$\pi$ does not lie in any quadratic extension of $\mathbb{Q}$

Question

Knowing that $\pi^2$ is irrational:

How can we prove that $\pi$ does not lie in any quadratic extension of $\mathbb{Q}$ ?

Without using that $\pi$ is transcendent.

Any hints would be appreciated.

Answer 1

If all we know about $\pi$ is that $\pi^2$ is irrational, then we can't conclude that $\pi$ isn't quadratic over $\Bbb{Q}$. After all, $(\sqrt{2}+1)^2$ is irrational, but $\sqrt{2}+1$ generates a quadratic extension.

Answer 2

Hints:

** Any finite extension of a field is an algebraic extension of that field

** Any element of an algebraic extension of a field is algebraic over that field

** Any element of an extension of a field of degree $\,n\,$ is a root of a non-zero polynomial over that field of degree at least $\,n\,$