a finite extension of fields is an algebraic extensionAFiniteExtensionOfFieldsIsAnAlgebraicExtension2003-09-11 17:15:232003-09-17 15:49:44Theoremfinite extensionalgebraicpolynomialfinite% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Rats}{\mathbb{Q}}\begin{thm}
Let $L/K$ be a finite field extension. Then $L/K$ is an algebraic
extension.
\end{thm}
\begin{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-zero
polynomial $p(x)\in K[x]$ 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
$$[L\colon K]=n$$
for some $n\in \Nats$.
Consider the following set of ``vectors'' in $L$:
$$\mathcal{S}=\{ 1, \alpha, \alpha^2,\alpha^3,\ldots,\alpha^n\}$$
Note that the cardinality of $S$ is $n+1$, one more than the
dimension of the vector space. Therefore, the elements of $S$ must
be linearly dependent over $K$, otherwise the dimension of $S$
would be greater than $n$. Hence, there exist $k_i\in K,\ 0\leq i
\leq n$, not all zero, such that
$$k_0+k_1\alpha+k_2\alpha^2+k_3\alpha^3+\ldots+k_n\alpha^n=0$$
Thus, if we define
$$p(X)=k_0+k_1X+k_2X^2+k_3X^3+\ldots+k_nX^n$$
then $p(X)\in K[X]$ and $p(\alpha)=0$, as desired.
\end{proof}
{\bf NOTE: } The converse is not true. See the entry ``algebraic
extension'' for details.