Let \(F\) be a field of characteristic zero, and let \(K\) be an algebraic extension of \(F\) that possesses the following property: every polynomial \(f\in F[x]\) has a root in \(K\). Show that \(K\) is algebraically closed.\
Hint: if \(K(\theta)/K\) is algebraic, consider \(F(\theta)/F\) and its normal closure; primitive elements might be of help.
<1>5. In fact \(\beta\in N\) and \(F(\beta)=N\).
<1>6. Therefore \(N\subseteq K\), and hence \(\theta\in K\).
<1>7. The field \(K\) is algebraically closed.
Solution
1. Let \(\theta\) be any element algebraic over \(K\). Then \(\theta\) is algebraic over \(F\).
The extension \(K/F\) is algebraic by hypothesis, and \(\theta/K\) is algebraic. Algebraicity is transitive, so \(\theta\) is algebraic over \(F\).
2. Let \(N/F\) be the normal closure of \(F(\theta)/F\). Then \(N/F\) is a finite Galois extension.
Since \(\theta\) is algebraic over \(F\), the extension \(F(\theta)/F\) is finite. Because \(\operatorname{char}F=0\), every algebraic extension of \(F\) is separable. Hence the normal closure of the finite separable extension \(F(\theta)/F\) is finite, normal, and separable, i.e. finite Galois.
3. There exists \(\alpha\in N\) such that\[ N=F(\alpha). \]
By the primitive element theorem, every finite separable extension is simple. Apply this to the finite Galois extension \(N/F\) from 2.
4. Let \(m_\alpha(x)\in F[x]\) be the minimal polynomial of \(\alpha\) over \(F\). By hypothesis, \(m_\alpha\) has a root \(\beta\in K\).
Work inside a common algebraic closure containing \(K\) and \(N\). Since \(N/F\) is normal and \(\alpha\in N\), the minimal polynomial \(m_\alpha\) splits completely over \(N\). Therefore every root of \(m_\alpha\), in particular \(\beta\), lies in \(N\).
Because \(m_\alpha\) is irreducible over \(F\) and \(\beta\) is one of its roots, \[ [F(\beta):F]=\deg m_\alpha=[F(\alpha):F]=[N:F]. \] But \(F(\beta)\subseteq N\), so equality of degrees forces \[ F(\beta)=N. \]
By 4, \(\beta\in K\), so \(F(\beta)\subseteq K\). By 5, \(F(\beta)=N\), hence \(N\subseteq K\). Since \(\theta\in F(\theta)\subseteq N\), we get \(\theta\in K\).
6. Thus every nonconstant polynomial over \(K\) has a root in \(K\), so \(K\) is algebraically closed.