# Simplify $\mathbb{Q}(\pi^3+\pi^2, \pi^8+\pi^5)$

My goal is to write $\mathbb{Q}(\pi^3+\pi^2, \pi^8+\pi^5)$ as $\mathbb{Q}(f(\pi))$ where $f \in \mathbb Q(X)$.

I have $(\pi^3+\pi^2)^2 = \pi^6+2\pi^5+\pi^4$ and then $2\pi^8 - \pi^6 - \pi^4$ is in my field. But I don't think it generates it.

• Use Euclidean algorithm on $\pi^3+\pi^2,\pi^8+\pi^5$. – Wolfram Feb 2 '17 at 20:07
• If $a=\pi^3+\pi^2$ and $b=\pi^8+\pi^5$, then $b^3 + (-12a^2 + 3a)b^2 + (8a^5 + 15a^4)b + (-a^8 - 3a^7 - 3a^6)=0$ and $b$ has degree $3$ over ${\mathbb Q}[a]$. – Ewan Delanoy Feb 2 '17 at 20:08
• $\frac{\pi^8 +\pi^5}{\pi^3+\pi^2} = \pi^3\frac{\pi^3+1}{\pi+1} = \pi^5 -\pi^4 + \pi^3$ – kotomord Feb 2 '17 at 20:21
• @EwanDelanoy Well, all fields are rings, I quite didn't got what you wanted to say. – Wolfram Feb 2 '17 at 20:22
• @Alphonse This is false. $\pi^3+\pi^2$ is clearly not in ${\mathbb Q}(\pi^3)$ (because the elements of ${\mathbb Q}(\pi^3)$ only contain monomials of the form $\pi^{3k}$, with the exponent divisible by $3$) – Ewan Delanoy Feb 2 '17 at 20:39

Let $a=\pi^3+\pi^2$ and $b=\pi^8+\pi^5$. Then
$$\pi=\frac{3b(1-a)+(a^4+3a^3)}{b(a-3)+(3a^3+3a^2)}$$
so you can take $f(\pi)=\pi$. How did I obtain this formula ? I tried to find the minimal polynomial of $\pi$ over ${\mathbb Q}(a,b)$, by applying the Euclidean algorithm to $X^3+X^2-a$ and $X^8+X^5-b$.
The last nonzero term produced by this algorithm is $3b(1-a)+(a^4+3a^3)-X(b(a-3)+(3a^3+3a^2))$.
Let $\alpha = \pi^3+\pi^2$ and $\beta = \pi^8+\pi^5$. It can be shown that if $f = \frac{p}{q}$ where $\gcd(p,q)=1$, we have $$[\mathbb{Q}(\pi):\mathbb{Q}(f(\pi))] = \max\{\deg(p), \deg(q)\}.$$ Therefore $[\mathbb{Q}(\pi):\mathbb{Q}(\pi^3+\pi^2)]=3$, and if $\pi \notin \mathbb{Q(\alpha, \beta)}$, we must have $\mathbb{Q(\pi)}:\mathbb{Q(\alpha, \beta)}=3$, since $$[\mathbb{Q}(\pi):\mathbb{Q}(\alpha)] = [\mathbb{Q}(\pi):\mathbb{Q}(\alpha, \beta)][\mathbb{Q}(\alpha, \beta):\mathbb{Q}(\alpha)].$$
This means that $\beta \in \mathbb{Q}(\alpha)$, or $\pi^8+\pi^5 = \frac{g}{h}$ where $g,h \in \mathbb{Q}[\alpha]$ and are relatively prime. So $\pi^5|g$ which is impossible, because if $\pi^k |g$, $k$ must be even. So $\pi \in \mathbb{Q}(\alpha, \beta)$.