Euclid’s theorem.There are infinitely many prime numbers $p \in \ZZ$.

Before showing you the proof, I recall some basic results from algebraic number theory.

A Dedekind domain $A$ is an integral domain that has well-behaved factorization properties. One definition is that every nonzero ideal $\mathfrak a \subseteq A$ is the product of maximal ideals $\mathfrak a = \mathfrak p_1 \dotsm \mathfrak p_s$; this decomposition is then unique (up to permutation). In Dedekind domains the containment relation is the divisibility relation, i.e. $\mathfrak b \supseteq \mathfrak a$ if and only if there is an ideal $\mathfrak c$ with $\mathfrak a = \mathfrak b \mathfrak c$.

Fact 1.The integral closure of $\ZZ$ in a finite field extension $K / \QQ$ (that is, all elements $\alpha \in K$ satisfying a monic polynomial equation $\alpha^n + a_{n-1}\alpha^{n-1} + \dots + a_1\alpha + a_0 = 0$, $a_i \in \ZZ$) is a Dedekind domain.

In fact, we need this result only for a particular quadratic extension of the rationals.

Fact 2.The integral closure of $\ZZ$ in $\QQ(\sqrt{-5})$ is $\ZZ[\sqrt{-5}]$.

This is not hard to see, for example by examining the minimal polynomial of a general element $\alpha = a+b \sqrt{-5}$, which is monic if and only if $\alpha$ is integral over $\ZZ$.

Dedekind domains need not be unique factorization domains, in fact **Dedekind + UFD = PID**. But we have the following result:

Fact 3.A Dedekind domain $A$ with finitely many maximal ideals is a principal ideal domain.

*Proof.* By the prime factorization it suffices that each of the finitely many maximal ideals $\mathfrak p_1,\dots,\mathfrak p_m$ is principal, let’s show this for $\mathfrak q = \mathfrak p_1$. By the Chinese Remainder Theorem applied to the comaximal ideals $\mathfrak q^2, \mathfrak p_2, \dots, \mathfrak p_m$, the map

is surjective. Pick $a \in \mathfrak q \setminus \mathfrak q^2$ and let $q \in A$ be any preimage of $(a,1,\dots,1)$. The prime factorization of $qA \subseteq \mathfrak q$ can not involve any of $\mathfrak p_2, \dots, \mathfrak p_m$ and also not $\mathfrak q^2$ by construction, hence $qA = \mathfrak q$. $\square$

Finally, we need the following easy fact

Fact 4.If $A \subseteq B$ is an integral extension and $\mathfrak M \subseteq B$ is a maximal ideal, then so is $\mathfrak m \coloneqq \mathfrak M \cap A \subseteq A$.

*Proof.* If $A \subseteq B$ is integral, then so is $A’ \coloneqq A/ \mathfrak m \subseteq B / \mathfrak M \eqqcolon K$. Thus it suffices to show that if $A’ \subseteq K$ is an integral extension with $K$ a field, then $A’$ is a field too. Let $0 \neq x \in A$, then $x^{-1} \in K$ is integral over $A’$, i.e. $x^{-n} + a_{n-1}x^{-(n-1)} + \dots + a_1x^{-1} + a_0 = 0$. Multiply by $x^{n-1}$ and rearrange to see

This implies that all maximal ideals in a finite extension $B \supseteq A$ of Dedekind domains are factors of $\mathfrak p B$ for $\mathfrak p \subseteq A$ maximal (Proof: $\mathfrak M \supseteq (\mathfrak M \cap A)B$).

We are ready to prove the infinitude of prime numbers!

*Proof of Euclid’s theorem.* Assume that $\ZZ$ has only finitely prime numbers. Then the Dedekind domain $\ZZ[\sqrt{-5}]$ (**Fact 1+2**) also has only finitely maximal ideals (**Fact 4** + remark after) and hence is a principal ideal domain (**Fact 3**). But the decomposition

shows that $\ZZ[\sqrt{-5}]$ is *not* an unique factorization domain! It is easy to see that $2$ is irreducible by considering the multiplicative norm $N(x) = \lvert x \rvert^2$, but $2$ does *not* divide $1\pm \sqrt{-5}$, so it is not prime. :zap:

Merry Christmas and happy holidays! :christmas_tree:

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