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Let $A$ be a rank 3 subgroup of the Euclidean plane, i.e. $A = \mathbb{Z} v_1 + \mathbb{Z} v_2 + \mathbb{Z} v_3$, where $v_1, v_2, v_3 \in \mathbb{R}^2$ are three $\mathbb{Q}$-linearly independent vectors. Let $S^1 = \{v \in \mathbb{R}^2 ; \, |v|=1\}$ denote the unit circle in the plane. Is it possible that $|A \cap S^1| = \infty$?

Remark: I could find rank 4 subgroups intersecting $S^1$ in infinitely many points. The idea was to take a complex algebraic integer $\alpha$ such that $|\alpha| = 1$ and $\alpha$ is not a root of unity. Then the additive group generated by $1, \alpha, \alpha^2, \dots$ has finite rank.

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  • $\begingroup$ The discussion at this MO Question might be helpful. $\endgroup$ – pgadey Feb 13 '14 at 15:01
  • $\begingroup$ I don't think you'll be able to get a solution to your problem via a similar construction using cubics. Theorem 4.1 of this note says that no such cubics will exist. $\endgroup$ – pgadey Feb 13 '14 at 15:21
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I am grateful to Edgardo whose answer supplied me with the Siegel theorem on integral points (I did not know it), and I am happy to see that my approach coincides with Edgardo's in many points.

The answer is negative: there is no abelian subgroup $A\subset{\mathbb R}^2$ of rank $3$ such that its intersection $I_A:={\mathbb S}^1\cap A$ with the unit circle ${\mathbb S}^1:=\big\{v\in{\mathbb R}^2\mid|v|=1\big\}$ is infinite.

In a few words: we show that such $I_A$ is a part of the intersection of two quadrics in ${\mathbb P}_{\mathbb C}^3$ defined over ${\mathbb Q}$, observe that this intersection in interesting cases is generic, hence, is an elliptic curve, and, finally, apply the Siegel theorem http://en.wikipedia.org/wiki/Siegel_theorem to an affine part of this curve.

Let $L_{\mathbb Q}:={\mathbb Q}A$ denote the ${\mathbb Q}$-linear space spanned by $A$. Besides the intersection $I_A$, we will also deal with the intersection $I_{\mathbb Q}:={\mathbb S}^1\cap L_{\mathbb Q}$. One can choose another ${\mathbb Q}$-basis $B'$ in $L_{\mathbb Q}$ and generate another $A'$. It is immediate that $A\subset\frac1nA'$ and $A'\subset\frac1nA$ for a suitable $n\in{\mathbb N}$. We conclude that $I_A$ is infinite for some $A$ iff, for any ${\mathbb Q}$-basis $B$ in $L_{\mathbb Q}$, there are infinitely many elements in $I_{\mathbb Q}$ whose denominators in terms of $B$ are limited.

Assuming that some $I_A$ is infinite, we can choose a ${\mathbb Q}$-basis of $L_{\mathbb Q}$ inside $I_{\mathbb Q}$. Indeed, it is easy to pick a couple of ${\mathbb R}$-linearly independent $b_1,b_2\in I_{\mathbb Q}$. They generate a discrete ${\mathbb Z}$-lattice $A_0\subset{\mathbb R}^2$; so, $\frac1nA_0$ is also discrete for any $n\in{\mathbb N}$. Consequently, the intersection ${\mathbb S}^1\cap\frac1nA_0$ is finite. Including $b_1,b_2$ in some basis of $L_{\mathbb Q}$, we can see that $I_{\mathbb Q}\subset{\mathbb Q}A_0$ would imply $I_A\subset\frac1nA_0$ for a suitable $n\in{\mathbb N}$. A contradiction. For a similar reason, any $2$-dimensional ${\mathbb Q}$-linear subspace in $L_{\mathbb Q}$ contains finitely many rational points from $I_{\mathbb Q}$ with limited denominators.

Let $b_1,b_2,b_3\in I_{\mathbb Q}$ be a ${\mathbb Q}$-basis in $L_{\mathbb Q}$. Its Gram matrix has the form $G:=\left[\begin{smallmatrix}1&c_3&c_2\\c_3&1&c_1\\c_2&c_1&1\end{smallmatrix} \right]$ with $-1<c_i<1$. Since the $b_i$'s are ${\mathbb R}$-linearly dependent, $\det G=0$. So, if $c_1,c_2,c_3\in{\mathbb Q}$, then, by means of the Gram-Schmidt orthogonalization in $L_{\mathbb Q}$ and by the Sylvester criterion, we could find, in view of $\det G=0$, an isotropic element $0\ne v\in L_{\mathbb Q}$, i.e., such that $\langle v,v\rangle=0$, which is impossible. Hence, $c_i\notin{\mathbb Q}$ for some $i$.

