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Why do elliptic curves have bad reduction at some point if they are defined over Q, but not necessarily over arbitrary number fields?

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6 Answers

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Here are two answers:

(a) If you try to write down an elliptic curve $y^2 = x^3 + a x + b$ with everywhere good reduction, you need to choose $a$ and $b$ such that $4a^3 + 27 b^2 = $ a unit. We can certainly solve this equation over some (lots!) of number fields, say if we set the unit equal to $1$ or $-1$, or a unit in some fixed base number field. But we can't solve it in ${\mathbb Q}$.

[Edit: As Bjorn intimates in his comment below, one has to be a little more careful than I am being here to be sure of good reduction mod primes above 2; the details are left to the interested reader (or, I imagine, can be found in Silverman in the section where he discusses the proof that there are no good reduction elliptic curves over $\mathbb Q$).]

(b) There are many non-trivial everywhere unramified extensions of number fields (e.g. $\mathbb Q(\sqrt{-5}, i)$ over $\mathbb Q(\sqrt{-5})$), but there are no everywhere unramified extensions of the particular number field $\mathbb Q$. The situation with elliptic curves is completely analogous.

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Matt: a typo in (b): everywhere <b>unramified</b> extension. – Qing Liu Feb 27 2010 at 14:06
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 Tate's example : over $K=\mathbb{Q}(\sqrt{29})$, the elliptic curve $E:y^2+xy+\varepsilon^2y=x^3$, where $\varepsilon=(5+\sqrt{29})/2$, has good reduction everywhere because its discriminant is a unit. – Chandan Singh Dalawat Feb 27 2010 at 15:01
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 In keeping with point (a) you might say that in a field with a finite number of units ($\mathbb{Q}$ or a quadratic imaginary field) you don't have much of a chance to solve $4a^3 + 27b^2 = $ unit. So, what is the situation for quadratic imaginary fields? – Victor Miller Feb 27 2010 at 19:08
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 Regarding (b): "completely analogous" concerns me a bit. E.g., for all but 9 imaginary quadratic fields K, there exists a nontrivial (abelian) extension of number fields of conductor 1. The question of which imaginary quadratic fields admit an elliptic curve with conductor 1 seems much more complicated: I vaguely remember a lot of special results. Is it even known whether this set is finite or infinite? [And what about real quadratic fields?] – Pete L. Clark Feb 27 2010 at 19:11
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 Nitpick: In (a), minor changes are required if you want good reduction also at primes above 2. – Bjorn Poonen Feb 28 2010 at 4:14
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It's also worth noting that "good reduction at $\mathfrak{p}$" is a local condition, so an elliptic curve may have everywhere good reduction, despite not having a Weierstrass equation that has good reduction at all primes. This is because over fields of class number greater than 1, there always exist elliptic curves that do not have global minimal Weierstrass equations. The existence, or lack of, a global minimal Weierstrass equation is governed by a certain ideal class (see Proposition VIII.8.2 in my Arithmetic of Elliptic Curves). The fact that if the class number is greater than 1, then there always exist curves with no global minimal Weierstrass equation is in the paper: "Weierstrass equations and the minimal discriminant of an elliptic curve", Mathematika 31 (1984), no. 2, 245–251. There is also a paper by Bekyel that describes the density of curves having (or not having) global minimal Weierstrass equations: "The density of elliptic curves having a global minimal Weierstrass equation", J. Number Theory 109 (2004), no. 1, 41–58. The moral is that you can produce curves with everywhere good reduction by writing down a specific Weierstrass equation, but to determine whether a given curve has everywhere good reduction is done via local calculations, and the associated elliptic scheme having everywhere good reduction may need to be patched together using more than one Weierstrass equation.

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I would like to stress something that, I think, has not been point out very explicitly in any of the nice answers above.

The question that you are asking, if I understand it correctly, is a local question. So let's start with, say, a $p$-adic field $K$, and with the equation $f(x,y)=0$ of (the affine piece of) an elliptic curve $E$ with coefficients in the ring of integers $R$ of $K$.

To define the reduction of $E$ modulo the maximal ideal $\mathfrak{m}$ of $R$, one cannot follow the naive approach of simply considering the reduction mod $\mathfrak{m}$ of the equation $f(x,y)=0$.

Instead one should look at what Tate called a minimal Weierstrass equation $f_{\rm min}(x,y)=0$ for $E$ (LNM 476 pag. 39), and define the reduction $\bar E$ of $E$ as the object obtained by considering the reduction mod $\mathfrak{m}$ of $f_{\rm min}(x,y)=0$.

The point now is that the minimal equation is NOT stable by taking field extensions: if $L/K$ is a finite extension then $f_{\rm min}(x,y)$ need not still be a minimal equation for the base change of $E$ to $L$ (but it is if $L/K$ is unramified). Therefore it might very well happens that while $f_{\rm K, min}(x,y)=0$ has a singular reduction mod $\mathfrak{m}$ the corresponding thing does not hold for $f_{\rm L, min}(x,y)=0$ (the meaning of the subscripts is obvious I hope).

