# Sum and product of algebraic numbers

I have encountered the following problem:

Let $L/K$ be a field extension and let $\alpha, \beta \in L$. Show that $\alpha$ and $\beta$ are algebraic over $K$ if and only if $\alpha + \beta$ and $\alpha\beta$ are algebraic over $K$.

For the forward implication I have already seen several ways to show it, but I cannot think of a way for the other direction.

• To be precise, you want to prove if $\alpha+\beta$ and $\alpha\beta$ are algebraic then $\alpha$ and $\beta$ are algebraic over $K$.. – user87543 Mar 10 '16 at 16:48
• Forwards, other...these are confusing terms: what exactly have you done and what you still have to do? – DonAntonio Mar 10 '16 at 17:00

Since $a$ and $b$ are the roots of the polynomial $$f(x)=x^2-(a+b)x+ab,$$ $[K(a,b):K(a+b,ab)]$ is finite. Hence we have $$[K(a,b):K]=[K(a,b):K(a+b,ab)] \cdot [K(a+b,ab):K].$$ Assuming that $ab$ and $a+b$ are algebraic, the right hand side is finite, and hence also the left hand side, so that $a,b$ are algebraic.
• Is this still valid if the field extensions of $K$ are infinite fields? – hhh Mar 11 '16 at 10:27
• Yes, think for example of $K=\mathbb{Q}$, which is an infinite field. – Dietrich Burde Mar 11 '16 at 13:14
$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$
$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$
$\alpha=\dfrac{\alpha+\beta+\alpha-\beta}{2}$
$\beta=\dfrac{\alpha+\beta-(\alpha-\beta)}{2}$