Prove that $k(\alpha+\beta)=k(\alpha,\beta)$

Question

I am trying to solve the following problem:

Let $k$ be a finite field and let $k(\alpha,\beta)/k$ be finite. If $k(\alpha)\cap k(\beta)=k$, prove that $k(\alpha,\beta)=k(\alpha+\beta)$.

What I already know:

1. $k(\alpha+\beta)\subset k(\alpha,\beta)$
2. $k(\alpha,\beta)/k$ is a Galois extension.
3. $[k(\alpha,\beta):k]=[k(\alpha):k][k(\beta):k]$.

Some thoughts in mind:

1. Prove that $\alpha\in k(\alpha+\beta)$ or $\beta\in k(\alpha+\beta)$.
2. Prove that $[k(\alpha,\beta):k(\alpha+\beta)]=1$
3. Try to express $[k(\alpha+\beta):k]$ in terms of $[k(\alpha):k]$ and $[k(\beta):k]$.
4. Try to use some Fundamental Theorem of Galois Theory.

However, I still have no idea how to move on. Any help?

Answer 1

Since they're Galois, you know that $$\text{Gal}(k(\alpha,\beta)/k)\cong\text{Gal}(k(\alpha)/k)\times\text{Gal}(k(\beta)/k).$$ But then, after the identification you can see that you can produce $[k(\alpha):k][k(\beta):k]$ different automorphisms of $\alpha+\beta$, by acting just on $\alpha$ while fixing $\beta$ from the subgroup

$$\{1\}\times\text{Gal}(k(\beta)/k)$$

and then applying automorphisms to all of those $[k(\alpha):k]$ distinct numbers from the

$$\text{Gal}(k(\alpha)/k)\times\{1\}$$

factor. Hence the degree of the extension it generates is at least this size--i.e. $[k(\alpha):k][k(\beta):k]$--but then

$$[k(\alpha):k][k(\beta):k]\le [k(\alpha+\beta):k]\le [k(\alpha,\beta):k]\le [k(\alpha):k][k(\beta):k]$$

so equality is everywhere.

Answer 2

Denote $k = \mathbb{F}_{p^n}$. We start with $k(\alpha,\beta)/\mathbb{F}_{p}$ is Galois. By Fundamental Theory of Galois Theory, $\text{Gal}(k(\alpha,\beta)/k) < \text{Gal}(k(\alpha,\beta)/\mathbb{F}_{p})$. Since $k(\alpha,\beta)/\mathbb{F}_{p}$ is finite, $\text{Gal}(k(\alpha,\beta)/\mathbb{F}_{p})$ is a cyclic group generated by Frobenius map. Therefore, $\text{Gal}(k(\alpha,\beta)/k)$ is a cyclic group which we denote the size of it as $m$.

Then we know, by Fundamental Theory of Galois Theory, there is a bijection between intermediate fields $\text{Int}(k(\alpha,\beta)/k)$ and subgroups of $\text{Gal}(k(\alpha,\beta)/k)$. Since $\text{Gal}(k(\alpha,\beta)/k)$ is a cyclic group, for each $d| |\text{Gal}(k(\alpha,\beta)/k)|$, there exists exactly one subgroup $H<\text{Gal}(k(\alpha,\beta)/k)$ s.t. $[G:H] = [k(\alpha,\beta)^H:k] = d$(The first equality is again by Fundamental Theory of Galois Theory). This means subgroups of $\text{Gal}(k(\alpha,\beta)/k)$ can be characterized by its index $[\text{Gal}(k(\alpha,\beta)/k):H]$. Then there is a bijection between $\text{Sub}(\text{Gal}(k(\alpha,\beta)/k))$ and divisors of $|\text{Gal}(k(\alpha,\beta)/k)| = c$, denote as $\text{Divisors}(c)$.

Then we can construct a bijection by two bijections we mentioned above: $$\begin{align*} \varphi: \text{Int}(k(\alpha,\beta)/k) &\to \text{Divisors}(m)\\ E &\mapsto [E:k] \end{align*}$$

Notice that $\text{Int}(k(\alpha,\beta)/k)$ is a lattice with subset as its partial order. For $K_1,K_2 \in \text{Int}(k(\alpha,\beta)/k)$, the Least Upper Bound is the compositum $K_1 \vee K_2$. It's easy to check $$ \begin{equation} k(\alpha,\beta) = k(\alpha) \vee k(\beta) = k(\alpha+\beta) \vee k(\alpha) = k(\alpha+\beta) \vee k(\beta) \end{equation} $$ The Greatest Lower Bound is the intersection $K_1 \cap K_2$.

Meanwhile, all divisors of $c$ also form a lattice with divide as its partial order. For $h_1,h_2 \in \text{Divisors}(m)$, the Least Upper Bound is $\text{g.c.d.}(h_1,h_2)$. The Greatest Lower Bound is $\text{l.c.m.}(h_1,h_2)$.

Therefore, $\varphi$ is not only a bijection, but also a order-preserving lattice isomorphism, which means $\varphi(a \wedge b) = \varphi(a) \wedge \varphi(b)$ and $\varphi(a \vee b) = \varphi(a) \vee \varphi(b)$. In this case, for $K_1,K_2 \in \text{Int}(k(\alpha,\beta)/k)$, we have: $$ [K_1 \cap K_2 : k] = \text{g.c.d.}([K_1 : k],[K_2 : k]) $$ $$ [K_1 \vee K_2 : k] = \text{l.c.m.}([K_1 : k],[K_2 : k]) $$ (Check it by yourself using the fact $\text{Gal}(E/(B \cap C)) = \text{Gal}(E/B) \vee \text{Gal}(E/C)$ and $\text{Gal}(E/(B \vee C)) = \text{Gal}(E/B) \cap \text{Gal}(E/C)$ given by Fundamental Theory of Galois Theory. Besides, don't forget $\text{Gal}(k(\alpha,\beta)/k)$ is cyclic!)

In the end, we finally can deal the original question by this order-preserving lattice isomorphism. Denote $[k(\alpha) : k] = a$ and $[k(\beta) : k] = b$. Since $k(\alpha) \cap k(\beta) = k$, we have $$ [k(\alpha) \cap k(\beta) : k] = \text{g.c.d.}([k(\alpha) : k],[k(\beta) : k]) = 1 $$ Then we have $a,b$ are coprime which gives: $$ [k(\alpha,\beta) : k] = [k(\alpha) \vee k(\beta) : k] = \text{l.c.m.}([k(\alpha) : k],[k(\beta) : k]) = ab $$

Therefore, applying the equation we mentioned above, we have: $$ [k(\alpha,\beta) : k] = [k(\alpha+\beta) \vee k(\alpha) : k] = \text{l.c.m.}([k(\alpha+\beta) : k],a) = ab $$ Since $\text{g.c.d.}(a,b) = 1$, $b|[k(\alpha+\beta) : k]$. Similarly, $$ [k(\alpha,\beta) : k] = [k(\alpha+\beta) \vee k(\beta) : k] = \text{l.c.m.}([k(\alpha+\beta) : k],b) = ab $$ This gives $a|[k(\alpha+\beta) : k]$. Therefore, $ab|[k(\alpha+\beta) : k]$.

Since $k(\alpha+\beta) \subseteq k(\alpha,\beta)$ and $[k(\alpha,\beta) : k] = ab$, we end up with the conclusion that $k(\alpha+\beta) = k(\alpha,\beta)$