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We know that a unit circle has dense rational points. But do spirals also have dense rational points, i.e. for two rational points:

$$(x_1,y_1) = (\sin(t_1) k t_1,\cos(t_1) k t_1) \in \mathbb Q^2$$

$$(x_2,y_2) = (\sin(t_2) k t_2,\cos(t_2) k t_2) \in \mathbb Q^2$$

with $t_1 < t_2$, is there a third rational point $(x_3,y_3)$ from a $t_3$, with $t_1 < t_3$ and $t_3 < t_2$? Is there a proof or disproof of rational density?

Please note: We do not make any rationality assumptions about the parameter $k$ except $k > 0$.

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  • $\begingroup$ .. .. ... what is $k$ ? $\endgroup$ – mercio Mar 25 '17 at 23:35
  • $\begingroup$ that's not really what dense means... at all. Do you count $(0,0)$ as a rational point belonging on all the spirals ? $\endgroup$ – mercio Mar 26 '17 at 0:11
  • $\begingroup$ well then most spirals (all but countably many) have $(0,0)$ as their only rational point, which means that according to your strange definition, they have dense rational points. $\endgroup$ – mercio Mar 26 '17 at 0:16
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The statement is false.

$\newcommand\Q{\mathbf{Q}}$ $\newcommand\Qb{\overline{\mathbf{Q}}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\p{\mathfrak{p}}$

Let's first consider the much wider class of non-zero points $P = (x,y)$ where $x$ and $y$ are merely assumed to be algebraic numbers (i.e. solutions to polynomials with rational coefficients).

Suppose that $P = (x,y) \in \Qb \times \Qb$ is such a point. Then

$$(x,y) = (\sin(t) kt, \cos(t) kt) \in \Qb \times \Qb.$$

If $x$ and $y$ are algebraic, then so is

$$\sqrt{x^2+y^2} = kt.$$

But then $x/kt = \sin(t)$ and $y/kt = \cos(t)$ are algebraic, and hence $e^{i t} = \cos(t) + i \sin(t)$ is also all algebraic.

Let us consider values of $t$ for which $e^{i t}$ and $kt$ are both algebraic. Suppoes this happens for two distinct points $s$ and $t$. Then $k s$ and $k t$ are algebraic, and so their ratio $\beta = s/t$ is also algebraic. But so are $\alpha = e^{i t}$ and $e^{i s}$. In particular,

$$\alpha^{\beta} = (e^{i t})^{s/t} = e^{i s}$$

is also algebraic. By the Gelfond–Schneider theorem, this implies that $\alpha = 0,1$ or $\beta$ is rational. Certainly $\alpha \ne 0$. If $\alpha = 1$, then we can reverse the roles of $s$ and $t$. If $e^{i s}$ is also $1$, then $s$ and $t$ are multiples of $2 \pi i$ and the conclusion that $\beta = s/t$ is rational still holds. Hence $s/t$ is rational for any two algebraic points on the spiral, and hence all the algebraic points must occur when $t$ is a rational multiple of some fixed value $\tau$.

Conversely, if $P = (x,y)$ is algebraic for $t = \tau$, then $P$ will also be algebraic for exactly the rational multiples of $\tau$. (We don't use this below, but it is easy).


Now consider the problem of rational points. Certainly rational points are algebraic, so we may deduce from the previous argument that $P$ is rational only for rational multiples of a fixed value $\tau$ of $t$. If $x$ and $y$ are rational for a fixed $\tau$,

$$\sqrt{x^2 + y^2} = k \tau$$

lives in an extension $E/\Q$ of degree at most two. But then

$$\alpha:=e^{i \tau} = \cos(\tau) + i \sin(\tau) \in E(i)$$

lives in an extension of degree at most four. We know that all the other algebraic points on the spiral have to occur for a rational multiple $r$ of this $\tau$. But for what rational numbers $r$ does

$$\alpha^r = e^{i r \tau}$$

live inside an extension of $\Q$ of degree at most $4$ when $\alpha$ itself has degree at most $4$? It's an elementary fact(* see below) from algebraic number theory that this will happen only for $r$ with denominator $D$ bounded absolutely in terms of $\alpha$. This implies that the rational points on the spiral will be discrete.


Here's a sketch of (*). First consider the case when $\alpha$ is a root of unity. Then $\tau$ is rational. But then $e^{i r \tau}$ is also root of unity, and this has degree at most four only if it is an $N$th root of unity with $\varphi(N) \le 4$, which occurs for only finitely many $N$. Thus we may assume $\alpha$ is not a root of unity, and it suffices to show that the degree of the fields:

$$E(\alpha), E(\alpha^{1/2}), E(\alpha^{1/3}), E(\alpha^{1/4}), \ldots $$

tends to infinity. (If $(r,s) = 1$, then $E(\alpha^{r/s}) = E(\alpha^{1/s})$.) Certainly $\alpha$ is not a $p$th power for sufficiently large $p$: either it has non-trivial valuation at some prime $\p$, in which case it's not a $p$th power for $p$ larger than the valuation at $\p$, or it's a unit, and the result follows from the structure of the unit group assuming that $\alpha$ is not a root of unity. Thus $X^p - \alpha$ is irreducible for large enough $p$, which implies that $E(\alpha^{1/p})$ has degree at least $p$ for those $p$. On the other hand, for a fixed $p$, the same argument shows that $\alpha$ is a $p^m$th power for only finitely many $m$, in which case the degree of $E(\alpha^{1/p^{m+1}})$ is at least $p$, and by repeating this argument, the degree of $E(\alpha^{1/p^n})$ tends to infinity with $n$.

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