Are $\mathbb{Q}(π)$ and $\mathbb{Q}(π^2)$ the same field?

Question

I've tried to check if $\mathbb{Q}(π)$ and $\mathbb{Q}(π^2)$ are equal fields.

I know there exists a field isomorphism between $\mathbb{Q}(π)$ and $\mathbb{Q}(π^2)$ since $π$ and $π^2$ are transcendental over $\mathbb{Q}$.

I also know that $π^2 = π\cdot π$, but I can't see the other implication to get the equality.

Thanks.

Answer 1

If $L/K$ is a field extension and $x \in L$ is trancendental over $K$, then $K(x^2)$ is a proper subfield of $K(x)$. In fact, otherwise $x \in K(x^2)$, so that we get an equation $x=f(x^2)/g(x^2)$ with $f,g \in K[T]$ (with $g \neq 0$). Then $x \cdot g(x^2)=f(x^2)$ and hence, since $x$ is transcendental, $T \cdot g(T^2)=f(T^2)$. The left side has odd degree, the right side even degree, contradiction.

Actually, a much stronger statement holds: When $p \in K(x) \setminus K$, then $K(x)$ is finite over $K(p)$, and the degree is $\max(\deg(p_1),\deg(p_2))$ when $p = p_1/p_2$ with coprime $p_1,p_2 \in K[T]$. Thus, we have $K(x)=K(p)$ iff $\max(\deg(p_1),\deg(p_2))=1$, which means that $p_1,p_2$ are constant or linear, where one has to be linear.