Do we know if all simple extensions of the field of rational numbers by transcendental numbers are not equal?

Question

I understand that $\mathbb{Q}(x) \cong \mathbb{Q}(u)$ for all transcendental $u$, where $\mathbb{Q}(x)$ is the field of rational forms over $\mathbb{Q}$ and thus that all simple extensions of the rational numbers by transcendental ones are isomorphic to one another, but is it true, proven, that these extensions are not all $equal$ to each other?

I understand that that would be an incredible thing to prove, and could simplify many problems in transcendental number theory if true, so is it possible for it to be? $Could$ these extensions all equal each other? Or is that, easily or otherwise, proven to be an impossible task? I recall that transcendental numbers are uncountable, whereas I would assume that $\mathbb{Q}(x)$ is, though I have not enough knowledge of set theory to settle that in my mind.

Answer 1

This is not possible. If it were possible then $\mathbb{R}$ would be a vector space of countable dimension over the algebraic closure of $\mathbb{Q}$, which is not possible because the algebraic closure of $\mathbb{Q}$ is countable.

Answer 2

Explicitly, $\mathbb Q(\pi)\ne\mathbb Q(\pi^2)$ as $\pi\notin\mathbb Q(\pi^2)$. The latter follows because $\pi=\frac{p(\pi^2)}{q(\pi^2)}$ would lead to the nontrivial polynomial equation $p(\pi^2)-\pi q(\pi^2)=0$.

Answer 3

There is a criterion under which $\mathbb Q(u) = \mathbb Q(v)$; it's just algebraic in nature, which makes it difficult to determine for analytically defined objects such as $\pi$ and $e$.

First of all, in order for the two to be equal, they must be contained in each other, which means each generator must be in the other set; specifically, let's say we write $v \in \mathbb Q(u)$, so that $v = p(u)/q(u)$ for polynomials $p$ and $q$ with no common factors.

Then $\mathbb Q(u) = \mathbb Q(v)$ if and only if $p$ and $q$ are both at most degree $1$, with at least one of them nonconstant.