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.
In the case that $L$ is a Galois extension of $K$ with $[L:K]$ not divisible by the characteristic, there is the following characterization. The trace form is the unique bilinear form $B:L\times L\to K$ which satisfies the following three properties:
- $B(ab,c)=B(b,ac)$ for all $a,b,c\in L$. (Multiplication by elements of $L$ is self-adjoint)
- $B(g(a),g(b))=B(a,b)$ for all $a,b\in L$ and $g\in Gal(L/K)$. (Galois-invariance)
- $B(1,1)=[L:K]$. (Normalization)
The proof is easy. Property (1) implies that $B(a,b)=B(1,ab)$, so $B$ is uniquely determined by the linear functional $T:L\to K$ given by $T(x)=B(1,x)$. Property (2) then says that $T$ is Galois-invariant, which implies $[L:K]T(x)$ is equal to the sum of the values of $T$ on the Galois-conjugates of $x$. But by linearity that sum is just $T(Tr(x))$, and then property (3) implies $T(Tr(x))=[L:K]Tr(x)$ so $T(x)=Tr(x)$ since we are assuming $[L:K]$ is not divisible by the characteristic.
Let me remark that I would expect something like property (1) to be crucial for any sort of characterization. In particular, you need some property that relates $B$ to the multiplication of $L$ (and not merely its structure as a module over the Galois group as in property (2)). The following example may be instructive. Suppose the characteristic is different from $2$ and $L=K(a)$ where $a^2\in L$ and $a\not\in L$. To define a bilinear form on $L$ we need to pick values for $B(1,1)$, $B(1,a)$, $B(a,1)$ and $B(a,a)$. Normalization will fix the value of $B(1,1)$, and Galois-invariance will force $B(1,a)$ and $B(a,1)$ to be $0$ since $a$ is conjugate to $-a$. But any value at all for $B(a,a)$ will still satisfy Galois-invariance, and any nonzero value will make $B$ symmetric and nondegenerate. To force $B(a,a)$ to be $2a^2$, you need to use some property like property (1) that knows what the value of $a\cdot a$ is using the multiplication of $L$.
Best Answer
For what it's worth, a complete proof of the following fact can be found in $\S 6$ of these notes.
Theorem: For a finite degree field extension $K/F$, the following are equivalent:
(i) The trace form $(x,y) \in K^2 \mapsto \operatorname{Tr}(xy) \in F$ is a nondegenerate $F$-bilinear form.
(ii) The trace form is not identically zero.
(iii) The extension $K/F$ is separable.
The specific answer to the OP's question appears in there, but here it is: always be careful to read the fine print when dealing with inseparable extensions. In this case, it turns out that when $K/F$ is inseparable, the trace from $K$ down to $F$ of $x \in K$ comes out as a power of $p$ (the characteristic of $F$) times the more familiar sum of Galois conjugates. But in characteristic $p$ multiplying something by a power of $p$ is the same as multiplying it by zero...so the trace is identically zero in inseparable extensions.
[Note also that the proof of another part of the theorem makes use of the Primitive Element Theorem, which is currently to be found in $\S 7$ of the notes, i.e., the following section. This is an obvious mistake which will have to be remedied at some point. There are plenty of other issues with these notes, which are still quite rough and incomplete.]