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. $\endgroup$ May 19, 2015 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, 2015 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. $\endgroup$ Jul 19, 2016 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, 2016 at 20:34

3 Answers 3


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.

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    $\begingroup$ 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. $\endgroup$ Dec 12, 2014 at 5:59
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    $\begingroup$ 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, 2014 at 6:40
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    $\begingroup$ 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 $2$. $\endgroup$ Dec 12, 2014 at 6:46
  • $\begingroup$ You're right of course; forgot about those other pesky covers. $\endgroup$ Dec 12, 2014 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, 2014 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. $\endgroup$ Dec 12, 2014 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, 2014 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.


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