simple field extensionSimpleFieldExtension2004-05-31 18:19:392008-03-12 17:37:53Definitionfield adjunctionadjunction% this is the default PlanetMath preamble. as your knowledge
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% define commands hereLet $K(\alpha)$ be obtained from the field $K$ via the \PMlinkescapetext{simple adjunction} of the element $\alpha$. \,We shall settle the \PMlinkescapetext{structure types} of the field $K(\alpha)$.
We consider the substitution homomorphism \,$\varphi: K[X] \rightarrow K[\alpha]$, where
$$\sum a_{\nu}X^\nu \mapsto \sum a_{\nu}\alpha^\nu.$$
According to the ring homomorphism theorem, the image ring $K[\alpha]$ is isomorphic with the residue class ring $K[X]/\frak{p}$, where $\frak{p}$ is the ideal of polynomials having $\alpha$ as their zero. \,Because $K[\alpha]$ is, as subring of the field $K(\alpha)$, an integral domain, then also $K[X]/\frak{p}$ has no zero divisors, and hence $\frak{p}$ is a prime ideal. \,It must be principal, for $K[X]$ is a principal ideal ring.
There are two possibilities:
\begin{enumerate}
\item $\frak{p} = (p(X))$, where $p(X)$ is an irreducible polynomial with \,$p(\alpha) = 0$. \,Because every non-zero prime ideal of $K[X]$ is maximal, the isomorphic image $K[X]/(p(X))$ of $K[\alpha]$ is a field, and it must give the \PMlinkescapetext{structure} of \,$K(\alpha) = K[\alpha]$. \,We say that $\alpha$ is {\em algebraic with respect to} $K$ (or {\em over} $K$). \,In this case, we have a finite field extension \,$K(\alpha)/K$.
\item $\frak{p} = (0)$. \,This means that the homomorphism $\varphi$ is an isomorphism between $K[X]$ and $K[\alpha]$, i.e. all expressions $\sum a_{\nu}\alpha^\nu$ behave as the polynomials $\sum a_{\nu}X^\nu$. \,Now, $K[\alpha]$ is no field because $K[X]$ is not such, but the isomorphy of the rings implies the isomorphy of the corresponding fields of fractions. \,Thus the simple extension field $K(\alpha)$ is isomorphic with the field $K(X)$ of rational functions in one indeterminate $X$. \,We say that $\alpha$ is {\em \PMlinkname{transcendental}{Algebraic} with respect to} $K$ (or {\em over} $K$). \,This time we have a simple infinite field extension \,$K(\alpha)/K$.
\end{enumerate}