Give an example of a finite extension of fields that has infinitely many intermediate fields.
Solution
Let \(p\) be prime, \(k=\FF_p(u,v)\) the rational function field in two indeterminates, and \(K=\FF_p(u^p,v^p)\subseteq k\).
1. \([k:K]=p^2\), with basis \(\{u^iv^j:0\le i,j\le p-1\}\).
Since \(u,v\) are algebraically independent over \(\FF_p\), \([K(u):K]=p\) with basis \(1,u,\dots,u^{p-1}\) and \([K(u,v):K(u)]=p\) with basis \(1,v,\dots,v^{p-1}\); the tower law gives \([k:K]=p^2\).
2. For \(c\in K\), the field \(E_c=K(u+cv)\) satisfies \([E_c:K]=p\).
In characteristic \(p\), \((u+cv)^p=u^p+c^pv^p\in K\), so \([E_c:K]\le p\). Since \(1,u,v\) are linearly independent over \(K\) by step 1, \(u+cv\notin K\), so \([E_c:K]>1\); and \([E_c:K]\) divides \(p^2\). Hence \([E_c:K]=p\).
3. If \(c_1\neq c_2\) in \(K\), then \(E_{c_1}\neq E_{c_2}\).
Suppose \(E=E_{c_1}=E_{c_2}\). Then \((u+c_1v)-(u+c_2v)=(c_1-c_2)v\in E\), and \(c_1-c_2\in K^\times\), so \(v\in E\) and then \(u=(u+c_1v)-c_1v\in E\). So \(E=k\), and \([E:K]=p^2\), contradicting step 2.
4. Q.E.D.
\(K\) is infinite (it contains the distinct elements \(u^{pn}\), \(n\ge1\)), so by step 3 the fields \(E_c\), \(c\in K\), are infinitely many distinct intermediate fields of the extension \(k/K\), which is finite of degree \(p^2\) by step 1.