Let \(F/K\) be a field extension. Let \(\mathrm{char} K = p \neq 0\) and let \(n\geq 1\) be an integer such that \((p, n) = 1\). If \(v\in F\) and \(nv \in K\), then \(v\in K\).
Solution
1. The element \(n\cdot1_K\) is nonzero in \(K\).
Since \(\operatorname{char}K=p\), one has \[ n\cdot1_K=0 \quad\Longleftrightarrow\quad p\mid n. \] The hypothesis \((p,n)=1\) excludes this, so \(n\cdot1_K\ne0\).
2. The element \(n\cdot1_K\) is invertible in \(K\).
Every nonzero element of a field is invertible. Apply this to 1.
3. If \(nv\in K\), then \(v\in K\).
Interpreting multiplication by the integer \(n\) as scalar multiplication by \(n\cdot1_K\), we have \[ nv=(n\cdot1_K)v. \] By hypothesis the left side belongs to \(K\), and by 2 the inverse \((n\cdot1_K)^{-1}\) belongs to \(K\). Therefore \[ v=(n\cdot1_K)^{-1}(nv)\in K. \]