a condition of algebraic extensionAConditionOfAlgebraicExtension2008-03-08 15:12:412008-03-08 16:27:03Theoremalgebraic extension% this is the default PlanetMath preamble. as your knowledge
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\textbf{Theorem.}\, A field extension $L/K$ is \PMlinkname{algebraic}{AlgebraicExtension} if and only if any subring of the extension field $L$ containing the base field $K$ is a field.\\
{\em Proof.}\, Assume first that $L/K$ is algebraic.\, Let $R$ be a subring of $L$ containing $K$.\, For any non-zero element $r$ of $R$, naturally\, $K[r] \subseteq R$,\, and since $r$ is an algebraic element over $K$, the ring $K[r]$ coincides with the field $K(r)$.\, Therefore we have\, $r^{-1} \in K[r] \subseteq R$,\, and $R$ must be a field.
Assume then that each subring of $L$ which contains $K$ is a field.\, Let $a$ be any non-zero element of $L$.\, Accordingly, the subring $K[a]$ of $L$ contains $K$ and is a field.\, So we have\, $a^{-1} \in K[a]$.\, This means that there is a polynomial $f(x)$ in the polynomial ring $K[x]$ such that\, $a^{-1} = f(a)$.\, Because\, $af(a)-1 = 0$,\, the element $a$ is a zero of the polynomial $xf(x)-1$ of $K[x]$, i.e. is algebraic over $K$.\, Thus every element of $L$ is algebraic over $K$.
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\bibitem{D.B.}{\sc David M. Burton}: {\em A first course in rings and ideals}. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).
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