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%This is a new version of the original question modified in the light of the answers and comments.

The word 'most' in the title is ambiguous. The following is one way of making it precise.

Question1: (This seems to be open. See Poonen's answer below)

A cubic projective curve over $\mathbb{Q}$ is given by ten relatively prime integers (the coefficients of its equation after clearing the denominators). Suppose we take a ten dimensional box $[-N,N]^{10}$ and choose points with integer coordinates with respect to the uniform measure and form the equation of the associated cubic curve. Suppose the number of points which give rise to a curve with a rational point is $E(N)$. Then what can we say about $E(N)/(2N+1)^{10}$ as $N\rightarrow \infty$?

Should the limit exist and if it does, should it be one, zero, or some other number?

Another question of interest is:

Question 2: (There is a satisfactory answer to this. See Voloch's response below.)

Are either of the sets {cubics with no rational point} and {cubics with at least one rational point} Zariski dense?

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Nice question! Let me point out something which confused me for a while, and might be confusing others. Any elliptic curve over Q can be embedded as a plane cubic with a rational flex. And I think it should be doable to show that almost all plane cubics over Q do not have rational flexes. However, this does not answer your question! These curves may still have rational points; it is just that they are not embedded in a way that puts those rational points at flexes. –  David Speyer Jan 10 '10 at 17:24
    
Yes, this is a nice point. Let me add that I am not quite satisfied with the formulation of my question. I think clearing denominators is not a natural thing to do here. I wonder if the condition of having a rational point or not is a (Zariski) dense condition. – Idoneal 2 mins ago –  Idoneal Jan 10 '10 at 17:40
    
@Idoneal, I think the question about Zariski density is a bit different and may very well have known answer (intuitively, more likely to be yes). So I think you should post it separately. –  Ilya Nikokoshev Jan 10 '10 at 19:56
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Is it possible to check this guess numerically? Say, with N= 1000, can one possibly get a lower bound on the number of such curves? –  Idoneal Jan 11 '10 at 5:08
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Checking all curves with N=1000 is unfeasible. One could try to do a Monte Carlo test choosing a bunch of random curves in that range. But it is going to be hard to distinguish between positive, but small, density in the limit and a density that goes to zero like 1/log N. –  Felipe Voloch Jan 12 '10 at 16:12
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4 Answers

Your question (as explained in the second paragraph) is not vague at all! In fact, it appears for instance after Conjecture 2.2 in http://www-math.mit.edu/~poonen/papers/random.pdf , which is Random diophantine equations, B. Poonen and J. F. Voloch, pp. 175–184 in: Arithmetic of higher-dimensional algebraic varieties, B. Poonen and Yu. Tschinkel (eds.), Progress in Math. 226 (2004), Birkhäuser.

The answer is not known, and the experts I've spoken to do not even have a convincing heuristic predicting an answer. Swinnerton-Dyer told me that he had a hunch that the answer was 0, and this is my hunch too, but we have little to back this up.

It is not even clear that the limit exists. One can prove, however, that the density (in your precise sense) of plane cubic curves that have points over $\mathbb{Q}_p$ for all $p \le \infty$ is a number strictly between $0$ and $1$ (Theorem 3.6 in the Poonen-Voloch paper), so the lim sup of the fraction of plane cubic curves with a rational point is at most this; in particular, it's not 1.

One could try to estimate the size of the Tate-Shafarevich group of a "random" elliptic curve, to get an idea of how often local solvability implies global solvability, but even if one does this it is not clear that this is counting curves in the same way.

