# a condition of algebraic extension

Theorem.  A field extension $L/K$ is algebraic (http://planetmath.org/AlgebraicExtension) if and only if any subring of the extension field $L$ containing the base field $K$ is a field.

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 $K$.

## References

• 1 David M. Burton: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).
Title a condition of algebraic extension AConditionOfAlgebraicExtension 2013-03-22 17:53:34 2013-03-22 17:53:34 pahio (2872) pahio (2872) 7 pahio (2872) Theorem msc 12F05 RingAdjunction FieldAdjunction Overring AConditionOfSimpleExtension SteinitzTheoremOnFiniteFieldExtension