TITLE: The intersection of a circle and a rank 3 subgroup of the plane
QUESTION [11 upvotes]: 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.

REPLY [4 votes]: 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.


TITLE: Can we determine if a reference frame is inertial or non-inertial by looking at an accelerating object?
QUESTION [0 upvotes]: if we find that in a given frame, an object being acted upon by a force and is accelerating with respect to the frame would that mean that the reference frame in which this object exist is inertial or non-inertial?

REPLY [0 votes]: I think of Newton's First law as a definition of an inertial frame

An inertial frame is one in which an object that experiences no net force moves with a constant velocity vector.

together with a statement that Newton's other laws only apply in inertial frames.
So, before analyzing a system using Newton II, one should do an "inertial frame test": consider an object with zero net force on it, e.g., a ball sitting on the ground (normal up, gravity down, net zero); if that ball does not accelerate, then you are in an inertial frame.  If you were in the back of an accelerating truck and performed the same experiement, you would see the ball mysteriously move from rest.  This "ghost force" indicates that you are not in an inertial frame.
So, to answer your question: just observing the acceleration of an object alone does not tell you a frame is inertial or non-inertial. But if an object appears to accelerate and there are no evident forces to explain that acceleration, then the frame is non-inertial.