2

I have a "simple" question about field extension, let me say, $\mathbb{Q}(\pi^2-\pi) \subsetneq\mathbb{Q}({\pi})??$

I know that $\pi^2-\pi \in \mathbb{Q}(\pi)$ so we have $\mathbb{Q}(\pi^2-\pi) \subset\mathbb{Q}({\pi})$, but why $\pi \notin \mathbb{Q}(\pi^2-\pi)$?

I've tried this way: If $\pi \in \mathbb{Q}(\pi^2-\pi)$ then $\mathbb{Q}(\pi^2) \subset \mathbb{Q}(\pi^2-\pi), $ but this is ok, since $\mathbb{Q}(\pi^2) \subset \mathbb{Q}(\pi)$.

Another way is: If $\pi \in \mathbb{Q}(\pi^2-\pi) $ then $\pi = \frac {p(\pi^2 - \pi)}{q(\pi^2-\pi)}$, what is the contradiction?

I cannot see the contradiction, but my professor told me that $[\mathbb{Q}(\pi): \mathbb{Q}(\pi^2-\pi)]=2$ (I have no ideia how to prove that).

Joãonani
  • 1,684

1 Answers1

2

Assume by contradiction that $\pi = \frac {p(\pi^2 - \pi)}{q(\pi^2-\pi)}$

Then $\pi$ is a root of $R(X)=XQ(X^2-X)-P(X^2-X) \mathbb Q[X]$. Since $\pi$ is transcendental over $\mathbb Q$ you get that $R=0$. Therefore $$ XQ(X^2-X)=P(X^2-X) $$ But this is not possible since the left-hand side has odd degree while the right-hand side has even degree.

N. S.
  • 132,525
  • ok, It makes sense. Do you have any idea about $[\mathbb{Q}(\pi): \mathbb{Q}(\pi^2-\pi)]=2?$ – Joãonani Dec 02 '20 at 02:46
  • @Joãonani That is "obvious". Indeed, $\pi$ is the root of $X^2-X - (\pi^2-\pi)$ which is a quadratic polynomial with coefficients in $\mathbb{Q}(\pi^2-\pi)$. This is irreducible since otherwise the degree of the extension would be 1. – N. S. Dec 02 '20 at 02:57
  • Oh right, I got it. – Joãonani Dec 02 '20 at 03:00