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Prove that $\xi \in \mathbb{C}$ is an algebraic number if and only if the set $\{1, \xi, \xi^2, \xi^3, \dots\}$ is linearly dependent over $\mathbb{Q}$.

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    $\begingroup$ And how do you define an algebraic number? $\endgroup$
    – Berci
    Oct 3, 2015 at 19:47
  • $\begingroup$ Presumably the standard way, as a number that is the root of a polynomial with integer coefficients. $\endgroup$
    – Robert Soupe
    Oct 4, 2015 at 3:33

2 Answers 2

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If the set $B = \{1,\xi,\xi^2,\xi^3,\dots\}$ is $\Bbb Q$-linearly dependent, then there is some (finite!) $\Bbb Q$-linear combination:

$a_0 + a_1\xi + a_2\xi^2 +\cdots + a_n\xi^n = 0$, with not all $a_j = 0$.

Without loss of generality, by choosing $n$ to be the smallest natural number such that $\{1,\xi,\xi^2,\dots,\xi^n\}$ is $\Bbb Q$-linearly dependent, we can ensure $a_n \neq 0$.

Then $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$ is a non-zero polynomial in $\Bbb Q[x]$ such that $f(\xi) = 0$, that is, $\xi$ is algebraic over $\Bbb Q$.

On the other hand if $f(\xi) = 0$, for some non-zero polynomial $f(x) = c_0 + c_1x + c_2x^2 + \cdots + c_nx^n \in \Bbb Q[x]$, it follows that:

$0 = c_0 + c_1\xi + c_2\xi^2 + \cdots + c_n\xi^n$ is a non-zero linear combination of $\{1,\xi,\xi^2,\dots,\xi^n\}$, so this set is $\Bbb Q$-linearly dependent, and thus so is $B = \{\xi^k: k\in \Bbb N\}$ which contains it.

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A number $\alpha$ is algebraic over $F$ if, and only if, has a polinomial $f$ in $F[x]$ in a such way that $f(\alpha) = 0$. In particular, fix $n$, and supposing that this set is L.I. then the only solutions for $a_0\xi + \ldots a_n\xi^n = 0$ is $a_i = 0$ for every $i = 0,1,\ldots,n.$ What is contradictory once $\xi$ is algebraic and at least one of these coordinates has to be non-null. Now, let f(x) be the polynomial of degree $n$ that has $\xi$ as root. Then, $g(x) = f(x) + a_n\xi^{n+1}$ has $\xi$ as root. Proceeding by induction the claim holds.

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