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 $a_0+a_{1}X+\cdots +a_n{X}^n\in \mathbb{F}[X]$ 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.