Actually, we study rational solutions of $$x_1^2+x_2^2+x_3^2+2c_1x_2x_3+2c_2x_3x_1+2c_3x_1x_2=1.\qquad(1)$$

As $c_3^2<1$, there are finitely many rational solutions of the form $[x_1,x_2,0]$ with limited denominators, because $(1)$ becomes $(x_1+c_3x_2)^2+(1-c_3^2)x_2^2=1$. In other words, we can assume that $x_1x_2x_3\ne0$ and rewrite $(1)$ in terms of $y_i:=x_i^{-1}$ as $$\frac{y_1y_2}{y_3}+\frac{y_2y_3}{y_1}+\frac{y_3y_1}{y_2}+2c_1y_1+2c_2y_2+2c_3y_3=y_1y_2y_3.\qquad(2)$$

The dimension of the ${\mathbb Q}$-linear space spanned by $[x_2x_3,x_3x_1,x_1x_2]$, where $[x_1,x_1,x_3]$ runs over all rational solutions of $(1)$, equals $2$. Indeed, it cannot be $3$ as, otherwise, $c_1,c_2,c_3\in{\mathbb Q}$. It cannot be $0$ because we have at least one rational solution. If the dimension equals $1$, then, for all rational solutions with $x_1x_2x_3\ne0$, the triples $[y_1,y_2,y_3]$ are all ${\mathbb Q}$-proportional. Looking at $(2)$, we see that the coefficient of proportionality should be $\pm1$, and we get only finitely many solutions of $(1)$ with limited denominators.

We conclude that $c_1=a_1c+d_1$, $c_2=a_2c+d_2$, $c_3=c$ for suitable $a_1,a_2,d_1,d_2\in{\mathbb Q}$. It follows from $\det G=0$ that $p(c)=0$, where $p(x):=\det\left[\begin{smallmatrix}1&x&d_2+a_2x\\x&1&d_1+a_1x\\d_2+a_2x&d_1+a_1x&1\end{smallmatrix}\right]$.

Now, $(1)$ takes the form $x_1^2+x_2^2+x_3^2+2d_1x_2x_3+2d_2x_3x_1+2(x_1x_2+a_1x_2x_3+a_2x_3x_1)c=1$. Since $c\notin{\mathbb Q}$, we deal in fact with the equations $$x_1^2+x_2^2+x_3^2+2d_1x_2x_3+2d_2x_3x_1=1,\quad x_1x_2+a_1x_2x_3+a_2x_3x_1=0.$$

We want to show that, for any integer $0\ne d\in{\mathbb Z}$, the equations $$x_1^2+x_2^2+x_3^2+2d_1x_2x_3+2d_2x_3x_1=d^2,\quad2x_1x_2+2a_1x_2x_3+2a_2x_3x_1=0$$ admit just finitely many integer solution.

The intersection $$x_1^2+x_2^2+x_3^2+2d_1x_2x_3+d_2x_3x_1-d^2x_0^2=0,\quad2x_1x_2+2a_1x_2x_3+2a_2x_3x_1=0$$ of two quadrics in ${\mathbb P}_{\mathbb C}^3$ is an elliptic curve if all the $4$ roots of the polynomial $q(\lambda):=\det(\lambda M_1+M_2)$ are pairwise distinct, where $M_1,M_2$ stand for the symmetric matrices of the quadrics (see, for example, http://archive.numdam.org/ARCHIVE/ASNSP/ASNSP_1980_4_7_2/ASNSP_1980_4_7_2_217_0/ASNSP_1980_4_7_2_217_0.pdf especially, from pages 221-222). Since $q(\lambda)=\det\left[\begin{smallmatrix}-d^2\lambda&0&0&0\\0&\lambda&1&d_2 \lambda+a_2\\0&1&\lambda&d_1\lambda+a_1\\0&d_2\lambda+a_2&d_1\lambda+ a_1&\lambda\end{smallmatrix}\right]=-d^2\lambda^4p(\lambda^{-1})$, in view of the Siegel theorem, it suffices to show that all the $3$ roots of the polynomial $\lambda^3p(\lambda^{-1})=0$ are $(\text{a})$ pairwise distinct and $(\text{b})$ all different from $0$. The polynomial $p(x)$ of degree $\le3$ has rational coefficients, $p(c)=0$, and $c\notin{\mathbb Q}$. Consequently, $p(x)$ cannot have multiple roots thus implying $(\text{a})$. If $(\text{b})$ is not true, the coefficient of degree $3$ in $p(x)$ vanishes, i.e., $2a_1a_2=0$. In this case, the second quadric is a couple of rational planes passing through the origin $0\in L_{\mathbb Q}$, and we already know that any $2$-dimensional ${\mathbb Q}$-linear subspace in $L_{\mathbb Q}$ contains finitely many rational points from $I_{\mathbb Q}$ with limited denominators.

