# How do I write down a curve with exactly one rational point

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.

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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.

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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).

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 $[-1:-1:1]$