algebraic extensionAlgebraicExtension2003-09-11 16:55:392008-04-01 13:35:40Definitionexamples of field extensiontranscendental extensionalgebraicroot of polynomial% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Rats}{\mathbb{Q}}\begin{defn}
Let $L/K$ be an extension of fields. $L/K$ is said to be an
\emph{algebraic extension} of fields if every element of $L$ is
algebraic over $K$. If $L/K$ is not algebraic then we say that it is a transcendental extension of fields.
\end{defn}
{\bf Examples: }
\begin{enumerate}
\item Let $L=\Rats(\sqrt{2})$. The extension $L/\Rats$ is an
algebraic extension. Indeed, any element $\alpha\in L$ is of the
form
$$\alpha=q+t\sqrt{2}\in L$$
for some $q,t\in\Rats$. Then $\alpha\in L$ is a root of
$$X^2-2qX+q^2-2t^2=0$$
\item The field extension $\Reals/ \Rats$ is not an algebraic
extension. For example, $\pi\in \Reals$ is a transcendental number
over $\Rats$ (see pi). So $\Reals/\Rats$ is a transcendental extension of fields.
\item Let $K$ be a field and denote by $\overline{K}$ the
algebraic closure of $K$. Then the extension $\overline{K}/K$ is
algebraic.
\item In general, a finite extension of fields is an algebraic
extension. However, the converse is not true. The extension
$\overline{\Rats}/\Rats$ is \emph{far} from finite.
\item The extension $\Rats(\pi)/\Rats$ is transcendental because $\pi$ is a transcendental number, i.e. $\pi$ is not the root of any polynomial $p(x)\in \Rats[x]$.
\end{enumerate}