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 $K$