Recall that a “perfect” field of characteristic \(p\) is one for which the Frobenius map \(\operatorname{Fr}: x \mapsto x^p\) is onto.
Let \(F\) be a perfect field, and \(E/F\) an algebraic extension. Show that \(E\) is perfect.
Solution
1. It is enough to prove that every \(\alpha\in E\) has a \(p\)th root in \(E\).
By the definition recalled in the problem, a field of characteristic \(p\) is perfect exactly when its Frobenius map \[ \operatorname{Fr}_E:E\longrightarrow E, \qquad x\longmapsto x^p, \] is surjective.
2. Fix \(\alpha\in E\) and put \(K=F(\alpha)\).Then \(K/F\) is finite.
The extension \(E/F\) is algebraic, so \(\alpha\) is algebraic over \(F\). Hence \[ [K:F]=[F(\alpha):F]<\infty. \]
3. If \(e_1,\ldots,e_n\) is an \(F\)-basis of \(K\), then \(e_1^p,\ldots,e_n^p\) is also an \(F\)-basis of \(K\).
It suffices to prove linear independence, because there are exactly \(n=[K:F]\) elements. Suppose \[ \sum_{i=1}^n c_i e_i^p=0, \qquad c_i\in F. \] Since \(F\) is perfect, Frobenius on \(F\) is surjective. Choose \(d_i\in F\) with \[ d_i^p=c_i. \] In characteristic \(p\), \[ 0 =\sum_{i=1}^n d_i^p e_i^p =\left(\sum_{i=1}^n d_i e_i\right)^p. \] A field has no nonzero nilpotents, so \[ \sum_{i=1}^n d_i e_i=0. \] Because the \(e_i\) are \(F\)-linearly independent, every \(d_i=0\), hence every \(c_i=0\). Thus \(e_1^p,\ldots,e_n^p\) is an \(F\)-basis of \(K\).
4. Frobenius on \(K\) is surjective.
Let \(y\in K\). By 3, write \[ y=\sum_{i=1}^n c_i e_i^p, \qquad c_i\in F. \] Again using perfection of \(F\), choose \(d_i\in F\) with \(d_i^p=c_i\). Then \[ y =\sum_{i=1}^n d_i^p e_i^p =\left(\sum_{i=1}^n d_i e_i\right)^p. \] Thus \(y\) is a \(p\)th power of an element of \(K\).
5. Therefore Frobenius on \(E\) is surjective, so \(E\) is perfect.
Apply 4 to \(y=\alpha\). There is \(\beta\in K\subseteq E\) such that \[ \beta^p=\alpha. \] Since \(\alpha\in E\) was arbitrary, every element of \(E\) has a \(p\)th root in \(E\). Hence Frobenius on \(E\) is surjective, and \(E\) is perfect.