Relative Algebraic Closure with Algebraically Closed Extension is Algebraic Closure

Theorem
Let $L$ be an algebraically closed field.

Let $L/K$ be a field extension.

Then the relative algebraic closure of $K$ contained in $L$ is an algebraic closure of $K$.

Proof
Let $K'$ denote the relative algebraic closure of $K$ contained in $L$.

By definition, $K' = \{\alpha \in L \mid \alpha \text{ is algebraic over } K\}$.

First we show that $K'$ is a field extension of $K$.

By Field is Algebraic over itself, $K \subseteq K'$.

Since $K$ is a field, it and (therefore) $K'$ are nonzero.

Let $\alpha, \beta \in K'$ be arbitrary.

By definition of generated field extension, $\alpha, \beta \in K(\alpha, \beta) \subseteq L$.

By the field axioms, $\alpha - \beta, \alpha\beta^{-1} \in K(\alpha, \beta) \subseteq L$.

By Finitely Generated Algebraic Extension is Finite and Finite Field Extension is Algebraic, we have that $K(\alpha, \beta)/K$ is algebraic.

Therefore both $\alpha - \beta$ and $\alpha\beta^{-1}$ are algebraic over $K$.

So, by definition, $\alpha - \beta, \alpha\beta^{-1} \in K'$.

Using the Subfield Test/Three-Step, we conclude that $K'$ is a subfield of $L$.

So $K'$ is a field containing $K$, ergo $K'/K$ is a field extension.

Furthermore, $K'$ is algebraically closed.

Indeed, let $p \in K'[X]$, where $K'[X]$ is the ring of polynomials with coefficients in $K'$.

Since $K'[X] \subseteq L[X]$ and, by hypothesis, $L$ is algebraically closed, $p$ has a root in $L$.

Let $\alpha \in L$ be a root of $p$.

Consider the field extension $K'(\alpha)/K'$.

By Finitely Generated Algebraic Extension is Finite and Finite Field Extension is Algebraic, $K'(\alpha)/K'$ is algebraic.

By definition of $K'$, the extension $K'/K$ is algebraic.

Therefore, by Transitivity of Algebraic Extensions, $K'(\alpha)/K$ is algebraic.

Thus $\alpha \in K'(\alpha)$ is algebraic over $K$.

By definition again, we conclude $\alpha \in K'$.

Thus $p$ has a root in $K'$.

We have that $K'/K$ is an algebraically closed algebraic field extension.

Therefore, by definition, $K'$ is an algebraic closure of $K$.

Example
Let $L =$ $\C$ and let $K$ be a subfield of $\C$.

Then the set $\{\alpha \in \C \mid \alpha \text{ is algebraic over } K\}$ is an algebraic closure of $K$.

Also see

 * Definition:Relative Algebraic Closure
 * Definition:Algebraic Closure
 * Definition:Algebraically Closed Field