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 fieldMathworldPlanetmath 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]R,  and since r is an algebraic element over K, the ring K[r] coincides with the field K(r).  Therefore we have  r-1K[r]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-1K[a].  This means that there is a polynomialMathworldPlanetmathPlanetmathPlanetmath 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.


  • 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
Canonical name AConditionOfAlgebraicExtension
Date of creation 2013-03-22 17:53:34
Last modified on 2013-03-22 17:53:34
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 7
Author pahio (2872)
Entry type Theorem
Classification msc 12F05
Related topic RingAdjunction
Related topic FieldAdjunction
Related topic Overring
Related topic AConditionOfSimpleExtension
Related topic SteinitzTheoremOnFiniteFieldExtension