I tried to come up with an explicit example illustrating this phenomena. I failed to find one in a reasonable amount of time. But I am sure they can be found (easily?) in the literature.

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 As Qing Liu has said above, to find an example of bad reduction at a particular place $p$ becoming good over an extension, you just need the $p$-adic valuation of the $j$-invariant to be positive. Examples of this abound. However, it is not true that the non-existence of elliptic curves with everywhere good reduction over $\mathbb{Q}$ is a local problem, since one can certainly find curves with good reduction at any given place. The question is a genuinely global one. – Alex Bartel Apr 7 2011 at 4:14
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 The question being asked is not local, but global. Over any local field, we can always find elliptic curves with good reduction. The question is why we can't find an elliptic curve over $\mathbb Q$ with good reduction at every prime, although we can find such curves over other number fields. In other words, what is special about the case of $\mathbb Q$ that makes it impossible to find such elliptic curves over this particular number field. – Emerton Apr 7 2011 at 4:17
Alex and Emerton, thanks for your comments, I agree with what you say, of course. I will just add that the original question above is quite succinct, and perhaps it is not so clear what the author had in mind. My interpretation of it was:"How come that there exist elliptic curves E over Q having primes P of bad reduction, but such that when regarded as elliptic curves over a suitable number field K they acquire good reductions at the primes of K above P?". And this is basically a local question, I think. – Tommaso Centeleghe Apr 7 2011 at 17:59
Ok, I should admit that reading again the question, with its emphasis on "if they are defined over Q", I see that my interpretation might just be wrong. – Tommaso Centeleghe Apr 7 2011 at 18:02
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For those with an algorithmic bent, you should look at the paper by Cremona and Lingham "Finding All Elliptic Curves with Good Reduction Outside a Given Set of Primes" In Experimental Math. Volume 16, Issue 3 (2007), 303-312. http://www.warwick.ac.uk/~masgaj/papers/egros.pdf

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Here is another answer, certainly overkill, but possibly interesting overkill.

By the elliptic modularity theorem, the result follows from the fact that the modular curve $X(1)$ has genus $0$: or equivalently, that the upper half-plane modulo $\operatorname{SL}_2(\mathbb{Z})$ is just the affine line $\mathbb{A}^1_{/\mathbb{C}}$. The fact that this curve has to have genus $0$ follows from Fontaine's theorem that there are no curves of positive genus with everywhere good reduction over $\mathbb{Q}$.

On the other hand, given a totally real field $F$ of narrow class number one with $[F:\mathbb{Q}] = 2n+1$, we may form the Shimura curve $X_F(1)$ corresponding to a quaternion algebra $B/F$ which is unramified at every finite place of $F$ is ramified at $2n$ out of the $2n+1$ real places; such a quaternion algebra exists because $2n$ is even. This curve does have everywhere good reduction over $F$. Moreover, there is a finite, known list of examples where the genus of this curve is equal to zero (as $F$ ranges over all totally real fields). So -- assuming that there are infinitely many such $F$ of strict class number one, which is certainly a widely believed conjecture -- one has infinitely many modular curves over totally real fields $F$ with everywhere good reduction. To get elliptic curves out of them one needs rank one factors splitting off from the Jacobian. Again, it seems very plausible that this will happen infinitely many times as we vary $F$ even over all totally real cubic fields of class number one.

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No, of course Fontaine's theorem is not necessary for this: it follows just by looking at the standard fundamental domain for SL_2(Z). However, given that the moduli space X(1) is smooth over Z, Fontaine's theorem gives a conceptual explanation for why it must have genus zero. – Pete L. Clark Feb 27 2010 at 19:03
Oops, sorry, I deleted my comment, to ask it in a better way. I will re-fill the comment in a moment. – Regenbogen Feb 27 2010 at 19:06
The deleted comment was the following: Is it really needed to use Fontaine's theorem to say that the quotient of $\mathbb H$ by $SL_2(\mathbb{Z})$ is $\mathbb{A}^1_{/\mathbb{C}}$? It is proved using the properties of $j$-function in Serre's "A course in Arithmetic", last chapter. – Regenbogen Feb 27 2010 at 19:09
Ok, apologies for messing things up(which I always seem to do), and now, might I ask some explanation for the "it seems very plausible that..." in the final sentence of your answer – Regenbogen Feb 27 2010 at 19:18
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There were some discussions on this question. This property is very specific to $\mathbb Z$.

To construct elliptic curves with everywhere good reduction over a number field, you can start with any elliptic curve over $\mathbb Q$ with integral $j$-invariant. Then it is well-kwnon that $E$ has good reduction everywhere over some finite extension $K$ of $\mathbb Q$ (it is actually enough to take $K$ be the extension generated by the $3$-torsion points of $E$). The existence of varieties with good reduction everywhere over number field holds also for abelian varieties and curves of any genus. This can be seen as existence theorem of integral points in some moduli schemes parametrizing abelian schemes and curves over $\mathbb Z$ (look for works of R. Rumely and of L. Moret-Bailly on Skolem properties).

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