Proof that finite extensions are algebraic

Proof

1. Let $L/K$ be a finite extension, with degree $n = [L : K]$. Let $a \in L$. We find a nonzero $f \in K[x]$ with $f(a) = 0$.
2. Case 1: $a^i = a^j$ for some $0 \le i < j \le n$. Then $f = x^j - x^i$ is nonzero and $f(a) = 0$.
3. Case 2: the elements $1, a, a^2, \dots, a^n$ are pairwise distinct. They form a set $S$ of $n + 1$ elements of $L$.
4. Suppose that $S$ were linearly independent over $K$. By extension to a basis, $S$ lies in a basis of $L$ over $K$ with at least $n + 1$ elements.
5. By the dimension theorem, every basis of $L$ over $K$ has $[L : K] = n$ elements, by the definition of dimension. This contradicts Step 4.
6. So there are $c_0, \dots, c_n \in K$, not all zero, with $c_0 + c_1 a + \dots + c_n a^n = 0$.
7. Then $f = c_0 + c_1 x + \dots + c_n x^n$ is a nonzero polynomial in $K[x]$ with $f(a) = 0$.
8. In both cases $a$ is algebraic over $K$. Since $a \in L$ was arbitrary, $L/K$ is algebraic.