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Thanks. What about Zariski denseness? I mean, is the complement of the set of points giving elliptic curves expected to be Zariski dense? –  Idoneal Jan 10 '10 at 17:52
    
Bjorn: I'm probably missing something here, but I don't see why you are saying that this density is strictly less than 1. Following the links in your paper, the limit density is prod c_p, where c_p is the proportion of cubics that have a root in Q_p. But the Weil bounds say that every smooth cubic has at least p - 2 \sqrt{p} + 1 > 0 points over F_p, and then Hensel tells us there is a solution in Q_p. Also, there is a real point because cubics have odd degree. I haven't though through the bad reduction case carefully, but it looks to me like there might be no local obstructions for cubics. –  David Speyer Jan 10 '10 at 18:00
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@Idoneal. If the set of cubics without points is not Zariski dense, then the proportion of cubics without points is zero. I expect that the set of cubics without points is Zariski dense. @David Look again at bad reduction and you'll see that $c_p < 1$. There is a positive proportion of cubics reducing mod p to three irrational lines not having a rational point. –  Felipe Voloch Jan 10 '10 at 18:28
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@David: There are local obstructions in the bad reduction case. For example, $x^3+2y^3+4z^3=0$ has no $2$-adic point (consider the $2$-valuations of the three monomials!) Any cubic polynomial congruent to this one mod 8 defines a curve with no $2$-adic point. This shows that a positive fraction of the curves is excluded. –  Bjorn Poonen Jan 10 '10 at 18:31
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Thanks, I see. Another example is a cubic which, over F_{p^3}, is three lines forming a triangle and where Frobenius rotates the lines cyclically. –  David Speyer Jan 10 '10 at 19:21
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Manjul Bhargava has answered this question yesterday :

A positive proportion of plane cubics fail the Hasse principle

Manjul Bhargava (Submitted on 5 Feb 2014)

When all ternary cubic forms over $\mathbf{Z}$ are ordered by the heights of their coefficients, we show that a positive proportion of them fail the Hasse principle, i.e., they have a zero over every completion of $\mathbf{Q}$ but no zero over $\mathbf{Q}$. We also show that a positive proportion of all ternary cubic forms over $\mathbf{Z}$ nontrivially satisfy the Hasse principle, i.e., they possess a zero over every completion of $\mathbf{Q}$ and also possess a zero over $\mathbf{Q}$. Analogous results are proven for other genus one models, namely, for equations of the form $z^2=f(x,y)$ where $f$ is a binary quartic form over $\mathbf{Z}$, and for intersections of pairs of quadrics in $\mathbf{P}_3$.

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Re: Zariski denseness. Let $Y$ be the set of cubics with a rational point and $N$ the set of cubics without a rational point, both viewed as subsets of $P^9$. I claim that both $Y$ and $N$ are Zariski dense. For $Y$, note that if you fix a point the plane, the set of cubics containing that point is a hyperplane in $P^9$ so the Zariski closure of $Y$ contains infinitely many hyperplanes and therefore is the whole of $P^9$. For $N$, we use Bjorn's example. Any curve congruent modulo 8 to $x^3+2y^3+4z^3=0$ is in $N$ and this set of curves is already Zariski dense.

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Felipe, I just asked Idoneal to make a separate post for a different question... do you think we could still do that, you posting an answer there? –  Ilya Nikokoshev Jan 10 '10 at 19:59
    
Do you want me to copy this answer in the new thread and delete it here? Sure, I can do that. –  Felipe Voloch Jan 10 '10 at 20:21
    
Although Zariski denseness is a separate question, it is in the same spirit and I think they should be together. However, I am going to modify my question accordingly. –  Idoneal Jan 11 '10 at 4:31
    
It's actually better for everyone when separate questions go to separate posts (easier to search for them, for instance), but I guess here it's simpler to leave it as it is. –  Ilya Nikokoshev Jan 11 '10 at 13:09
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"One could try to estimate the size of the Tate-Shafarevich group of a "random" elliptic curve, to get an idea of how often local solvability implies global solvability, but even if one does this it is not clear that this is counting curves in the same way."

Bhargava has reportedly proven the 3-Selmer group has average size 4. The assumption of a minimal rank (1/2 average) and Parity conjecture would account that 2 of the 4 come from rank, and 2 of the 4 come from Sha, so 50%. His counting is by $|c_4| < X^2 $ and $|c_6| < X^3$ I think. The workers on 3-descent have some bounds that relate the invariants to the coefficient size.

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Bhargava and Shankar have now announced that they can prove along these lines that a positive proportion of cubics fail the Hasse principle and thus have no point, despite having points locally everywhere, and that a positive proportion of of cubics have a point. –  Felipe Voloch May 27 '11 at 19:05
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