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As the title says. I think a proof by contradiction is the most natural thing. Suppose $\pi \in Q(\pi^3)$. Then \begin{equation} \pi = \frac{a_n(\pi^3)^n+\cdots+a_1\pi^3+a_0}{b_m(\pi^3)^m+\cdots+b_1\pi^3+b_0} \text{.} \end{equation} Not sure how to proceed from here though.

I also have a related question, that is to show $\sqrt{2} \notin Q(\pi)$. I think if I can solve either one of them, I can solve the other. Any help would be much appreciated.

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Multiply the denominator over, and show that there is no rational coefficient $c$ such that $\pi^{3m+1} = c \pi^{3n}$ for any $m,n \in \Bbb Z$. –  Arkamis Dec 17 '12 at 21:54
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Are you familiar with the fact that $\pi$ is transcendental? To solve this exercise you must know that, me thinks. –  Jyrki Lahtonen Dec 17 '12 at 21:54
    
@JyrkiLahtonen Yes, I do know that $\pi$ is transcendental. –  Aden Dong Dec 17 '12 at 21:58
    
@EdGorcenski Why is this sufficient for showing the problem? –  Aden Dong Dec 17 '12 at 21:58
    
Ok. Then you can follow Hagen's solution (+1) below. –  Jyrki Lahtonen Dec 17 '12 at 22:00
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3 Answers 3

up vote 15 down vote accepted

From $$\pi = \frac{a_n\pi^{3n}+\ldots+a_1\pi^3+a_0}{b_m\pi^{3m}+\ldots +b_1\pi^3+b_0}$$ with $a_n\ne0$, $b_m\ne0$ we obtain a polynomial equation for $\pi$: $$\tag1(b_m\pi^{3m+1}+\ldots +b_1\pi^4+b_0\pi)-(a_n\pi^{3n}+\ldots+a_1\pi^3+a_0)=0.$$ If $n>m$, this is of degree $3n$ with leading coefficient $-a_n\ne 0$, if $n\le m$ this is of degree $3m+1$ with leading coefficient $b_m\ne0$. Hence $(1)$ shows that $\pi$ is algabraic, which it isn't.


From $$\sqrt 2=\frac{a_n\pi^{n}+\ldots+a_1\pi+a_0}{b_m\pi^{m}+\ldots +b_1\pi+b_0},$$ we obtain $$2=\frac{a_n^2\pi^{2n}+\ldots+2a_0a_1\pi+a_0^2}{b_m^2\pi^{2m}+\ldots +2b_0b_1\pi+b_0^2},$$ hence $$\tag2(a_n^2\pi^{2n}+\ldots+a_0^2)-2(b_m^2\pi^{2m}+\ldots +b_0^2)=0.$$ Since $\pi$ is transcendental, this must be the zero polynomial, i.e. everything cancels. Especially, we must have $n=m$ and $a_n^2-2b_m^2=0$. But then $\sqrt 2=\left\vert\frac{a_n}{b_m}\right\vert\in\mathbb Q$.

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Awesome! This is such a simple proof that I can't believe I missed it. Thanks a lot. –  Aden Dong Dec 17 '12 at 22:01
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The following general result might interest you.
If $k$ is a field and if $\phi(x)=\frac {f(x)}{g(x)}\in k(x)$ is a rational function with $f(x),g(x)\in k[x]$ relatively prime polynomials (not both constant), then the extension of fields $k(\phi (x))\subset k(x)$ has degree $$[k(x):k(\phi (x))]=\text {max}\:(\text {deg} \; f(x),\text {deg} \; g(x))$$ This of course implies (if you know that $\pi$ is transcendental and thus may play the role of the indeterminate $x$) that $[\mathbb Q(\pi):\mathbb Q(\pi^3]=3$ and thus a fortiori that $\pi \notin \mathbb Q(\pi^3)$ .

Bibliography
The displayed formula can be found in Theorem 8.36, page 614 of Jacobson's Basic Algebra II.

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I will prove the problem as a simple case of what Georges describes.

Let $k$ be a field. Let $L = k(x)$ be the rational function field with one variable $x$(we may take $x = \pi$ when $k = \mathbb{Q}$). Let $K = k(x^3)$. Then $L = K(x)$. Let $X^3 - a \in K[X]$, where $a = x^3$. Clearly $X^3 - a$ cannot have a root in $K$ considering the degree of a root if any. Hence it is irreducible over $K$. Hence $[L\colon K] = 3$.

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