Let \(F/K\) be a field extension. Show that
\([F: K] = 1\) iff \(F = K\).
If \([F: K]\) is prime, then there are no intermediate fields between \(F\) and \(K\).
If \(u\in F\) has degree \(n\) over \(K\), then \(n\) divides \([F: K]\).
Solution
1. One has \([F:K]=1\) if and only if \(F=K\).
If \(F=K\), then \(\{1\}\) is a \(K\)-basis of \(F\), so \([F:K]=1\).
Conversely, suppose \([F:K]=1\). Since \(1\ne0\), the singleton \(\{1\}\) is a linearly independent subset of the one-dimensional \(K\)-vector space \(F\), hence is a basis. Therefore every \(x\in F\) has the form \(x=a\cdot1\) for some \(a\in K\), so \(F\subseteq K\). Since \(K\subseteq F\) by hypothesis, \(F=K\).
2. If \([F:K]\) is prime, there is no proper intermediate field\(K\subsetneq L\subsetneq F\).
Let \(K\subseteq L\subseteq F\). The tower law gives \[ [F:K]=[F:L][L:K]. \] If \([F:K]=p\) is prime, then one of the positive integer factors on the right is \(1\). If \([F:L]=1\), then \(F=L\) by 1; if \([L:K]=1\), then \(L=K\). Thus no proper intermediate field exists.
3. If \(u\in F\) has degree \(n\) over \(K\), then\[ [F:K]=[F:K(u)]\,n. \]
By definition of the degree of an algebraic element, \[ [K(u):K]=n. \] Applying the tower law to \[ K\subseteq K(u)\subseteq F \] gives \[ [F:K]=[F:K(u)][K(u):K]=[F:K(u)]n. \] When \([F:K]\) is finite, \([F:K(u)]\) is a positive integer, so \(n\) divides \([F:K]\).