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  • $\begingroup$ Yes, our proofs seem to be basically the same, but your treatment of the degenerate cases is cleaner than mine. Thanks. $\endgroup$ – Edgardo Feb 16 '14 at 15:33
  • $\begingroup$ @Edgardo For me, it was the lack of the Siegel theorem. For you, the standard criterion providing a couple of quadrics in ${\mathbb P}_{\mathbb C}^3$ to be generic, i.e., their intersection to be smooth, hence, elliptic curve. I voted in your solution and for the proper question, of course. $\endgroup$ – Sasha Anan'in Feb 16 '14 at 15:50
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Here's at least a method that probably solves the problem. I say "probably" because I'm not entirely confident in my treatment of degenerate cases. Anyway, the main idea is to try to construct a second quadratic function that vanishes at all the points of your $S^1 \cap A$, thus reducing the problem to finiteness of integral points on a certain curve.

Let $q(x,y,z) = \|x v_1 + y v_2 +z v_3\|^2$, a quadratic form in $x,y,z$. Note that $q(x,y,z)$ has a one-dimensional radical. We assume that the set $S = \{(x,y,z) \in \mathbb{Z}^3: q(x,y,z) =1\}$ is infinite and (try to) derive a contradiction.

For $\lambda \in \mathbb{Q}$, consider the set $\mathcal{Q}_{\lambda}$ of all quadratic forms $Q$ in $x,y,z$ which have the property that $Q = \lambda$ on $S$. This is an affine subspace of the $5$-dimensional real vector space of all quadratic forms in $x,y,z$. Moreover, it is defined over $\mathbb{Q}$, since it is defined by affine equations with rational coefficents.

$\mathcal{Q}_{1}$ cannot consist of the single point $\{q\}$, because otherwise $q$ would have rational coefficients, and then the radical of $q$ would contain a rational vector $(x,y,z)$, contradicting the assumed independence of $v_1, v_2, v_3$.

Now $S$ is contained in the intersection of the affine quadric surfaces $Q=1$ for $Q \in \mathcal{Q}_1$. Call $X$ that intersection. Because $\mathcal{Q}_1 \neq \{q\}$ we see that $X$, considered as an algebraic variety, is of dimension $\leq 1$. If $X$ is of dimension $0$ there is nothing to show; we will show that if $X$ has dimension $1$ it is (an open subset of) an elliptic curve. Then Siegel's theorem on integral points yields a contradiction.

$\mathcal{Q}_0$ contains a nondegenerate form $R$, for otherwise we may pick degenerate $Q \in \mathcal{Q}_0$ with rational coefficients. In that case, the set of zeroes of $Q$ is contained in the union of two $\mathbb{Q}$-rational proper subspaces of $(x,y,z) \in \mathbb{Q}^3$. So $S$ is contained in the union of two sublattices of $\mathbb{Z}^3$ of rank $2$. That's easily seen to be impossible.

We have $q \in \mathcal{Q}_1$ and $R \in \mathcal{Q}_0$. If $X$ has dimension $1$, it must be the intersection of $R=0$ and $q=1$ (these equations are not defined over $\mathbb{Q}$, but $X$ is anyway...) Now the intersection of two quadrics is an elliptic curve under a certain nonsingularity condition. I THINK that, in our situation, the nonsingularity condition amounts to asking that no linear combination $aR+bq$ of $R,q$ is a square of a linear form $\ell$ (could certainly be wrong about this!). But if that were so, $S$ would be contained in the union of two affine planes $\ell = \pm \sqrt{b}$, and that is again easily seen to be impossible.

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  • $\begingroup$ There's no answer here -- just a (somewhat confusing) reinterpretation. $\endgroup$ – J. Martel Feb 16 '14 at 14:36

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