Every Field Extension Which is Finite is Algebraic

There is a one way relationship between the notions of a finite and algebraic field extension, given by the following two theorems.

Proof

Let \(\mathbb{K}\) be a finite extension of \(\mathbb{F}\) with \([\mathbb{K} : \mathbb{F}] = n\). Let \(\alpha \in \mathbb{K}\). Then \(\mathbb{F}(\alpha) \subseteq \mathbb{K}\). Consider then the elements \(\{1, \alpha, \alpha^2, \dots, \alpha^n\}\). This is a list of \(n + 1\) elements in an \(n\) dimensional vector space, and thus is linearly dependent. Therefore there exists \(a_0, \dots, a_n \in \mathbb{F}\) not all zero such that

This however implies that \(a_0 + a_1 X + \dots + a_n X^n \in \mathbb{F}[X]\) is a polynomial with \(\alpha\) as a root, and thus \(\alpha\) is algebraic over \(\mathbb{F}\).

Given that \(\alpha\) was arbitrary, this implies that all elements of the field \(\mathbb{K}\) are algebraic.

Proof

This is just the contrapositive.

The converse of these results however does not hold.

Proof

Clearly every generator is algebraic, since \(\sqrt[n]{2}\) is a root of \(X^n - 2 \in \mathbb{Q}[X]\). Then, a field generated by finitely many algebraic elements is algebraic.

This degree however is not finite, since using the above polynomials, we have that \([\mathbb{Q}(\sqrt[n]{2}): \mathbb{Q}] = n\) given that \(X^n - 2\) is irreducible by Eisenstein's criterion.

Since \(\mathbb{Q}(\sqrt[n]{2})\) is a subfield of \(\mathbb{K}\) for all \(n\), we therefore have that \([\mathbb{K} : \mathbb{Q}] > [\mathbb{Q}(\sqrt[n]{2}) : \mathbb{Q}] = n\) for all \(n\). This implies that the degree of \(\mathbb{K}\) over \(\mathbb{Q}\) cannot be finite, and is therefore infinite.