a finite extension of fields is an algebraic extension

Theorem 1.

Let $L\mathrm{/}K$ be a finite field extension. Then $L\mathrm{/}K$ is an algebraicextension.

Proof.

In order to prove that $L/K$ is an algebraic extension, we need to show that any element$\alpha \in L$ is algebraic, i.e., there exists a non-zeropolynomial $p(x)\in K[x]$ such that $p(\alpha )=0$.

Recall that $L/K$ is a finite extension of fields, by definition,it means that $L$ is a finite dimensional vector space over $K$.Let the dimension be

$$[L:K]=n$$

for some $n\in \mathbb{N}$.

Consider the following set of “vectors” in $L$:

$$𝒮=\{1,\alpha ,{\alpha }^2,{\alpha }^3,\mathrm{\dots },{\alpha }^n\}$$

Note that the cardinality of $S$ is $n+1$, one more than thedimension of the vector space. Therefore, the elements of $S$ mustbe linearly dependent over $K$, otherwise the dimension of $S$would be greater than $n$. Hence, there exist $k_i\in K,\mathrm{\hspace{0.25em}0}\le i\le n$, not all zero, such that

$$k_0+k_1\alpha +k_2{\alpha }^2+k_3{\alpha }^3+\mathrm{\dots }+k_n{\alpha }^n=0$$

Thus, if we define

$$p(X)=k_0+k_{1}X+k_2{X}^2+k_3{X}^3+\mathrm{\dots }+k_n{X}^n$$

then $p(X)\in K[X]$ and $p(\alpha )=0$, as desired.

∎

NOTE: The converse is not true. See the entry “algebraicextension” for details.