For each prime number \(p\) and each positive integer \(n\), how many elements \(\alpha\) are there in \(\mathbb F_{p^n}\) such that \(F_p(\alpha)=F_{p^6}\)?
Solution
1. If \(6\nmid n\), there are no such elements.
Suppose \(\alpha\in\mathbb F_{p^n}\) satisfies \[ \mathbb F_p(\alpha)=\mathbb F_{p^6}. \] Then \[ [\mathbb F_p(\alpha):\mathbb F_p]=6. \] Since \(\mathbb F_p(\alpha)\) is a subfield of \(\mathbb F_{p^n}\), the tower law gives \[ 6\mid n. \] Therefore no such \(\alpha\) exists when \(6\nmid n\).
2. Assume \(6\mid n\). Then \(\mathbb F_{p^6}\) is the unique subfield of \(\mathbb F_{p^n}\) having \(p^6\) elements.
A finite field \(\mathbb F_{p^n}\) has a subfield of order \(p^d\) exactly when \(d\mid n\), and that subfield is unique. Since \(6\mid n\), the stated subfield exists and is unique.
3. Under the assumption \(6\mid n\), the desired elements are exactly the elements of \(\mathbb F_{p^6}\) having degree \(6\) over \(\mathbb F_p\).
For \(\alpha\in\mathbb F_{p^6}\), \[ \mathbb F_p(\alpha)=\mathbb F_{p^6} \] if and only if \[ [\mathbb F_p(\alpha):\mathbb F_p]=6. \] By 2, every element satisfying the original condition lies in this distinguished copy of \(\mathbb F_{p^6}\) inside \(\mathbb F_{p^n}\).
4. The number of elements of degree exactly \(2\) over \(\mathbb F_p\) is\[ p^2-p, \]and the number of elements of degree exactly \(3\) is\[ p^3-p. \]
The only proper divisor of \(2\) is \(1\), so the elements of \(\mathbb F_{p^2}\) having degree less than \(2\) are exactly the \(p\) elements of \(\mathbb F_p\). Thus the degree-\(2\) count is \(p^2-p\).
Likewise, the only proper divisor of \(3\) is \(1\), so the elements of \(\mathbb F_{p^3}\) having degree less than \(3\) are exactly the elements of \(\mathbb F_p\). Thus the degree-\(3\) count is \(p^3-p\).
5. The number of elements of degree exactly \(6\) over \(\mathbb F_p\) is\[ p^6-p^3-p^2+p. \]
For any \(\alpha\in\mathbb F_{p^6}\), the degree \[ d=[\mathbb F_p(\alpha):\mathbb F_p] \] divides \(6\), so \(d\in\{1,2,3,6\}\). These four degree classes are disjoint and exhaust \(\mathbb F_{p^6}\). The degree-\(1\) class has \(p\) elements, while 4 gives the degree-\(2\) and degree-\(3\) counts. Therefore, if \(N_6\) denotes the degree-\(6\) count, \[ p^6=p+(p^2-p)+(p^3-p)+N_6. \] Solving gives \[ N_6=p^6-p^3-p^2+p. \]
6. Hence the requested number is\[ \boxed{\begin{cases} 0,&6\nmid n,\\[2mm] p^6-p^3-p^2+p,&6\mid n. \end{cases}} \]
Combine 1 with 3 and 5.