Let \(F\) be a field of order \(5^{20}\). List the orders of all the subfields of \(F\).
Does \(F\) have more than one subfield of order \(25\)? Explain.
Solution
The subfield orders are \[ 5,\quad 5^2,\quad 5^4,\quad 5^5,\quad 5^{10},\quad 5^{20}. \] There is exactly one subfield of each listed order. In particular, there is exactly one subfield of order \(25\).
1. Every subfield has order \(5^d\) with \(d\) a positive divisor of \(20\).
The characteristic of a finite field is a prime dividing its order: the additive order of \(1\) divides the additive group order, and a composite characteristic would give nonzero zero divisors. Here that prime must be \(5\). Thus \(F\) contains its prime field \(\mathbb F_5\). If \([F:\mathbb F_5]=n\), a basis counts its elements as \(5^n\), so \(n=20\).
Every subfield \(K\) contains \(\mathbb F_5\). Writing \(d=[K:\mathbb F_5]\), the tower law gives \[ 20=[F:\mathbb F_5]=[F:K]d. \] Consequently \(d\mid20\) and \(|K|=5^d\). The positive divisors of \(20=2^2\cdot5\) are exactly \(1,2,4,5,10,20\).
2. For each \(d\mid20\), the set\(F_d=\{a\in F:a^{5^d}=a\}\) is a subfield with exactly \(5^d\) elements.
Put \(h_d(T)=T^{5^d}-T\). First, \(h_d\) divides \(h_{20}\) in \(\mathbb F_5[T]\). Indeed, in the quotient ring by \((h_d)\) the class \(t\) of \(T\) satisfies \(t^{5^d}=t\). Repeatedly taking the \(5^d\)th power gives \(t^{5^{kd}}=t\) for every positive integer \(k\). Taking \(k=20/d\) shows that \(h_{20}\) has zero class in this quotient, which is exactly the asserted divisibility.
Every nonzero element of \(F\) satisfies \(a^{5^{20}-1}=1\) by Lagrange’s theorem in its multiplicative group [@DF04]. Zero also satisfies \(h_{20}(0)=0\). Thus \[ h_{20}(T)=\prod_{a\in F}(T-a), \] a product of distinct linear factors. A monic divisor of this polynomial is itself a product of distinct linear factors in \(F[T]\), by unique factorization in a polynomial ring over a field [@DF04]. Hence \(h_d\) has exactly its degree, \(5^d\), distinct roots in \(F\).
It remains to check that those roots form a field. The map \(a\mapsto a^{5^d}\) is a field homomorphism, as follows by iterating the characteristic-five identity \((a+b)^5=a^5+b^5\) and using multiplicativity. Its fixed elements contain \(0,1\) and are closed under addition, subtraction, and multiplication. For a nonzero fixed element \(a\), \[ (a^{-1})^{5^d}=(a^{5^d})^{-1}=a^{-1}, \] so they are also closed under inversion. This proves that \(F_d\) is a subfield, with the cardinality already established.
3. A subfield of a given listed order is unique.
If \(K\subseteq F\) has order \(5^d\), every nonzero \(a\in K\) satisfies \(a^{5^d-1}=1\) by Lagrange’s theorem in \(K^\times\). Including zero, every element of \(K\) satisfies \(a^{5^d}=a\). Thus \(K\subseteq F_d\). Since both fields have \(5^d\) elements, they are equal. Together with steps 1 and 2, this proves that the displayed list is exhaustive and each order is realized by exactly one embedded subfield. For \(d=2\) this says \[ \{a\in F:a^{25}=a\} \] is the unique subfield of order \(25\), answering part (b).