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