If \(F\) is algebraically closed and \(E\) is the set of all elements in \(F\) that are algebraic over a field \(K\), then \(E\) is an algebraic closure of \(K\).
Solution
Assume \(K\subseteq F\), with \(F\) algebraically closed, and set \[ E=\{a\in F:a\text{ is algebraic over }K\}. \]
1. The set \(E\) is a subfield of \(F\) containing \(K\).
Every element of \(K\) is algebraic over \(K\), so \(K\subseteq E\). If \(a,b\in E\), then \(K(a,b)/K\) is finite. Hence every element of \(K(a,b)\) is algebraic over \(K\). In particular, \[ a-b,\quad ab\in E, \] and, if \(a\ne0\), also \(a^{-1}\in E\). Thus \(E\) is a subfield of \(F\).
2. The extension \(E/K\) is algebraic.
This is immediate from the definition of \(E\): every element of \(E\) is algebraic over \(K\).
3. Every nonconstant polynomial in \(E[x]\) has a root in \(E\).
Let \[ f(x)=a_0+a_1x+\cdots+a_nx^n\in E[x], \qquad n>0. \] The field \[ K'=K(a_0,\ldots,a_n) \] is a finite extension of \(K\), since each coefficient is algebraic over \(K\).
Because \(F\) is algebraically closed, \(f\) has a root \(\alpha\in F\). The element \(\alpha\) is algebraic over \(K'\) because it satisfies the nonzero polynomial \(f\in K'[x]\). Hence \(K'(\alpha)/K'\) is finite. Since \(K'/K\) is finite, the tower law shows that \(K'(\alpha)/K\) is finite. Therefore \(\alpha\) is algebraic over \(K\), so \(\alpha\in E\).
4. The field \(E\) is algebraically closed.
By 3 every nonconstant polynomial in \(E[x]\) has a root in \(E\). Factoring off a linear factor and repeating by induction on degree shows that every polynomial in \(E[x]\) splits into linear factors over \(E\). Thus \(E\) is algebraically closed.
5. Hence \(E\) is an algebraic closure of \(K\).
By 2, \(E/K\) is algebraic, and by 4, \(E\) is algebraically closed. These are exactly the defining properties of an algebraic closure of \(K\).