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.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Create your website with WordPress.com
Get started
%d bloggers like this: