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This question seems easy and I feel like it should be true, but as of yet I have not been able to convince myself one way or the other, so I figure someone around here knows for sure.

Question: Let $\alpha = U+iV$ be a root of some $P\in\mathbb{Q}[z]$ so that $\alpha$ is algebraic. Is it true that $U$ and $V$ are necessarily algebraic as well?

Writing $z=x+iy$, one has $P(z) = R(x,y)+iS(x,y)$ for some $R,S\in\mathbb{Q}[z,w]$ and in particular:

$R(U,V) = S(U,V) = 0$

From this it is clear that if either $U$ or $V$ is algebraic then both are, but it is not clear whether it is possible for both $U$ and $V$ to be algebraically dependent transcendentals.

In particular, any algebraic number has a unique monic minimal polynomial; is there an analogous statement that arbitrary sets of algebraically dependent numbers have a unique minimal polynomial? If so, then unless I am mistaken such a result answers the main question in the affirmative.

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1 Answer 1

up vote 14 down vote accepted

Algebraic numbers form a field and they are closed under conjugation, therefore $U=(\alpha+\bar\alpha)/2$ and $V=(\alpha-\bar\alpha)/(2i)$ are algebraic.

Reference: Lang's Algebra (Springer 2002), Chapter V, Proposition 1.6

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I knew there had to be some too-simple argument that I was missing. Thanks. –  ARupinski Apr 18 '11 at 3:57

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