# Could Furstenberg's Argument Prove the Infinitude of Primes in Number Fields?

I have a somewhat unconventional view of the Prime Number Theorem as a "quantification" of the infinitude of primes. Here I recall the argument of Furstenberg. Define a topology $\mathcal{X}$ on $\mathbb{Z}$.

• $\varnothing \in \mathcal{X}$
• $a \, \mathbb{Z} + b \in \mathcal{X}$

By unique factorization, there is a finite set that is the complement of these open set:

$$\mathbb{Z} \backslash \{ -1, 1 \} = \bigcup_p S(p,0)$$

This could probably work for any number field. Let's see if I write it correctly. Is it clear?

$$K \backslash \{ \text{units} \} = \bigcup_p S(p,0)$$

There's few literature I could find on the Furstenberg Topology itself  Perhaps it has a generalization that is more commonly discussed? Does the Furstenberg topology have any deeper meaning that we know about?

Later in his life he should prove Szemeredi Theorem which is entirely also about arithemetic sequences.

About two years ago, I asked for infinite Gaussian primes in segments. In standard algebraic number theory these are all merged into questions about Dirichlet and Hecke characters. One person responded that we could use sieves. and filter out the primes that we need.

Elementary Proof of Infinitely many primes $\mathfrak{p} \in \mathbb{Z}[i]$ in the sector $\theta < \arg \mathfrak{p} <\phi$

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• The Fürstenberg topology is the one induced by $\hat{\mathbb{Z}}$. – M.G. Jul 13 '17 at 1:10
• @July But this doesn't a priori answer the question. – Fan Zheng Sep 12 '17 at 18:34
• @FanZheng: which is why it's only a comment :-) However, it does (somewhat) address the question "Does the Fürstenberg topology have any deeper meaning that we know about?" in the body. – M.G. Sep 12 '17 at 20:41

## 1 Answer

Yes. Just replace arithemetic progressions by translations of a nonzero (integral) ideal. The same argument shows that the union on the RHS is closed, so the set of units $K^\times$ has to be open. Since 1 is a unit, $K^\times$ must contain $1+\frak a$ for some nonzero (integral) ideal $\frak a$. Take a nonzero element $a\in\frak a$. Then $1+na\in K^\times$ for all integer $n$. But $N(1+na)$ is a degree $d=[K:\mathbb Q]$ polynomial in $n$ with leading coefficient $N(a)\neq 0$, so it cannot be $\pm1$ for all integer $n$.