# Finite Field Extension is Algebraic

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## Theorem

Let $L / K$ be a finite field extension.

Then $L / K$ is algebraic.

## Proof

Let $x \in L$ be arbitrary.

Let $n = \left[{L : K}\right]$ be 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 $\left\{{1, \ldots, x^n}\right\}$ 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.

$\blacksquare$