# Let $\alpha, \beta \in \mathbb{C}$, such that $\alpha + \beta$, and $\alpha\beta$ are algebraic. Show that $\alpha$ and $\beta$ are algebraic.

Let $\alpha, \beta \in \mathbb{C}$, such that $\alpha + \beta$, and $\alpha\beta$ are algebraic. Show that $\alpha$ and $\beta$ are algebraic.

attempt: Suppose $\alpha, \beta \in \mathbb{C}$ such that $\alpha + \beta$, and $\alpha\beta$ are algebraic.

Let $\alpha + \beta$, $\alpha\beta$ are algebraic be algebraic over $F$. And let $\alpha, \beta \in K$, where $K$ is an extension of $F$. then $\alpha + \beta$ and $\alpha\beta$ lie in the extension $K = F(\alpha,\beta)$, which is finite over $F$.

so $[F(\alpha, \beta) : F] = [F(\alpha, \beta): F(\alpha + \beta, \alpha\beta)][F(\alpha + \beta, \alpha\beta) : F] \leq 2[F(\alpha + \beta, \alpha \beta) : F]$, since $\alpha + \beta$ and $\alpha \beta$ are roots .

So $[F(\alpha + \beta) : F]$ is finite and in the extension $F(\alpha,\beta)/F$ the elements must be algebraic over $F$, then $\alpha , \beta$ are algebraic.

Can someone please verify this? Or help me give a better approach. Thank you!

• Another simpler approach, assuming that you already know that the algebraic numbers are 'closed' (in the sense that a value that's "algebraic over algebraics" is algebraic itself): what are the roots of $P(x)=x^2-(\alpha+\beta)x+\alpha\beta$? – Steven Stadnicki Nov 29 '15 at 6:34
• The roots would be $\alpha$ and $\beta$. – Mahidevran Nov 29 '15 at 6:49

As long as the characteristic of the field $F$ is not $2$, there is a simpler? argument.
Aside from addition and multiplication, the set of numbers algebraic over a field $F$ is closed under square root. This is because if $P(\gamma) = 0$ for some $P \in F[x]$, then $Q(\sqrt{\gamma}) = 0$ for $Q(x) = P(x^2) \in F[x]$.
This implies up to the sign, $\alpha-\beta = \sqrt{(\alpha+\beta)^2 - 4\alpha\beta}$ is algebraic over $F$. As a corollary, $\alpha = \frac12 \left((\alpha+\beta)+(\alpha-\beta)\right)$ and $\beta = \frac12 \left((\alpha+\beta)-(\alpha-\beta)\right)$ are algebraic over $F$.
Hint: $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$.
• Your approach is fine, though excessively complicated. Your comment also finishes the problem, more efficiently than mine, but I could not resist mentioning the identity. Your complicated approach implicitly uses the fact that $\alpha$ and $\beta$ are roots of the quadratic, so why not just go with that? – André Nicolas Nov 29 '15 at 6:22
• Just go with $\alpha$ and $\beta$ are the roots of the quadratic $x^2-(\alpha+\beta)x+\alpha\beta=0$. Here I am assuming that you have the theorem that algebraic over algebraic is algebraic. – André Nicolas Nov 29 '15 at 6:33