Is there an elliptic curve over $\mathbb{C}[t, t^{-1}]$ that has a nonconstant $j$-invariant? What is an equation for such a curve, if it exists?

  • 1
    $\begingroup$ I'm probably going to write a short paper inspired this question (extending it to much more general families, i.e. Abelian schemes/geometric variations of Hodge structure). I'd like to acknowledge you by name; please email me if that's OK. My email is in my MO user page. $\endgroup$ – Daniel Litt May 19 '15 at 5:35
  • $\begingroup$ @DanielLitt: Sounds great, I'll take a look at it once it's done. Don't worry, no acknowledgement for me is necessary. $\endgroup$ – Lisa S. May 24 '15 at 3:56
  • $\begingroup$ There's a draft of the paper ready--I've cited this question, of course. Let me know if you'd like to take a look before I put it on the ArXiV. $\endgroup$ – Daniel Litt Jul 19 '16 at 1:48
  • $\begingroup$ @DanielLitt: Sorry for a late reply. Thanks for mentioning your paper, I've seen it on the arXiv $\endgroup$ – Lisa S. Sep 6 '16 at 20:34

Let me give an analytic argument, to complement Noam's algebraic one. Suppose $E\to \mathbb{C}^*$ is an elliptic curve. Then pulling back $E$ along the universal covering map (also known as the exponential map) $\mathbb{C}\to \mathbb{C}^*$, one obtains an elliptic curve $\tilde E\to \mathbb{C}$. Choosing a basis for the first homology of $\tilde E$ induces a holomorphic map $$\mathbb{C}\to \mathbb{H},$$ where $\mathbb{H}$ is the upper half-plane, viewed as the moduli space of elliptic curves with a homology basis for $H_1$. But any such map must be constant by Liouville's theorem. $\blacksquare$

Added Later. Since the OP enjoyed the sketch algebraic version in the comments (corrected by user74230), let me give some more details (and in particular, say what happens in characteristic $p>3$). WLOG $k$ is algebraically closed. Suppose $E\to \mathbb{G}_m/k$ is an elliptic curve. If we can find some $n$ so that the pullback of $E$ along $\mathbb{G}_m\overset{[n]}\longrightarrow \mathbb{G}_m$ has trivial $\ell$-torsion, with $(\ell, p)=1$ and $\ell \gg 0$, we're done, because choosing a trivialization we get a map from $\mathbb{G}_m$ to a high genus modular curve, which must be constant as $\mathbb{G}_m$ is rational.

To do this, we must show that for infinitely many $\ell$, the map $E[\ell]\to \mathbb{G}_m$ has tame ramification at $0, \infty$. It suffices to find $\ell$ so that $GL_2(\mathbb{Z}/\ell\mathbb{Z})$ has order prime to $p$. But this order is $(\ell^2-1)(\ell^2-\ell)=\ell(\ell-1)^2(\ell+1).$ But by Dirichlet's theorem on primes in arithmetic progressions, there are infinitely many $\ell$ so that $\ell(\ell-1)^2(\ell+1)$ is prime to $p$, if $p>3$. $\blacksquare$

As Noam observes in the comment, the result is false in characteristic $2$ and $3$; he gives examples of non-isotrivial families in these characteristics.

  • 3
    $\begingroup$ One can make an algebraic version of this argument as follows, which works in arbitrary characteristic. Some finite etale cover of $\mathbb{G}_m$ trivializes the $\ell$-torsion of $E$, and so after picking a trivialization induces a map from $\mathbb{G}_m$ to a modular curve. But the genus of modular curves tends to infinity, so for large $\ell$, such a map must be constant. $\endgroup$ – Daniel Litt Dec 12 '14 at 5:59
  • 4
    $\begingroup$ This algebraic argument might have problems in characteristics 2 and 3. The covering map $\ell:E \rightarrow E$ over $\mathbf{G}_m$ is generally not a Galois covering, just finite etale, corresponding to a representation of $\pi_1(\mathbf{G}_m)$ into ${\rm{GL}}_2(\mathbf{F}_{\ell})$ that might be wildly ramified at $0$ and $\infty$ in those low positive characteristics, so the corresponding cover (which depends on $\ell$) might not be a genus-0 curve. $\endgroup$ – user74230 Dec 12 '14 at 6:40
  • 10
    $\begingroup$ Indeed $y^2 = x^3 + x^2 - t$ has discriminant $t$ and $j$-invariant $1/t$ in characteristic $3$. Likewise $y^2 + xy = x^3 + t$ in characteristic $2$. $\endgroup$ – Noam D. Elkies Dec 12 '14 at 6:46
  • $\begingroup$ You're right of course; forgot about those other pesky covers. $\endgroup$ – Daniel Litt Dec 12 '14 at 7:17
  • $\begingroup$ @DanielLitt: Thank you. Of the four given proofs, I like the one that you give in your comment most, so I am going to accept this answer. $\endgroup$ – Lisa S. Dec 12 '14 at 22:56

There is no such curve.

