I think it should be a standard procedure to construct such things, can anyone give a reference or give a hint? Can this be done over any base scheme?
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$\begingroup$ Perhaps it would be best if you clarify what you mean by an isotrivial family. I take this to mean (after shrinking the base to get an actual relative elliptic curve, i.e., discarding singular fibers), the j-invariants of all the fibers coincide. $\endgroup$– Pete L. ClarkCommented Jan 21, 2010 at 5:52
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$\begingroup$ yes, that's what I mean. $\endgroup$– naturaCommented Jan 21, 2010 at 5:57
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2 Answers
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Hint: use quadratic twists.
Edit: So as not to drag things out, I hope it's okay if I just give you a standard example. Let
$E_0: y^2 = x^3 + Ax + B$
be your favorite elliptic curve over $\mathbb{Q}$ (i.e., any will do). Consider the
elliptic curve
$E: t y^2 = x^3 + Ax + B$
over the rational function field $\mathbb{Q}(t) = \mathbb{Q}(\mathbb{P}^1)$. Spreading this out as a scheme over $\mathbb{P}^1_{/\mathbb{Q}}$, we see that there are two singular fibers, at $t = 0$ and $t = \infty$. Discarding these we get an elliptic curve over $\mathbb{A}^1 \setminus \{0\}$ which is isotrivial -- the $j$-invariant over every fiber is $j(E_0)$ -- but nontrivial: the isomorphism classes of the fibers are in bijection with $H^1(\mathbb{Q},\mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Q}^{\times}/\mathbb{Q}^{\times 2}$.
It's a good bet that you'll find this example somewhere in the chapter on elliptic surfaces in Silverman's Advanced Topics in the Arithmetic of Elliptic Curves.
Can it be done over any base scheme? Unless I misunderstand, of course not, e.g. not over the spectrum of a field.
answered Jan 21, 2010 at 5:44
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$\begingroup$ Is this right? Consider the family of elliptic curve $y^2=x^3+tx$ defined over Spec(k[t]), every fiber (except for t=0) is an elliptic curve with the same j-invariant 1728. but then how to demonstrate that this family is not a trivial family (i.e. tensor product of Spec(k[t]) and a single elliptic curve)? $\endgroup$– naturaCommented Jan 21, 2010 at 7:03
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$\begingroup$ I think I get what you mean. Those fibers are isomorphic over C but not over Q, right? My example seems to also work out, except that the fibers seem to be more than 2 kinds (over Q)?? Thank you! $\endgroup$– naturaCommented Jan 21, 2010 at 7:18
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$\begingroup$ There are infinitely many kinds of fibres! Q^x/(Q^x)^2 is an infinite group: it's a vector space over Z/2Z with basis -1 and the prime numbers. $\endgroup$ Commented Jan 21, 2010 at 7:49
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1$\begingroup$ @basic: Your example uses "quartic twists" rather than quadratic twists (for that, see Chapter 10 of Silverman's Arithmetic of Elliptic Curves) but it is otherwise the same (and there are still infinitely many isomorphism classes of fibers over k, if k is any number field). So yes, that works too. $\endgroup$ Commented Jan 21, 2010 at 8:44
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2$\begingroup$ @basic: if your $y^2=x^3+tx$ were a trivial family, then it would be equal to the family y^2=x^3-x. The latter curve has all its 2-torsion, so if your family were trivial then the 2-torsion in your curve over S:=Spec(k[t][t^-1]) (considered as a global object) would be four copies of S. But the two-torsion in the t-curve isn't four copies of S, as you can easily check (you are taking a square root of -t). $\endgroup$ Commented Jan 21, 2010 at 12:37
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(Edit: added motivation for my "answer" even though it isn't exactly what was asked for. It shows the moduli space of genus 1 curves cannot be fine - see last paragraph.)
The Hopf surface is a (non-algebraic) example of a non-trivial family of isomorphic elliptic curves. Here I use "elliptic curve" to mean "smooth complex curve of genus 1", which is probably not what you mean. (In particular my curves have no distinguished base point.)
The Hopf surface $X$ is a quotient of $\mathbb{C}^2\setminus0$ by the action of $\mathbb Z$ generated by $z \mapsto 2z$. Since this action commutes with the action of $\mathbb{C}^*$ there is a map $X \to \mathbb{CP}^1$. Each fibre is a copy of $\mathbb{C}\setminus\{0\}$ divided by the action $z\sim 2z$. So all the fibres are isomorphic elliptic curves. (If you want base points on each curve, this example doesn't work because the fibration $X \to \mathbb{CP}^1$ doesn't have a section.) On the other hand, its easy to see that $X$ is diffeomorphic to $S^1\times S^3$ and so you can't trivialise the fibration even topologically. Replacing $z\mapsto 2z$ by $z\mapsto \lambda z$ for $\lambda\in \mathbb{C}$ non-zero, I guess you can get any genus-1 curve to appear as all fibres of a topologically non-trivial family.
Even though this is not eaxctly what you're looking for, I thought it worth giving the example anyway, because it's a very simple way to see that the moduli space of genus 1 curves cannot be fine. (Fine moduli spaces carry a universal family and all other families are pulled back from the universal family via the map to moduli space - in particular a family of isomorphic objects in such a moduli space is pulled back by the constant map and so must be trivial. One often sees that objects cannot be parametrised by a fine moduli space by giving examples of special objects with "extra" automorphisms. The Hopf surface is an alternative to that approach in this situation.)
