If $[F(\alpha):F]=p$ and $[F(\beta):F]=q$, $p$ and $q$ distinct primes, then $[F(\alpha, \beta):F]=pq$

Question

I'm reading David R. Finston and Patrick J. Morandi's book Abstract Algebra: Structure and Application and hit this exercise at Chapter 5 that I'm a bit lost.

Let $p$ and $q$ be distinct primes. If $[F(\alpha):F]=p$ and $[F(\beta):F]=q$, prove that $[F(\alpha, \beta):F]=pq$.

I suppose this shall use the Dimension Formula Theorem (5.35 in the book):

Let $K$ be a field extension of $F$ and $L$ a field extension of $K$. Then $[L:F] = [L:K][K:F]$.

This leads to $$[F(\alpha, \beta):F] = [F(\alpha, \beta):F(\alpha)][F(\alpha):F] = p[F(\alpha,\beta):F(\alpha)]$$

However, how can I prove $[F(\alpha,\beta):F(\alpha)]$ is $[F(\beta):F]$, which is $q$? Shall I construct some kind of isomorphism?

Answer 1

Indeed, the comment of 伽罗瓦 to the question itself provides the key to the solution, which I merely flesh out here.

We are given that

$[F(\alpha):F] = p \in \Bbb P, \tag 1$

$[F(\beta):F] = q \in \Bbb P, \tag 2$

and

$q \ne p; \tag 3$

then we have

$[F(\alpha, \beta):F]$ $= [F(\alpha, \beta):F(\alpha)] [F(\alpha):F] = [F(\alpha, \beta):F(\alpha)]p \Longrightarrow p \mid [F(\alpha, \beta):F], \tag 4$

and

$[F(\alpha, \beta):F]$ $= [F(\alpha, \beta):F(\beta)] [F(\beta):F] = [F(\alpha, \beta):F(\beta)]q \Longrightarrow q \mid [F(\alpha, \beta):F]; \tag 5$

by virtue of (3), (4) and (5) in concert yield

$pq \mid [F(\alpha, \beta):F], \tag 6$

since $p$ and $q$ are distinct primes, and this in turn implies

$pq \le [F(\alpha, \beta):F]; \tag 7$

now consider

$[F(\alpha, \beta):F]$ $= [F(\alpha)(\beta):F] = [F(\alpha)(\beta):F(\alpha)] [F(\alpha):F] = [F(\alpha)(\beta):F(\alpha)]p; \tag 8$

we thus have

$\deg m_\alpha(x) = [F(\alpha)(\beta):F(\alpha)], \tag 9$

where

$m_\alpha(x) \in F(\alpha)[x] \tag{10}$

is the minimal polynomial of $\beta$ over $F(\alpha)$; letting

$m(x) \in F[x] \tag{11}$

be the minimal polynomial of $\beta$ over $F$, and so

$\deg m(x) = q = [F(\beta):F]; \tag{12}$

thus,

$\deg m_\alpha(x) \le \deg m(x), \tag{13}$

since $m_\alpha(x)$ is minimal for $\beta$ over $F(\alpha)$ but

$m(x) \in F[x] \subset F(\alpha)[x], \; m(\beta) = 0. \tag{14}$

We may bring together (8), (9), (12) and (13), and voila!:

$[F(\alpha, \beta):F] = [F(\alpha)(\beta):F(\alpha)]p \le qp; \tag {15}$

together with (7) we find

$pq \le [F(\alpha, \beta):F] \le pq, \tag {16}$

so at last,

$ [F(\alpha, \beta):F] = pq, \tag {17}$

the requisite result. $OE\Delta$.

Special regards to 伽罗瓦 for insightful comments.