インテジャーズ

インテジャーズ

数、特に整数に関する記事。

Two Proofs of Euclid's Theorem via Valuation Theory

Euclid's Theorem. There are infinitely many prime numbers.

Valuation Theory

Ostrowski's theorem. Every non-trivial valuation on the rational number field is equivalent to either the usual absolute value or a p-adic valuation for some prime number p.

By this beautiful theorem, we see that the infinitude of prime numbers is equivalent to the infinitude of non-trivial valuations on the rational number field. In this note, we give two valuation theoretic proofs of Euclid's theorem by the approximation theorem. Note that we don't use Ostrowski's theorem. On the other hand, we have to consider the infinite place.

From now on, we denote p as a prime number or the symbol \infty. Let | \ |_p be the p-adic valuation normalized as \left|p\right|_p=p^{-1} for a prime number p and | \ |_{\infty} the usual absolute value. Let \mathbb{Q}_{\infty} = \mathbb{R}.

First, we recall the following well-known facts:

Approximation Theorem. Let | \ |_1, \dots, | \ |_n be pairwise inequivalent non-trivial valuations of the rational number field and let a_1, \dots, a_n be given rational numbers. Then, for every \varepsilon > 0, there exists a rational number q such that

\left|q-a_i\right|_i < \varepsilon \ \ for all i = 1, \dots, n.

Product Formula. For every non-zero rational number a, one has

\displaystyle \prod_p \left|a\right|_p = 1,
where p varies over all prime numbers and \infty.

Proposition. Let \mathbb{A}_{\mathbb{Q}}=\prod^{\prime}_p\mathbb{Q}_p be the adele ring of the rational number field \mathbb{Q}. Then, the subgroup \mathbb{Q} under the diagonal embedding is discrete in \mathbb{A}_{\mathbb{Q}}.

Remark. We don't need the infinitude of prime numbers to prove the above facts. (But we use the notion of prime factorizations.)

The Proof by Product Formula

We assume that there are only finitely many prime numbers. For every p, take a rational number a_p such that \left|a_p\right|_p > 1. By the approximation theorem, we can take a rational number q such that \left|q\right|_p > 1 for every p. Then, \prod_p\left|q\right|_p > 1 holds. This contradicts to the product formula. Q.E.D.

Adelic Proof

We assume that there are only finitely many prime numbers. Then, \mathbb{A}_{\mathbb{Q}} = \prod_p\mathbb{Q}_p holds by definition and the topology coincides with the product topology. Since \mathbb{Q} is dense in \mathbb{Q}_p for each p, \mathbb{Q} is also dense in \mathbb{A}_{\mathbb{Q}} by the approximation theorem. Hence, we have \mathbb{Q} = \mathbb{A}_{\mathbb{Q}} by Proposition and a fact that every discrete subgroup of a Hausdorff topological group is closed*1. This is clearly impossible. Q.E.D.

This proof is also a topological proof*2.

*1:integers.hatenablog.com

*2:cf. Furstenberg's proof of Euclid's theorem.