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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?

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    $\begingroup$ In your situation where L = K, P(d) and Q(d) are not defined over K (unless d is a square). See the answers to mathoverflow.net/questions/76413/… $\endgroup$ – Ari Shnidman Oct 12 '15 at 21:20
  • $\begingroup$ The "stacky version" of $Y(3)$ 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 have nothing to 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 $\mathbf{A}^1$). $\endgroup$ – nfdc23 Oct 13 '15 at 0:19
  • $\begingroup$ @nfdc23 Are you claiming that the fiber of $Y(3)$ over $y$ is not a scheme of level structures on some elliptic curve with $j$-invariant $y$? If not, then what is it? I mean, I'm sure it has some moduli-theoretic meaning... $\endgroup$ – Will Chen Oct 13 '15 at 3:13
  • $\begingroup$ The reason that it might not be true that $F^d$ has a natural follow-your-nose moduli-theoretic interpretation is that it depends on a point in $Y(1)$ and $Y(1)$ is not representing iso classes of ell curves, so $F^d$ inherits some issues. On the other hand it's most certainly true that all the $K$-points $(E,P,Q)$ in the fiber have isomorphic $E$'s: all you're doing here is letting $GL(2,3)$ act on $P$ and $Q$ and leaving $E$ alone. The scheme $Y(3)$ and the action of $GL(2,3)$ are fine, it's the quotient that has problems. I no longer know what, if anything, you are confused about. $\endgroup$ – eric Oct 13 '15 at 14:11
  • $\begingroup$ PS I might also comment that the idea that an arbitrary elliptic curve over a field $K$ has a "special" twist which only depends on the isomorphism class of the curve over an algebraic closure of $K$ does not sit well with me at all. What is true is that once the ground field is big enough and you choose two points of order 3 defined over the ground field there 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 $K$. $\endgroup$ – eric Oct 13 '15 at 14:16
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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.)

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  • $\begingroup$ Thanks for your answer, though I believe $Y(3)$ does distinguish $K$-isomorphism classes, since it is a fine moduli scheme (this is why I chose $Y(3)$ and not, say, $Y(2)$)... $\endgroup$ – Will Chen Oct 12 '15 at 22:47
  • $\begingroup$ @oxeimon You might be right. But I thought points of $Y(N)$ classified equivalence classes of triples $(E,P,Q)$ with $P$ and $Q$ generating $E[N]$, where $(E,P,Q)\sim(E',P',Q')$ if there is an isomorphism $f:E\to E'$ defined over $\overline K$ such that $f(P)=P'$ and $f(Q)=Q'$. Being a fine moduli space doesn't affect the fact that the equivalence relation is via isomorphisms over an algebraically closed field. (Also, $Y(N)$ actually classifies $(E,P,C)$, where $P$ has order $N$ and $C$ is cyclic of order $N$ with $C+\mathbb Z P=E[N]$. The moduli space for triples $(E,P,Q)$ is disconnected.) $\endgroup$ – Joe Silverman Oct 13 '15 at 2:49
  • $\begingroup$ Dear Professor Silverman, I'm actually quite sure that $Y(N)$ parametrizes $(E/K,P,Q)$ with isomorphisms defined over $K$ (not $\overline{K}$). This is because the stack of elliptic curves equipped with a full $N$-structure is representable, which means that its S-valued points correspond to triples $(E/S,P,Q)$, and isomorphisms are isomorphisms of schemes - a nontrivial twist of E is not isomorphic to E as a scheme. Also surely $(P,Q)$ is not equivalent to $(P,C)$? I believe the moduli space for full N-structures over $Q(\zeta_N)$ has $\phi(N)$ components, each isomorphic to the $Y(N)$.. $\endgroup$ – Will Chen Oct 13 '15 at 3:05
  • $\begingroup$ ...and each corresponding to a possible Weil pairing $e_N(P,Q)$. Thus, if you fix a primitive $N$-th root of unity $\zeta_N$, then I believe $Y(N)$ classifies triples $(E,P,Q)$ such that $e_N(P,Q) = \zeta_N$. (This is what I meant by symplectic basis - I hope I'm using "symplectic basis" correctly here) $\endgroup$ – Will Chen Oct 13 '15 at 3:08
  • $\begingroup$ In this situation an isomorphism over the algebraic closure implies an isomorphism defined over the ground field, because the obstruction to descending the isomorphism lies in a Galois cohomology group associated to the automorphisms of $(E,P,Q)$, which are zero, so the cohomology group is also zero. Joe Silverman is absolutely right that in general one has to worry about isomorphisms over the alg closure for a general moduli space, especially $Y_0(N)$, but oxeimon is right that for full level 3 it's not an issue. $\endgroup$ – eric Oct 13 '15 at 12:36
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$\newcommand{\QQ}{\mathbb{Q}}$ $\newcommand{\ZZ}{\mathbb{Z}}$

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 $F^d = K(\sqrt{d})$.

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