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Let $g(x)$ be a polynomial in $\mathbb{Q}[x]$ such the $deg(g)>1$. I wish to show that $\mathbb{Q}$ are the only algebraic elements in $\mathbb{Q(g(\pi))}$.

I tried to show that $g(\pi)$ is not algebraic, yet even if I assume it, I don't know how to explain why the only elements in $\mathbb{Q}(\alpha)$ when $\alpha$ is not algebraic are $\mathbb{Q}$.

user5721565
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    Suppose that $g(\pi) $ is algebraic. Then there is a polynomial $f$ over the rationals from which is zero. Therefore $f(g(\pi)) =0$, which contradicts the fact of that $\pi$ is not algebraic. – Math.mx Jan 31 '19 at 17:35

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Hint: If $\alpha$ is a complex number that is transcendental over $\mathbb{Q}$ then $\mathbb{Q}(\alpha)$ is isomorphic to $\mathbb{Q}(x)$ i.e., to the field of rational functions over $\mathbb{Q}$.

Transcendental Extensions. $F(\alpha)$ isomorphic to $F(x)$

Dietrich Burde
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