Galois Theory | Random Math

In this note I will give a Galois-theoretic proof that for a prime ${p}$ and positive integer ${n}$ ,

$\displaystyle n \mid \frac{\varphi(p^n-1)}{\varphi(p-1)}.$

I’d love to see a more elementary proof if you can come up with one.
First we need the following:

Lemma 1 Let ${Z_n}$ be the cyclic group with ${n}$ elements. Let ${m}$ be a positive divisor of ${n}$ , and consider ${Z_m}$ as a subgroup of ${Z_n}$ . Then the number of automorphisms of ${Z_n}$ which fix ${Z_m}$ pointwise is equal to ${\varphi(n)/\varphi(m)}$ (which, in particular, is an integer).

Proof of the Lemma: Note that any automorphism of ${Z_n}$ fixes ${Z_m}$ , though not necessarily pointwise: indeed ${Z_n}$ has a unique subgroup of order ${m}$ , and thus any automorphism of ${Z_n}$ must take this subgroup to itself. Thus we have a group homomorphism ${\text{Aut}(Z_n) \rightarrow \text{Aut}(Z_m)}$ which is easily seen to be surjective; its kernel is precisely the subgroup consisting of those automorphisms of ${Z_n}$ which fix ${Z_m}$ pointwise. The statement follows by comparing orders. ${\square}$

Now to prove the initial claim, consider the field extension ${\mathbf{F}_{p^n}/\mathbf{F}_p}$ . Basic Galois theory tells that this is a Galois extension of degree ${n}$ . Consider the canonical homomorphism

$\psi: \displaystyle \text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p) \rightarrow \text{Aut}(\mathbf{F}_{p^n}^\times)$

which restricts an ${\mathbf{F}_p}$ -automorphism ${\sigma}$ to the group of units of ${\mathbf{F}_{p^n}}$ . Clearly it is an injective homomorphism since ${\sigma}$ is completely determined by where it sends the units. Moreover for any ${\sigma \in \text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p)}$ , $\psi(\sigma)$ lies in the subgroup of ${\mathbf{F}_{p^n}^\times}$ of those automorphisms fixing pointwise the cyclic subgroup ${\mathbf{F}_{p}^\times}$ of order ${p-1}$ , because the Galois group consists of $\mathbf{F}_p$ -homomorphisms. By the lemma the subgroup of these automorphisms has order ${\frac{\varphi(p^n-1)}{\varphi(p-1)}}$ , whereas ${\text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p)}$ has order ${n}$ . This does it.

This problem was given to me by my research supervisor. This is the problem and my solution:

Let $k$ be a field of characteristic zero, and $k(x)$ the rational function field in one variable over $k$ . Suppose $F_1$ and $F_2$ are subfields of $k(x)$ such that $[k(x):F_1]$ and $[k(x):F_1]$ are finite. Is it possible for $[k(x):F_1\cap F_2]$ to be infinite? Indeed, it is. Note that $\mbox{Gal }(k(x)/k) = \mbox{PSL}_2(k)$ , the projective special general linear group over $k$ (we identify its elements with Möbius transformations). Let $S, T$ denote the generators of the modular group (or rather, their image in $\mbox{PSL}_2(k)$ ). Note that they are of finite order. Hence, taking $F_1=k(x)^{<S>}$ and $F_2=k(x)^{<T>}$ , we have $[k(x):F_i]<\infty$ for $i=1,2$ . However, $F_1 \cap F_2$ is fixed by the whole modular group, which is infinite since $k$ has characteristic zero. Hence $k(x)$ cannot be of finite index over $F_1 \cap F_2$ .