M}_{1,1}$ to $Y(1)$ is degree "one half").

If we do a little relabeling and say that the scheme of full level-$N$ structures on $E^d/\QQ$ is a galois $\QQ$-algebra $B^d$ of degree $2d_N$, then $B = \prod_{i=1}^{k_y} F^d[x]/(x^e)$, then again we find that since there must be a unique elliptic curve $E'$ over $F^d$ (up to $F^d$-isomorphism) whose full level-$N$ structures occupy all the $F^d$ points of $Y(N)_y$, that this elliptic curve must be $E^d$, and thus $F^d = K(\sqrt{d})$.