Is there any intermediate fields between these two fields？ Is there any intermediate fields between the algebraic closure of the rational field and the complex field？
Can I get such a field by adding a transcendental number and then taking the algebraic closure over the complex field？
If so，how can I prove it？
 A: Hint: what is the cardinality of the algebraic closure of the rationals? And therefore, what is the cardinality of an extension of this field by a sinlge transcendental element? And the cardinality of the algebraic closure of the new obtained field?
On the other hand, what is the cardiniality of the complex numbers?
A: Yes, of course there are such fields.
Let $L = \overline ℚ ⊆ ℂ$ be the algebraic closure of $ℚ$ within $ℂ$. Then $L$ is algebraically closed and algebraic over $ℚ$. For any transcendental number $α ∈ ℂ$, the algebraic closure $L_α = \overline ℚ(α) ⊆ ℂ$ within $ℂ$ of $ℚ(α)$ is an algebraically closed field with $ℚ ⊆ L_α$. So $L ⊆ L_α$. Since $α$ is transcendental $α \notin L$, so the inclusion is strict.
You can do this construction with every subset $A ⊆ ℂ$ instead of just some $α ∈ ℂ$. Whenever $A$ contains transcendental elements, you end up with a proper field extension $L_A$ of $L = \overline ℚ$. In fact, you can get every algebraically closed intermediate extension $E / L$ this way, because you may take $A = E$, so $L_E = E$. However, there might be extensions $E / L$ that are not algebraically closed …
However, you can’t get all intermediate extensions by only doing this with single elements $α$ (or even finite or countable sets $A ⊆ ℂ$), as pointed out / hinted at by Pedro.
An alternative way to see that $L_α ≠ ℂ$ is to consider algebraically independent elements. Say, if $α, β ∈ ℂ$ are algebraically independent over $ℚ$, then $β \notin L_α$.
