Let \(p\) be a prime number and let \(K\), \(L\) be fields of orders \(p^m\), \(p^n\) respectively, where \(m < n\). When is \(K\) isomorphic to a subfield of \(L\)?
Solution
1. If \(K\) is isomorphic to a subfield of \(L\), then \(m\mid n\).
Suppose \(K\) is identified with a subfield of \(L\). Then \(L\) is a finite-dimensional vector space over \(K\), so the tower law gives \[ [L:\mathbb F_p]=[L:K][K:\mathbb F_p]. \] Since \[ [L:\mathbb F_p]=n, \qquad [K:\mathbb F_p]=m, \] we obtain \[ n=[L:K]m. \] Hence \(m\mid n\).
2. If \(m\mid n\), then \(K\) is isomorphic to a subfield of \(L\).
Write \[ n=rm. \] Every element \(x\in L=\mathbb F_{p^n}\) satisfies \[ x^{p^n}=x. \] Consider the set \[ F=\{x\in L:x^{p^m}=x\}. \] This is a subfield of \(L\): if \(x,y\in F\), then in characteristic \(p\), \[ (x+y)^{p^m}=x^{p^m}+y^{p^m}=x+y, \] and \[ (xy)^{p^m}=x^{p^m}y^{p^m}=xy. \] Also, if \(x\ne0\), then \[ (x^{-1})^{p^m}=(x^{p^m})^{-1}=x^{-1}. \]
It remains to show that \(F\) has exactly \(p^m\) elements. Since \(m\mid n\), \[ p^m-1\mid p^n-1. \] The multiplicative group \(L^\times\) is cyclic of order \(p^n-1\), so it has exactly \(p^m-1\) solutions to \[ x^{p^m-1}=1. \] Together with \(x=0\), these are exactly the roots of \[ x^{p^m}-x. \] Hence \[ |F|=p^m. \] There is, up to isomorphism, a unique field with \(p^m\) elements, so \[ F\cong K. \] Thus \(K\) is isomorphic to a subfield of \(L\).
3. Therefore\[ K\hookrightarrow L \quad\Longleftrightarrow\quad m\mid n. \]
Combine 1 and 2.