Is $\mathbb{Q}(\pi)$ a simple extension of $\mathbb{Q}\left(\frac{\pi^3}{1+\pi}\right)$?

Question

In the case of an algebraic extension, I could think easier than this case. But I got stuck in this problem. I know that the dimension of $\mathbb{Q}(\pi)$ over $\mathbb{Q}\left(\frac{\pi^3}{1+\pi}\right)$ is $3$. So I tried to find the irreducible polynomial $$\operatorname{irr}\left(a,\mathbb{Q}\left(\tfrac{\pi^3}{1+\pi}\right)\right)$$ where the dimension is equal to $\deg\left(a,\mathbb{Q}\left(\frac{\pi^3}{1+\pi}\right)\right)$. But I've failed.

Answer 1

Let $K=\mathbb{Q}(\frac{\pi^3}{1+\pi})$. Clearly $\pi$ satisfies the relation $$\frac{x^3}{x+1}-\frac{\pi^3}{\pi+1}=0$$ Therefore $\pi$ is a root of the following irreducible polynomial of degree $3$: $$x^3+\left(\frac{\pi^3}{1+\pi}\right)x+\left(\frac{\pi^3}{1+\pi}\right)\in K[x]$$ And to answer the question in the title of your post, yes, $\mathbb{Q}(\pi)$ is a simple extension of $K$ because there exists an $a\in\mathbb{Q}(\pi)$ such that $$\mathbb{Q}(\pi)=K(a)$$ One obvious example is $a=\pi$, though there are infinitely many other $a$ that work too.