Minimal polynomial of $\pi$ over a field

Question

This is a question from my past Qual exams.

"Determine the minimal polynomial of $\pi$ over the field $\mathbb{Q}((\pi^2+1)/(\pi-1))$."

I do not know how to approach this kind of problem. If the problem is find polynomial of $a$ over the field $\mathbb{Q}(b)$ where $a,b$ are algebraic, I will solve it by taking $[\mathbb{Q}(a,b):\mathbb{Q}]/[\mathbb{Q}(b):\mathbb{Q}]$. Of course, this way is not effective here.

I read books of Dummit, Foote and Escofier. They do not mention this kind of problem.

Can you tell me how to approach this?

Answer 1

Let $\alpha=(\pi^2+1)/(\pi-1)$ and let $R=\mathbb Q(\alpha)$. Observe that $\pi$ is a root of the quadratic polynomial $$ p(x)=x^2-\alpha(x-1)+1\in R[x]. $$ Thus, the minimal polynomial of $\pi$ (over $R$) divides $p(x)$. On the other hand, it is straightforward to show that $\pi\not\in R$, and thus the minimal polynomial of $\pi$ has degree greater than $1$, which implies that it equals $p(x)$.

Answer 2

Let $r=\frac {\pi^2+1}{\pi-1} $. By "undoing" this number you can get that $$p (x)=r (x-1)-1-x^2$$ satisfies $p (\pi)=0$. It will be minimal if you see that the minimal polynomial cannot be of degree one, which amounts to say that $\pi $ is not in your field.