TITLE: Prove that there exists no algebraically closed field $F$ such that $\bar{\mathbb{Q}} \subsetneq F \subsetneq \overline{\mathbb{Q}(\pi)}$
QUESTION [2 upvotes]: I'm studying chapter 7.6 Splitting field and Algebraic closure in Abstract Algebra by S. Lovett.
The above is one of its exercises and I'm not sure that I can prove it only with the materials that I have read.
I know that the above exercise is easy if we consider transcendence degree, but we didn't learn anything about it.
Here is one of my attempts.
Suppose that such $F$ exists.
Then there exists a transcendental element $\beta \in F$ over $\mathbb{Q}$, and $\mathbb{Q}(\beta) \subseteq F$.
Now we have an isomorphism $\mathbb{Q}(\pi) \rightarrow \mathbb{Q}(\beta)$.
Considering their algebraic closures, we have $\overline{\mathbb{Q}(\pi)}$ an algebraic closure of $\mathbb{Q}(\pi)$ and an embedding $f:\mathbb{Q}(\pi) \rightarrow \overline{\mathbb{Q}(\beta)}$ given by the above isomorphism.
So there exists an embedding $\lambda : \overline{\mathbb{Q}(\pi)} \rightarrow \overline{\mathbb{Q}(\beta)}$ extending $f$.
Since $\overline{\mathbb{Q}(\pi)}$ is algebraically closed and $\overline{\mathbb{Q}(\beta)}$ is algebraic over $f(\mathbb{Q}(\pi))={\mathbb{Q}(\beta)}$,
$\lambda (\overline{\mathbb{Q}(\pi)})$ is algebraically closed and $\overline{\mathbb{Q}(\beta)}$
is algebaic over $\lambda (\bar{\mathbb{Q}(\pi)})$.
So as an algebraic extension of an algebraically closed field $\overline{\mathbb{Q}(\beta)}=\lambda (\overline{\mathbb{Q}(\pi)})$.
Therefore $\lambda$ is actually an isomorphism.
I once thought that I have found the reverse inclusion $\overline{\mathbb{Q}(\pi)} \subseteq \bar{F}=F$, but I just found that $\overline{\mathbb{Q}(\beta)}$ and $\overline{\mathbb{Q}(\pi)}$ are isomorphic, not equal, so it does not mean anything.
I know that not all of you can access to the textbook that I have, so I just want to learn various ways to prove this without using transcendence degree, or a proof that is accessible for an undergraduate student who just started to learn about field extensions.

REPLY [1 votes]: What you are doing (showing $\overline{\Bbb Q(\pi)}$ is isomorphic to a subfield of $F$) is not what the question is asking.
Suppose $\overline {\Bbb Q} \subset F \subset \overline{\Bbb Q(\pi)}$. 
You want to show that $\overline{\Bbb Q(\pi)} \subset F$ or $F \subset \overline {\Bbb Q}$. 
Since $F$ is algebraically closed, to show $\overline{\Bbb Q(\pi)} \subset F$ it is enough to show $\pi \in F$.
Suppose $F \neq \overline{\Bbb Q}$. Then there is an element $x \in \overline{\Bbb Q(\pi)}$ such that $x \in F$ and $x \notin \overline {\Bbb Q}$. And since $F$ might as well just be $\overline{\Bbb Q(x)}$, to show that $\pi \in F$ is the same as showing that $\pi \in \overline{\Bbb Q(x)}$.
However, given that $\pi$ and $x$ aren't in $\overline {\Bbb Q}$, $\pi \in \overline{\Bbb Q(x)}$ and $x \in \overline {\Bbb Q(\pi)}$ are both equivalent to the existence of a polynomial $P$ in two variables with rational coefficients such that $P(x,\pi)= 0$, and so they are equivalent.
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