Finite Field Extension is Algebraic

Theorem
Let $L/K$ be a finite extension of fields.

The $L/K$ is algebraic.

Proof
Let $x \in L$ be arbitrary, and $n = [L : K]$ the degree of $L$ over $K$.

Since any set of $n+1$ vectors in $L$ is linearly dependent, there is a $K$-linear combination of $\{1,\ldots,x^n\}$ equal to $0$.

Say $a_n x^n + \cdots + a_1 x + a_0 = 0$, $a_i \in K$, $i = 0, \ldots, n$.

Therefore $x$ satisfies a polynomial with coefficients in $K$.

That is, $x$ is algebraic.

Since $x \in L$ was chosen arbitrarily, $L/K$ is algebraic.