# I want to show how many intermediate fields there are between $GF(3^{12})$ and $GF(3^4)$.

So $E=GF(3^{12})$ is an extension of $F=GF(3^4)$, and I wish to know how many intermediate fields $F \subsetneq K \subsetneq E$ there are.

By a result in Escofier's Galois Theory I have that $G={\rm Gal}(E/F)$ is cyclic and of order $3$, so $G\cong \mathbb{Z}_3$. Since the fundamental theorem gives a bijective correspondence between intermediate fields $F \subset K \subset E$ and subgroups of $G$, and since $\mathbb{Z}_3$ has no proper non-trivial subgroups, there shouldn't be any intermediate fields $F \subsetneq K \subsetneq E$. Am I missing something or is this correct? It seemed a bit too easy, especially since I often get confused when doing Galois theory and expect to get stuck.

• The argument seems fine. May 30, 2016 at 13:19
• So you expect to get stuck and confused, and now you are confused because you don't? May 30, 2016 at 13:20
• Any (finite) extension of $F$ must have a power of $3^4$ for its number of elements, and any subfield of $E$ must have a number of elements some power of which is $3^{12}$. The only numbers $n$ such that $n$ is a power of $3^4$ and $3^{12}$ is a power of $n$ are $3^4$ and $3^{12}$, and we're done, sans Galois. May 30, 2016 at 13:26
• @MarcvanLeeuwen Yes. It seems I'm confused either way. What a mess. Thanks for confirming my argument though :) May 30, 2016 at 13:44

Yes, your argument is correct. In general, the lattice of subfields of $GF(p^n)$ is isomorphic to the lattice of divisors of $n$. For example, the lattice of subfields of $GF(3^{12})$ will look exactly like the lattice of divisors of 12 which contains the elements 12, 6, 4, 3, 2 and 1. In this lattice, there will not be any elements between 4 and 12 because the ratio 12/4 is a prime.