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?
[Math] Algebraically closed field extension can be not algebraic
extension-fieldfield-theory
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I’d like to expand on @tomasz”s answer in several ways.
There are really three concepts here: (1) an algebraically closed field; (2) an algebraic closure of a field; and (3) the algebraic closure of one field in another.
I’m going to sweep some delicate points under the rug by choosing a very handy definition of an algebraically closed field: Def. A field $F$ is algebraically closed if it has no proper algebraic extensions. Now, given a field $k$, an algebraic closure of $k$ is an algebraically closed field that is algebraic over $k$. Finally, if $L\supset k$ is an extension of fields, the algebraic closure of $k$ in $L$ is the set of elements of $L$ that are algebraic over $k$. For instance, the algebraic closure of $\mathbb Q$ in $\mathbb R$ is the set (field, really) of all real algebraic numbers. The algebraic closure of $k$ in $L$ is not generally algebraically closed, nor is it an algebraic closure of $k$.
Notice that there is no unique algebraic closure of a field $k$, and that it’s necessary to prove that any two such are isomorphic (as fields containing $k$).
Now to your question. It’s to take $F\subset L$, where $L$ is algebraically closed, and to define $K$ to be the algebraic closure of $F$ in $L$, and show that $K$ is also algebraically closed. Well: $K$ is certainly an algebraic extension of $F$ (all elements are algebraic over $F$). So, let $K'$ be an algebraic extension of $K$. We have $F\subset K\subset K'$, both inclusions being algebraic. But algebraic over algebraic is still algebraic, so $K'$ is algebraic over $F$, i.e. every element of $K'$ is algebraic over $F$, so every element of $K'$ is in $K$, and you’ve shown that $K$ has no proper algebraic extensions.
Yes, $K$ is also algebraically closed. Consider a polynomial $f(x) \in K[x]$, then let $\alpha$ be a root of $f(x)$,then $\alpha$ generates an algebraic extension over $K$, call it $K(\alpha)$, then it follows that $K(\alpha)$ is algebraic over $F$, because $K$ is algebraic over $F$. Thus $\alpha$ is algebraic over $F$, but then $\alpha \in K$, because $K$ contains all the roots of $F$. It follows that $K$ is algebraically closed.
And yes, a corollary is that the complex numbers are algebraically closed.
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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 $\mathbb Q$.
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.