Set of Algebraic Numbers is Countable

Theorem

There are countably many algebraic integers.

This result is particularly useful in providing a proof for the existence of transcendental numbers, without the need to construct any particular example. For the proof we will use the fact that a countable union of countable sets is countable, and thus we first desire the following lemma.

Lemma

\(\mathbb{Q}[X]\) is countable.

Proof

Write

\[ \mathbb{Q}[X] = \bigcup_{n = 0}^\infty \{ f \in \mathbb{Q}[X] : \deg f = n \}.\]

Each set in this union is countable, because we can construct a natural bijection between the set of polynomials of degree \(n\) and \(\mathbb{Q}^{n + 1}\) by

\[ (a_0, a_1, \dots, a_n) \mapsto a_0 + a_1 X + \dots a_n X^n\]

and \(\mathbb{Q}^{n + 1}\) is countable because a finite cartesian product of countable sets is countable, and \(\mathbb{Q}\) is countable. Therefore \(\mathbb{Q}[X]\) is written as a countable union of countable sets, and is therefore countable.

We can now prove the main theorem.

Proof

We may write the set of algebraic integers as

\[ \bigcup_{f \in \mathbb{Q}[X]} \mathrm{roots}(f)\]

and by the fundamental theorem of algebra, since \(\mathbb{Q}\) is a subfield of \(\mathbb{R}\), we have that

\[ |\mathrm{roots}(f)| \leq \deg f < \infty.\]

Thus we have written the set of algebraic integers as the countable union of countable sets, which is countable.