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Let $F = \mathbb{Q}(\pi^3)$. Find a basis for $F(\pi)$ over $F$.

How can I solve this. Can anyone help me please. Thanks.

agdum
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  • Similar question: https://math.stackexchange.com/questions/1771469/show-mathbbq-pi-mathbbq-pi3-is-finite – 19021605 Jun 24 '25 at 08:16

3 Answers3

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Hint: The number $\pi$ is transcendental. So this is the same problem as finding a basis for $F(x)$ over $F(x^3)$.

André Nicolas
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  • yes,I know that. but I am not sure how to find a basis for $F(x)$ over $F(x^3)$. Is it {$1,x,x^2$} ? – agdum Apr 25 '13 at 05:24
  • Yes, that's the most natural one. – André Nicolas Apr 25 '13 at 05:25
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    You are welcome. Good, so you had no trouble with the $\pi$ part. That was intended to be the tricky part of the problem. – André Nicolas Apr 25 '13 at 05:28
  • @AndréNicolas can u pls explain me how does the basis of F(x) over F(x^3) is {1,x,x^2}. I don‘t know the ans. And what is the basis of F(x^3) over F(x) also. Is there any? Pls explain in detail both the two cases. Thanking u. – avijit kundu Jun 11 '15 at 04:07
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Note that $\pi$ is algebraic over $\mathbb Q(\pi^3)$ with minimal polynomial $x^3-\pi^3$ (why?). Thus $\frac{\mathbb Q(\pi^3)[x]}{<x^3-\pi^3>}$ $\cong$ $\mathbb Q(\pi^3)(\pi)$ and $\{1,\pi,\pi^2\}$ is a basis for $\mathbb Q(\pi^3)(\pi)$ over $\mathbb Q(\pi^3)$.

The theorem behind this is as followed:

" Let $\mathbb F$ be a field and let $p(x)\in\mathbb F[x]$ be irreducible over $\mathbb F$. If $a$ is a zero of $p(x)$ in some extension $\mathbb E$ of $\mathbb F$, then $\mathbb F(a)$ is isomorphic to $\frac{\mathbb F[x]}{<p(x)>}$. Furthermore, if $deg\{p(x)\}=n$, then every member of $\mathbb F(a)$ can be uniquely expressed as $$c_{n-1}a^{n-1}+c_{n-2}a^{n-2}+......c_{1}a+c_{0},$$

where $c_{0},c_{1},....,c_{n-1}\in\mathbb F$ "

So you can say that if $a$ is a zero of an irreducible polynomial over $\mathbb F$ of degree $n$, then the set $\{1,a,...,a^{n-1}\}$ is a basis for $\mathbb F(a)$ over $\mathbb F$.

Mansi Tyagi
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Finding a basis for $F(x)$ over $F(x^3)$.

This means you need to find a basis $v_1(x),\cdots,v_n(x)$ such that $p_1(x)v_1(x)+p_2(x)v_2(x)+\cdots+p_n(x)v_n(x)$ generates any polynomial of degree $n\in\mathcal{N}$ in $F(x)$. But $p_i(x)\in F(x^3)$ for any $i$, then $p_i(x)=a_nx^{3n}+a_{n-1}x^{3n-3}+\cdots+a_0$ where $a_i\in F$.

It is clear that $\{v_1(x),v_2(x),v_3(x)\}=\{1,x,x^2\}$ is the easiest answer.

yzlstat
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