# algebraic sum and product

Let $\alpha,\,\beta$ be two elements of an extension field of a given field $K$.  Both these elements are algebraic over $K$ if and only if both $\alpha\!+\!\beta$ and $\alpha\beta$ are algebraic over $K$.

Proof.  Assume first that $\alpha$ and $\beta$ are algebraic.  Because

 $[K(\alpha,\,\beta):K]=[K(\alpha,\,\beta):K(\alpha)]\,[K(\alpha):K]$

and both here are finite (http://planetmath.org/ExtendedRealNumbers), then $[K(\alpha,\,\beta):K]$ is finite.  So we have a finite field extension $K(\alpha,\,\beta)/K$ which thus is also algebraic, and therefore the elements $\alpha\!+\!\beta$ and $\alpha\beta$ of $K(\alpha,\,\beta)$ are algebraic over $K$.  Secondly suppose that $\alpha\!+\!\beta$ and $\alpha\beta$ are algebraic over $K$.  The elements $\alpha$ and $\beta$ are the roots of the quadratic equation$x^{2}-(\alpha\!+\!\beta)x+\alpha\beta=0$  (cf. properties of quadratic equation) with the coefficients in $K(\alpha\!+\!\beta,\,\alpha\beta)$.  Thus

 $[K(\alpha,\,\beta):K]=[K(\alpha,\,\beta):K(\alpha\!+\!\beta,\,\alpha\beta)]\,[% K(\alpha\!+\!\beta,\,\alpha\beta):K]\leqq 2[K(\alpha\!+\!\beta,\,\alpha\beta):% K].$

Since  $[K(\alpha\!+\!\beta,\,\alpha\beta):K]$  is finite,  then also  $[K(\alpha,\,\beta):K]$ is, and in the finite extension (http://planetmath.org/FiniteExtension)  $K(\alpha,\,\beta)/K$  the elements $\alpha$ and $\beta$ must be algebraic over $K$.

 Title algebraic sum and product Canonical name AlgebraicSumAndProduct Date of creation 2013-03-22 15:28:03 Last modified on 2013-03-22 15:28:03 Owner pahio (2872) Last modified by pahio (2872) Numerical id 8 Author pahio (2872) Entry type Theorem Classification msc 11R32 Classification msc 11R04 Classification msc 13B05 Synonym sum and product algebraic Related topic FiniteExtension Related topic TheoryOfAlgebraicNumbers Related topic FieldOfAlgebraicNumbers