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# 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 factors here are finite, 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 $K(\alpha,\,\beta)/K$ the elements $\alpha$ and $\beta$ must be algebraic over $K$.

## Mathematics Subject Classification

11R32*no label found*11R04

*no label found*13B05

*no label found*

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