One way to see this is via the action of ${\rm Gal}\bigl(\overline{{\mathbb C}(t)}\big/{\mathbb C}(t)\bigr)$ on the group $E[p]$ of $p$-torsion points of a putative elliptic curve $E / {\mathbb C}(t)$ that has good reduction at all $t \neq 0, \infty$. The image of Galois would be abelian, because the coordinates of $E[p]$ would generate an extension of ${\mathbb C}(t)$ unramified above all $t \in {\mathbb C}^*$, and ${\mathbb C}^*$ has abelian fundamental group. On the other hand, once $p$ exceeds the order of a pole of the $j$-invariant $j_E^{\phantom.}$, the image of Galois includes a $p$-cycle ramified above that pole. A $p$-cycle in ${\rm SL}_2({\mathbb Z}/p{\mathbb Z})$ generates its own centralizer, so the image of Galois would be a $p$-element subgroup of ${\rm SL}_2({\mathbb Z}/p{\mathbb Z})$. It would follow that $E$ has $p$ nontrivial $p$-torsion points rational over ${\mathbb C}(t)$ (because a $p$-cycle in ${\rm SL}_2({\mathbb Z}/p{\mathbb Z})$ fixes a 1-dimensional space of $({\mathbb Z}/p{\mathbb Z})^2)$. But this cannot happen for infinitely many $p$, QED.

The result can also be obtained by tracking down the cases when Szpiro's inequality is sharp: a nonconstant elliptic curve over $E / {\mathbb C}(t)$ has discriminant degree at least $12$, and therefore conductor degree at least $\frac{12}{6} + 2 = 4$ by Szpiro; equality can hold in several ways, but in each case $j_E^{\phantom.}$ turns out to be constant. (With good reduction away from $t=0$ and $t=\infty$, the conductor degree is at most $2+2=4$.)

  • $\begingroup$ There might not be an entirely simple proof of this result. The nonconstant elliptic curves over ${\mathbb C}^*$ are quadratic twists $y^2 = x^3 + at x^2 + bt^2 x + ct^3$, cubic twists $y^2 = x^3 + at^2$ and $y^2 = x^3 + at^4$, quartic twists $y^2 = x^3 + at x$ and $y^2 = x^3 + at^3 x$, and sextic twists $y^2 = x^3 + at$ and $y^2 = x^3 + at^5$; any argument must account for all of these. The proof I gave is conceptual but advanced; the Szpiro path is elementary (Szpiro is basically Mason = polynomial ABC) but requires case analysis. $\endgroup$ – Noam D. Elkies Dec 12 '14 at 5:11
  • $\begingroup$ Thank you. Why does the image of Galois induce a $p$-cycle ramified above the pole of $j_E$ (and what does that mean)? $\endgroup$ – Lisa S. Dec 12 '14 at 19:48

It is an older topic, and very nice answers have been given. But I am new in MO and let me give another answer.

The reason for nonexistence of such an elliptic curve is the same as for nonexistence of such an elliptic curve over $S:=$Spec$(\mathbb{C}[t])$. In order to have a universal elliptic curve one must remove the points from $S$ which correspond to elliptic curves with automorphism group larger than $\{ \pm1\}$, and these fibers have $j$-invariants 0 and 1728. This is explained in the book Arithmetic Moduli of Elliptic Curves by Katz and Mazur. So we must remove at least two points from $S$, i.e. inverting $t$ in $\mathbb{C}[t]$ is not enough.

This argument works if we replace $\mathbb{C}$ by any algebraically closed field of characteristic $p \geq 5$. But for $p=2$ or 3 we have $0 = 1728$, so it is enough to remove only one point from $S$ in these cases and this explains the existence of examples given by Elkies above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.