Every element in $\mathbb{Q}[\sqrt[5]{2}]$ has a multiple in $\mathbb{Q}_+[\sqrt[5]{2}]$

Question

I need help understanding this theorem. My professor gave it to me and I can't find anything about it online. The topic is Algebraic numerical fields (I couldn't find the tag in the list of topics).

Let $a= \sqrt[5]{2}$, that is, $a^5 = 2$.

Consider the algebraic number field $$F = \mathbb Q[a] = \{\alpha_0 + \alpha_1 a + \alpha_2a^2 + \alpha_3a^3 +\alpha_4a^4 \mid \alpha_i \in \mathbb Q\}$$ (check that F is a subfield of R)

and $$P = \{\alpha_0 + \alpha_1 a + \alpha_2a^2 + \alpha_3a^3 +\alpha_4a^4\mid \alpha_i \in \mathbb{Q}_+\}.$$ P is a subfield of F

Prove the theorem: for any $f \in F$, there exists a $g \in P$ such that $fg \in P$ or $-fg ∈ P$.

I tried solving this on my own but I'm failing to even start. Any assistance will be appreciated.

Answer 1

A variation on Batominovski's comment:

First note that there are rational numbers $r$ arbitrarily close to $a$.

Then inspect the coefficients of $1, a,a^2,a^3\text{ en } a^4$ for $(1+\frac ar+\frac{a^2}{r^2}+\frac{a^3}{r^3}+\frac{a^4}{r^4})\,f$ and their limits as $r\to a$.