Algebraically closed field extension can be not algebraic?

Question

By definition, one says that an algebraic closure $K$ of a given field $F$ is an algebraic extension of $F$ who is algebraically closed. This makes me curious about whether there exists an extension $K'$ of $F$ who is algebraically closed, while $K'$ is not an algebraic extension?

Answer 1

Sure, this occurs naturally. Consider $F = \mathbb Q$ and its algebraic closure $K = \bar{\mathbb Q}$. Now consider $\mathbb Q$ as a subfield of $K' = \mathbb C$.

$\mathbb C$ is algebraically closed, its an extension of $\mathbb Q$ but it is not algebraic over $\mathbb Q$.

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edit: Here is another example. Let $F$ be a finite field and let $K$ be its algebraic closure. Let $F \subseteq F'$ be an extension of $F$ of sufficiently large cardinality (i.e. uncountable) and let $K'$ be the algebraic closure of $F'$. (*)

$K'$ is a field extension of $F$ that is algebraically closed but it cannot be algebraic over $F$ because its cardinality is too large. I.e. there are only countable many polynomial roots over $F[X]$ but $K'$ contains uncountably many elements - most of which cannot be algebraic over $F$. In this fashion you can get an arbitrarily large 'gap' between the cardinality of the algebraic closure of $F$ and an algebraically closed extension. (The fact that $F$ is finite is irrelevant here - I simply thought it might help to make this example more approachable.)

(*) Note that $F'$ exists by the existence of the infinite field $K$ and the upward Löwenheim-Skolem Theorem, but you can also just view it as a field extension $F' = F(x_i \mid i \in I)$ with indepenent $x_i$ and $I$ uncountable.

Answer 2

Let $K'$ be an algebraic closure of $F(x)$, where $F(x)$ is the field of rational functions over $F$. Then $K'$ is an extension of $F$, and is algebraically closed, but is not an algebraic extension of $F$.

Answer 3

Sure. $\mathbb{C}$ contains $\pi$.

Let $F = \mathbb{Q}$. $\mathbb{C}$ is algebraicly closed (although it is not the algebraic closure of $\mathbb{Q}$, which is just the set of algebraic numbers) and is an extension of $\mathbb{Q}$. But $\mathbb{C}$ contains $\pi$, which is not algebraic over $\mathbb{Q}$, so $\mathbb{C}$ is an algebraically closed transcendental extension of $\mathbb{Q}$.