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Let $Y_1(n)$ (for $n\ge 4$) be the fine moduli scheme over $\mathbb{Q}$ parametrizing elliptic curves with a rational point of order $n$. Let $\mathbb{A}^1_j$ be the $j$-line over $\mathbb{Q}$, the coarse moduli scheme of the moduli stack of elliptic curves $\mathcal{M}_{1,1}$ (over $\mathbb{Q}$). Let $\mathcal{M}_{1,1}^\circ$ be the open substack classifying elliptic curves with $j$-invariant $\notin\{0,1728\}$. Let $Y_1(n)^\circ$ be the open subscheme obtained by removing the fibers above $j = 0,1728$, and let $\mathbb{A}_j^{1\circ}$ be the $j$-line minus $0,1728$.

One may consider the fiber products $$A := Y_1(5)^\circ\times_{\mathcal{M}_{1,1}^\circ}Y_1(7)^\circ,\qquad B := Y_1(5)^\circ\times_{\mathbb{A}_j^{1\circ}}Y_1(7)^\circ$$

Both $A,B$ are schemes. $Y_1(7)^\circ$ is finite etale over $\mathcal{M}_{1,1}^\circ$ of degree $[SL_2(\mathbb{Z}):\Gamma_1(7)] = 48$, in the sense that the fiber category above a geometric point $\text{Spec }k\rightarrow\mathcal{M}_{1,1}^\circ$ is a groupoid with 48 objects, but with isomorphisms between them (48 should be the degree according to the definition of the Galois category of an algebraic stack). Thus, since we're staying away from $j = 0,1728$, the resulting scheme theoretic fiber should have degree 24 over $\text{Spec }k$. For the same reason, $A$ should be finite etale over $Y_1(5)^\circ$ of degree 24.

Similarly, $Y_1(7)^\circ$ is finite etale over $\mathbb{A}_j^{1\circ}$, and so the pullback $B$ is finite etale over $Y_1(5)^\circ$, also of degree 24.

The universal property of fiber products provides a map $A\rightarrow B$ (commuting with the finite etale maps to $Y_1(5)^\circ$). This map must then also be finite etale, and by comparing degrees, must be an isomorphism.

Is this correct? (That $A\cong B$?)

In particular, this would seem to say that if we have an elliptic curve $E_1/K$ with a point $P_1$ of order 5, and an elliptic curve $E_2/K$ with a point $P_2$ of order 7, such that $j(E_1) = j(E_2) \ne 0,1728$, then this pair $((E_1,P_1),(E_2,P_2))$ would correspond to a $K$-point of $B$, and hence a $K$-point of $A$, but by definition of the fiber product of stacks, giving such a $K$-point of $A$ would imply that in fact $E_1\cong E_2$ over $K$ - which says more than just $j(E_1) = j(E_2)$ - ie, $E_1$ is not a twist of $E_2$.

If so this seems quite striking. In particular, this would seem to imply that if an elliptic curve $E/K$ with automorphism group $\{\text{id},[-1]\}$ has a $K$-point of order $n\ge 4$ (so that $Y_1(n)$ is a fine moduli scheme), then none of the twists of $E/K$ can have $K$-points of order $\ge 4$. Ie, amongst the set of all twists of $E$, at most one can admit a $K$-rational point of order $\ge 4$.

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    $\begingroup$ The ramification does not really cancel: for this to happen one must take the normalisation of the fibre product. (The fibre product will never be smooth in such a situation.) $\endgroup$ – ulrich May 21 '17 at 6:23
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    $\begingroup$ Take X and Y to be points and $\mathcal M =BG $ for a dramatic failure of your guess in the last paragraph. $\endgroup$ – Dan Petersen May 21 '17 at 6:27
  • $\begingroup$ @ulrich That's a great point actually. However the issue should go away if one avoids elliptic curves of $j$-invariant 0,1728 right? $\endgroup$ – Will Chen May 21 '17 at 6:30
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I think your degree calculation is bogus. You write:

$Y_1(7)^\circ$ is finite etale over $\mathcal{M}_{1,1}^\circ$ of degree $[SL_2(\mathbb{Z}):\Gamma_1(7)] = 48$, in the sense that the fiber category above a geometric point $\text{Spec }k\rightarrow\mathcal{M}_{1,1}^\circ$ is a groupoid with 48 objects, but with isomorphisms between them [...] The resulting scheme theoretic fiber should have degree 24 over $\text{Spec }k$.

But the map $Y_1(7)^\circ \to \mathcal{M}_{1,1}^\circ$ is representable, and this means in particular that the fiber category over $\text{Spec }k$ (or over any scheme for that matter) is an algebraic space, not a stack. So the only isomorphisms in your 48-object category are identities, and the "scheme theoretic fiber" should have degree 48 over $\text{Spec }k$.

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  • $\begingroup$ ... and we find that $A$ is finite étale over $Y_1(5)^\circ$ of degree 48, not 24. $\endgroup$ – S. Carnahan May 22 '17 at 7:54
  • $\begingroup$ Ahh...okay I see my mistake. I was thinking that the fiber category is fibered in setoids with 48 objects, but 24 isomorphism classes. Now I see where I went wrong. Thanks! $\endgroup$ – Will Chen May 22 '17 at 17:33
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I can't say for sure whether or not $A \cong B$, but it's not true that if $E_{1}/K$ and $E_{2}/K$ are elliptic curves with $j(E_{1}) = j(E_{2}) \not\in \{ 0, 1728 \}$ and $P_{1} \in E_{1}(K)$ and $P_{2} \in E_{2}(K)$ both have order $\geq 4$ then necessarily $E_{1} \cong E_{2}$.

In particular, take $E_{1} : y^{2} + xy = x^{3} - 540x + 777615$. This is the "default" elliptic curve with $j = -1/15$, which forces $E_{1}$ to have a cyclic 4-isogeny defined over $\mathbb{Q}$. Let $E_{2}$ be the quadratic twist of $E_{1}$ by $161$ - then $E_{2}$ has a $\mathbb{Q}$-rational 4-torsion point.

Now, let $K/\mathbb{Q}$ be the degree 24 extension obtained by adjoining to $\mathbb{Q}$ the coordinates of a $5$-torsion point of $E_{1}$. Magma tells me that the torsion subgroup of $E_{1}/K$ is isomorphic to $\mathbb{Z}/10\mathbb{Z}$. In particular, $E_{1}$ and $E_{2}$ are not isomorphic over $K$, but $j(E_{1}) = j(E_{2}) = -1/15$ and $E_{1}(K)$ and $E_{2}(K)$ both contain torsion points of order $\geq 4$.

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  • $\begingroup$ Interesting... in fact according to lmfdb, there are two nonisomorphic elliptic curves over Q with j-invariant -1/15 each admitting a point of order 4, so what I said is definitely wrong. Though...I'm still not sure where the flaw in the argument is. $\endgroup$ – Will Chen May 22 '17 at 5:58

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