Sum and product of algebraic numbers

Question

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 $K$.. — , 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
Does this answer your question? How to prove that the sum and product of two algebraic numbers is algebraic? — user557, Mar 30 '21 at 20:37

score 6 · Accepted Answer · answered Mar 10 '16 at 19:48

6

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.

answered Mar 10 '16 at 19:48

Dietrich Burde

130,978

1

Is this still valid if the field extensions of $K$ are infinite fields? – hhh Mar 11 '16 at 10:27
2

Yes, think for example of $K=\mathbb{Q}$, which is an infinite field. – Dietrich Burde Mar 11 '16 at 13:14

score 1 · Answer 2 · answered Mar 10 '16 at 17:01

1

$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$

$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$

$\alpha=\dfrac{\alpha+\beta+\alpha-\beta}{2}$

$\beta=\dfrac{\alpha+\beta-(\alpha-\beta)}{2}$

answered Mar 10 '16 at 17:01

Sum and product of algebraic numbers

2 Answers2

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