Showing a $\mathbb{Z}$-module is not a discrete subset of the complex plane.

Question

The following is an algebraic number theory assignment problem from 2020 that I didn't know how to solve. The lecturer didn't provide solution either. So I'm asking here. Hints are also welcomed.

Let $z_1, z_2, z_3 \in \mathbb{C}$ be linearly independent over $\mathbb{Z}$. Denote $M:=z_1 \mathbb{Z} + z_2\mathbb{Z} + z_3 \mathbb{Z}$ the $\mathbb{Z}$-module generated by the $z$'s. Show that $M$ is not discrete in $\mathbb{C}$.

Thanks in advance.

Answer 1

Without loss of generality, $$z_3=\alpha_1z_1+\alpha_2z_2$$ for some real numbers $\alpha_1,\alpha_2.$

By Dirichlet's simultaneous approximation theorem, for every $\epsilon>0,$ there are integers $q\ne0$ and $p_1,p_2$ such that $$|q\alpha_i-p_i|<\epsilon$$ (for $i=1,2$).

Therefore, there exist sequences of integers $q_n\ne0,p_{1,n},p_{2,n}$ such that $$qz_3-p_{1,n}z_1-p_{2,n}z_2\to0,$$ which proves the claim.

Edit: obviously, @C-RAM's later generalization can be proved the same way.

Answer 2

Case $1$: $z_1$, $z_2$ and $0$ are aligned (but we could have chosen $z_1,z_3,0$), this is equivalent to the same problem but with $z_i \in \mathbb{R} $ and it is way easier since it defines an additive subgroup of $\mathbb{R}$ which is either isomorphic to $\mathbb{Z}$ or dense thus in this case it is dense and $0$ is not an isolated point, thus $M$ is not discrete.

Case $2$: not case $1$ which means that up to a sign multiplication, $z_1,z_2,z_3$ form a triangle that contains $0$.
Let us set $$r = \inf_{z \in M} |z|$$ Inspiration: subgroups of $\mathbb{R}$. If $r=0$ $M$ is not discrete since $0$ is not isolated. Let us suppose that $r >0$. If you observe closely, you will easily deduce that there exist $z_0\in M$ such that $|z|=r$ since you can only fit a finite number of disjoint balls or radius $r$ with center in the ball of radius $2r$.

With the same argument, there is a finite number of points of $M$ in the triangle $(z_1,z_2,z_2)$, so we just need to show that this is impossible. For that, we just need to show that there exist $z_4$ in the triangle, that is different from each vertex, then we reiterate with the submodule $M'$ given by $z_2,z_3,z_4$...

For now I can always do that but given the points coordinates, I'm working on it but I think this is already a big progress...

Ok, I have it, let us suppose $|z_1| \geq |z_2| \geq |z_3|$, we set $z'$ as the intersection of the lines $(0,z_3)$ and $(z_1,z_2)$. $$z' = z_1 + t (z_2 -z_1)$$ with $0<t<1$.

If we consider the projection $\pi$ of $\mathbb{Z} z_1 \oplus \mathbb{Z} z_2 $ on $(z_1,z_2)$ parallel to $(0,z_3)$, this forms an additive subgroup of $\mathbb{R}$ which is necessary dense (see first part), thus we can approach $\pi(z')$ with $\pi(z'')$ up to $\epsilon$ with $z'' \in \mathbb{Z} z_1 \oplus \mathbb{Z} z_2$

Now, since $|z'-z_3|\geq \min(|z_2-z_3|,|z_2-z_3|) \geq 2|z_3|$, if we choose $\epsilon$ small enough, there is $m \in \mathbb{Z}$ such that $z''+mz_3 $ belongs to the triangle.

We have our desired point which concludes the proof!

Answer 3

Here's a proof of the more general result:

Let $m<n$, and $z_1,z_2 \cdots,z_n\in\mathbb{R}^m$ be linearly independent over $\mathbb{Z}$. Denote $M:=\mathbb{Z}z_1+\mathbb{Z}z_2+\cdots+\mathbb{Z}z_n$ the $\mathbb{Z}$-module generated by the $z$'s, then $M$ is not discrete in $\mathbb{R}^m$.

First, denote $S=\text{span}_\mathbb{R}\{z_1,z_2,\cdots,z_n\}$ the real subspace generated by the $z$'s. Let $k=\dim_\mathbb{R}(S)$, then since $k\leq m<n$, then we may pick indices $j_1,j_2,\cdots,j_k,\ell$ such that $z_{j_1},z_{j_2},\cdots,z_{j_k}$ is an $\mathbb{R}$-basis for $S$ and $z_\ell$ is a vector not in that basis. Since $z_{j_1},z_{j_2},\cdots,z_{j_k},z_\ell$ are linearly dependant over $\mathbb{R}$, for every positive integer $p$ we may write $pz_\ell=a_1^{(p)}z_{j_1}+a_2^{(p)}z_{j_2}+\cdots+a_k^{(p)}z_{j_k}$ for some coefficients $a_i^{(p)}\in\mathbb{R}$. We may then define \begin{equation} \begin{split} x^{(p)}&=pz_\ell-\lfloor a_1^{(p)}\rfloor z_{j_1}-\lfloor a_2^{(p)}\rfloor z_{j_2}-\cdots-\lfloor a_k^{(p)}\rfloor z_{j_k}\\ &= \{a_1^{(p)}\} z_{j_1}+\{a_2^{(p)}\} z_{j_2}+\cdots+\{a_k^{(p)}\} z_{j_k}\\ \end{split} \end{equation} where $\{b\}$ denotes the fractional part of $b$. Note that by definition $x^{(p)}\in M\cap K$, where $K$ is the bounded set defined by $K=\{b_1z_{j_1}+b_2z_{j_2}+\cdots+b_kz_{j_k}:b_i\in [0,1)\text{ for all } i\}$. Now, suppose that $p,q$ are positive integers such that $x^{(p)}=x^{(q)}$, then \begin{equation} \begin{split} (p-q)z_\ell&=pz_\ell-qz_\ell\\ &=(a_1^{(p)}-a_1^{(q)})z_{j_1}+(a_2^{(p)}-a_2^{(q)})z_{j_2}+\cdots+(a_2^{(p)}-a_2^{(q)})z_{j_k}\\ &=(\lfloor a_1^{(p)}\rfloor -\lfloor a_1^{(q)}\rfloor)z_{j_1}+(\lfloor a_2^{(p)}\rfloor -\lfloor a_2^{(q)}\rfloor)z_{j_2}+\cdots+(\lfloor a_2^{(p)}\rfloor - \lfloor a_2^{(q)}\rfloor)z_{j_k}\\ \end{split} \end{equation} and since $z_{j_1},z_{j_2},\cdots,z_{j_k},z_\ell$ are linearly independent over $\mathbb{Z}$, then $p=q$. In other words, the sequence $(x^{(p)})_{p=1}^\infty$ is an infinite sequence of distinct elements in $M\cap K$ (which is bounded), therefore by the Bolzano-Weierstass theorem, it has convergent subsequence $(x^{(p_i)})_{i=0}^\infty$. This means that the sequence $(x^{(p_{i+1})}-x^{(p_i)})_{i=0}^\infty$ is a sequence in $M\backslash\{\boldsymbol{0}\}$ which converges to $\boldsymbol{0}$. In other words, $\boldsymbol{0}$ is a non-isolated point in $M$, and thus $M$ is not discrete in $\mathbb{R}^m$.