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?

is$Y(3)$. For an elliptic curve $E$ over $\mathbf{Q}$ and $y_E: {\rm{Spec}}(\mathbf{Q}) \rightarrow M_{1,1}$ corresponding to $E$, by definition of fiber product of stacks $Y(3)_{y_E}$ is the finite etale $\mathbf{Q}$-scheme of full level-3 structures on $E$ (over variable $\mathbf{Q}$-schemes); these havenothingto do with each other as $E$ varies (even with fixed $j$-invariant) and the argument with Hilbert irreducibility doesn't make sense (and what you suspect based on that is false) since $M_{1,1}$ isn't even a scheme (let alone $\mathbf{A}^1$). $\endgroup$what is it? I mean, I'm sure it hassomemoduli-theoretic meaning... $\endgroup$istrue is thatonce the ground field is big enough and you choose two points of order 3 defined over the ground fieldthere is now a "special" twist, which is just the curve itself. Let me again stress that if $K$ is a field and $(E,P,Q)$ and $(E',P',Q')$ are two $K$-points which are isomorphic over $\overline{K}$ then they're isomorphic over $K$. $\endgroup$5more comments