A (mostly) purely algebraic proof of the infinitude of prime numbers

The following proof was provided/explained to me by @grassmanian on twitter. I thank him again for the support. Unfortunately, neither he nor I have found a way to show that each maximal ideal of $\mathbb{Z}$ contains a unique prime number without using the fact that $\mathbb{Z}$ is a PID, hence the mostly in the title.

We begin with a lemma:

$\textbf{Lemma:}$ If $A$ is an integral domain which has finitely many units, and finitely many maximal ideals, then it is a field.

$\textbf{Proof:}$ It is known that any finite integral domain is a field, so we shall assume that $A$ is infinite. Consider the Jacobson radical of A, denoted Jac(A). This is defined as follows:

$\text{Jac}(A) := \{x \mid 1 + xy \text{ is a unit for all }y\in A\}$

We observe the following: because $A$ is an integral domain, we have that for any nonzero $x\in A$ , and for any $y_1\neq y_2 \in A$ , $1 + xy_1 \neq 1 + xy_2$ . $A$ is infinite, so if Jac(A) contains any nonzero elements, then $A$ will have infinitely many units. This is forbidden by hypothesis, so we are forced to conclude that $\text{Jac}(A) = \{0\}$ .

Now, because $A$ is an integral domain (in particular because $A$ is commutative), we have the following equivalence of definitions:

$\text{Jac}(A) = \bigcap_{M\in\text{MSpec}(A)}M$

Where $\text{MSpec}(A)$ denotes all maximal ideals of $A$ .

By hypothesis, $A$ has only finitely many maximal ideals. That is, $\text{MSpec}(A) = \{M_1, M_2, \dots, M_n\}$ for some $n\in \mathbb{N}$ . Any two maximal ideals are coprime, so we may conclude the following by the Chinese Remainder Theorem for rings:

$A = A/(0) = A/(M_1\cap M_2\cap \dots \cap M_n) \cong \prod_{i=1}^n A/M_i$

Because each $M_i$ is maximal, we have that each $A/M_i$ is in fact a field, and thus $A$ is a product of fields. However, the product of two or more fields is never an integral domain. By hypothesis $A$ is an integral domain, so we are forced to conclude that there is only one such $M_i$ , meaning that $A$ is indeed a field.

With that lemma out of the way, we proceed.

$\textbf{Proof of the main claim:}$ Because $\mathbb{Z}$ is an integral domain with only finitely many units ( $\{-1, 1\}$ ), but is not a field, we may conclude that $\mathbb{Z}$ has infinitely many maximal ideals. $\mathbb{Z}$ is a principal ideal domain, so we have that each maximal ideal is of the form $(p)$ for some prime $p\in\mathbb{Z}$ . Hence, there are infinitely many primes.

Share this:

Like this:

Leave a Reply Cancel reply