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Show $[\mathbb{Q}(\pi):\mathbb{Q}(\pi^3)]$ is finite and give a basis for $\mathbb{Q}(\pi)$ over $\mathbb{Q}(\pi^3)$.

I'm not sure how to approach this problem. I know that if $u$ is algebraic of a field $F$ then $[F(u),F]=n$ and a basis is $\{1,u,\dots,u^{n-1}\}$. But I don't think I can apply that here since clearly $\mathbb{Q}\ne \mathbb{Q}(\pi^3)$. So how can I show this?

Burgundy
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    Which element do you need to adjoint to $\mathbb{Q}(\pi^3)$ such that you get $\mathbb{Q}(\pi)$? As soon as you got this, you can apply the theory you've written. – Severin Schraven May 04 '16 at 15:51
  • Oh! $\mathbb{Q}(\pi^3)(\pi) = \mathbb{Q}(\pi)$ so I can apply that theorem, thank you – Burgundy May 04 '16 at 16:00
  • Similar question: https://math.stackexchange.com/questions/1713841/extension-of-mathbbq-over-which-pi-is-algebraic-of-degree-3 – 19021605 Jun 24 '25 at 08:14

2 Answers2

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Hint: $\pi$ is the root of $X^3-\pi^3\in Q(\pi^3)[X]$ so$Q(\pi)$ is a finite algebraic extension of $Q(\pi^3)$ since it is a subfield of the splitting field of $X^3-\pi^3\in Q(\pi^3)[X]$.

Tsemo Aristide
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Actually, in order to prove that $[\mathbb Q(\pi):\mathbb Q(\pi^3)]=3$ you need some facts: specifically that, if $\alpha$ is transcendental over $K$ field, then $K[\alpha]\cong K[X]$, the ring of polynomials with coefficients in K, and $K(\alpha)\cong K(X)$, its field of fractions; so, since $\pi^3$ is transcendental over $\mathbb Q$ (otherwise, $\mathbb Q(\pi^3)/\mathbb Q$ would be finite, and $\mathbb Q(\pi)/\mathbb Q$ is finite => $\mathbb Q(\pi)/\mathbb Q$ finite => $ \pi$ algebraic over $\mathbb Q$, absurd), this gives you the isomorphism $\mathbb Q(\pi^3)\rightarrow\mathbb Q(X),\pi^3\rightarrow X$, which extends to an isomorphism $\phi:\mathbb Q(\pi)\rightarrow \mathbb Q(\xi),\pi\rightarrow\xi, $where $\xi$ is a cubic root of X in an algebraic closure of $\mathbb Q(X)$. So, $\pi\in\mathbb Q(\pi^3)\iff \xi\in\mathbb Q(X)$; but, if that was the case, you should have $\xi$=P/Q, with P,Q$\in \mathbb Q[X]$, $P,Q\neq 0$ (if P=0, then $\pi=0$ through $\phi$,absurd), and so deg(P)=n, deg(Q)=m, and $X=\xi^3=P^3/ Q^3$ $\iff XQ^3=P^3$ $\Rightarrow 3m+1=3n$, which is absurd. So, $\pi\notin\mathbb Q(\pi^3)$, and , since 3 is prime, it must be that $[\mathbb Q(\pi): \mathbb Q(\pi^3)]=3$, thus $(1,\pi,\pi^2)$ is a basis of $\mathbb Q(\pi)/\mathbb Q(\pi^3)$.

PaulTaylors
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