Is $\Bbb Q[\alpha]$ a field ? Here $\alpha \in \Bbb C$ is such that $\alpha + \pi = {\pi}^2 \alpha $ .
I tried to argue by contradiction. Tried to come up with tricks commonly used for concluding that if $\theta$ is transcendental, then so is $\theta^2$, so is $\sqrt{\theta}$ and so on. But couldn't come up with a suitable one!
Any hint will be really appreciated.
Given that $\alpha+\pi=\pi^2\alpha$, by basic algebra $\alpha=\frac{\pi}{\pi^2-1}$. This is transcendental because $\pi$ is. After all, if $\alpha$ were algebraic then $f(\alpha)=0$ for some $f\in\Bbb{Q}[X]$, but then $\pi$ is a root of $$(X^2-1)^nf(X)\in\Bbb{Q}[X],$$ where $n:=\deg f$, contradicting the fact that $\pi$ is transcendental.