Infinitely many primes for Christmas

As my first ever blog post on here and also as a mathematical Christmas gift, I wanted to share my favorite proof of the infinitude of prime numbers, only using commutative algebra (in the guise of the theory of Dedekind domains).

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

\[A \to A/\mathfrak q^2 \times A/\mathfrak p_2 \times \dotsm \times A/\mathfrak p_m\]

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

\[x^{-1} = -a_{n-1} - a_{n-2}x - \dots - a_1 x^{n-2} - a_0 x^{n-1} \in A'. \tag*{$\square$}\]

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

\[2\cdot 3 = 6 = (1+\sqrt{-5}) \cdot (1-\sqrt{-5})\]

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.

Merry Christmas and happy holidays!