Could there exist some subfield of $\mathbb{R}$ such that $\pi$ is algebraic over that subfield?

Question

Could there exist a subfield $F$ of $\mathbb{R}$ such that $\pi$ is algebraic of, say degree 5 over $F$?

We know that $\pi$ is transcendental over $\mathbb{Q}$, hence $\mathbb{Q}(\pi)$ is not a field. right?

So, could there exist some other subfield of $\mathbb{R}$ such that $\pi$ is algebraic over that subfield?

Answer 1

Yes, consider the field $\mathbb{Q}[\pi^2]$, the subfield generated by $\pi^2$, it does not contain $\pi$ since $\pi$ is trancendental, $\pi$ is the solution of $X^2-\pi$.

Answer 2

The symbol $\mathbb Q(\pi)$ is a field by definition. It is the smallest field inside $\mathbb R$ that contains all of $\mathbb Q$ and also $\pi$. More concretely, you can think of it as all rational expressions $\frac{f(\pi)}{g(\pi)}$ where $f$ and $g$ are polynomials in one variable with coefficients in $\mathbb Q$ and $g$ is not the zero polynomial. We can be sure that the bottom is never zero for any choice of $g$ because $\pi$ is transcendental over $\mathbb Q$. So the inverse of $\frac{f(\pi)}{g(\pi)}$ is given by $\frac{g(\pi)}{f(\pi)}$ and you can see that it is a field.

The element $\pi$ is algebraic of degree $5$ over $\mathbb Q(\pi^5)$, just like an indeterminate $x$ is algebraic of degree $5$ over $\mathbb Q(x^5)$. As far as $\mathbb Q$ is concerned, it "understands" $\pi$ as little as it understands $x$. In it's mind, both satisfy no polynomial relations involving $\mathbb Q$.

Answer 3

By Eisenstein's criterion $X^5-\pi^5$ is irreducible in $\mathbb Q[\pi^5][X]$.

By Gauss's Lemma, $X^5-\pi^5$ is irreducible in $\mathbb Q(\pi^5)[X]$.

This implies that $\pi$ is algebraic of degree $5$ over $\mathbb Q(\pi^5)$.