Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $g\geq 1$. I would like to write down (for all $g$) a smooth projective geometrically connected curve $X$ over $\mathbf{Q}$ of genus $g$ with precisely one rational point.

Is this possible?

For which $g$ is this possible?

I think for $g=1$ this is possible. I just don't know an explicit equation, but I should be able to find it. (We just write down an elliptic curve without torsion of rank zero over $\mathbf{Q}$.)

For $g\geq 2$ things get more complicated for me.

I would really like the curve to be of gonality at least $4$, but I'll think about that later.

share|cite|improve this question
What is $g$ supposed to have to do with the curve? – Chris Eagle Aug 3 '12 at 9:08
That would be the genus. – Harry Aug 3 '12 at 9:15
Dear Harry, what do you mean by "curve" and "genus" ? – Georges Elencwajg Aug 3 '12 at 10:40
Curve = smooth projective geometrically connected curve over $\mathbf{Q}$. The genus is the arithmetic genus (or geometric genus because of our definition of curves). – Harry Aug 3 '12 at 10:41

The smooth plane projective curve $E$ defined over $\mathbb Q$ by the equation $y^2z=x^3+2z^3$ has its point at infinity $[0:1:0]$ as its only rational point: $E(\mathbb Q)=\lbrace [0:1:0]\rbrace $. Indeed:

a) The torsion group of the curve $y^2z=x^3+az^3 $ is zero as soon as $a$ is a sixth-power free integer which is neither a square nor a cube nor equal to $-432$.
(Despite appearences I'm not making this crazy theorem up, but I am quoting theorem (3.3) of Chapter 1 in Husemöller's Elliptic Curves !).

b) On the other hand the curve $E$ has rank $0$, which means that its group of rational points is torsion (this is stated in the table following the theorem I just quoted).

The two results a) and b) prove the assertion in my introductory sentence.

share|cite|improve this answer
Thank you for your answer! This is the elliptic curve I was hinting at in my question. Do there exist similar examples in higher genus? – Harry Aug 3 '12 at 13:36
Dear Georges, I think I have an idea on how to get more curves with just one rational point. For $n\geq 1$, let $X_n$ be the curve given by $(y^2z)^n = x^{3n}+2z^{3n}$. This curve is singular at its only rational point $(0:1:0)$ if $n\geq 2$. Its normalization will probably have more rational points, so this won't work. But maybe this gives an idea on how to construct non-trivial branched covers of $X_1 = E$ with just one rational point? – Harry Aug 3 '12 at 14:35
I forgot to say that $X_n \to E$ via $(x:y:z)\mapsto (x^n:y^n:z^n)$. The fibre over the $\mathbf{Q}$-rational point $(0:1:0)$ contains the point $(0:1:0)$. Note that for $n>2$, this map is unramified over $(0:1:0)$. There is exactly one rational point on $X_n(\mathbf{Q})$, but there might be more on its normalization. – Harry Aug 3 '12 at 15:04

I believe that there is a class of curve that has a single rational point on it, it is an hyperbola given by (x^2 + A)/(B - x). The conditions to give a single rational point would be for B^2 + A = a prime number, e.g. B= 6 A = 5 the rational point would be at (5, 30). The conditions for 2 rational points would be that B^2 + A = N, where N = pq, e.g. B = 114 A = 203, the ratioanl points are at (47, 36) and (113, 12972).

Hope that this may help you.

share|cite|improve this answer
I don't think this works. Maybe you're writing down curves with only one integer point? In fact, any rational value for $x$ will give a value for $y$ which is a rational number. I think that the projective curve induced by your equation is isomorphic to the projective line. – Harry Aug 3 '12 at 10:40

I can't comment, but I'm afraid that the answer of George is incorrect. Indeed the equation has at least 2 solutions, namely $[0:1:0]$ and $[-1:-1:1]$

share|cite|improve this answer

Harry When x is rational, the y coordinates are not rational for the majority of the x values. If you look at (x^2 + 5) mod (6 - x) the value is only 0, at x = 5; similarly for the other example (x^2 + 203) mod (114 -x) only as 0 at x = 47 and x = 113. It is the same for larger numbers(100 of digits).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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