If $K$ is a field extension of $F$ and $\alpha, \beta\in K$, with $\alpha, \beta$ algebraic over $F$ and $\deg(\alpha)=m$, $\deg(\beta)=n$, such that $m,n$ are coprime, then $[F(\alpha,\beta):F]=mn$.
I don't actually know what it means to say that $\deg(\alpha)=m$. Does this mean that the degree of the minimal polynomial whose root is $\alpha$ is $m$? If so then I'm not sure how to proceed.
Would appreciate some clarification.
On
Yes, $\deg(\alpha) = m$ means that the minimum polynomial of $\alpha$ has degree $m$. It is equivalent to $[F(\alpha):F] = m$.
To proceed, use the fact that in a tower of (algebraic) extensions $F \subseteq K \subseteq L$, $[L:F] = [L:K] [K:F]$.
On
Hint: $F(\alpha, \beta) \supset F(\alpha) \supset F$ and $F(\alpha, \beta) \supset F(\beta) \supset F$ are two towers of finite extensions.
On
We know $F(\alpha,\beta)\supseteq F(\alpha)\supseteq F$ and $F(\alpha,\beta)\supseteq F(\beta)\supseteq F$. So, We can say the following about degrees
$$[F(\alpha,\beta):F] = [F(\alpha,\beta):F(\alpha)][F(\alpha):F] = [F(\alpha,\beta):F(\alpha)]\cdot m$$
$$[F(\alpha,\beta):F] = [F(\alpha,\beta):F(\beta)][F(\beta):F] = [F(\alpha,\beta):F(\beta)]\cdot n$$
Now, we know that $[F(\alpha,\beta):F] \le mn$ (As the basis of $F(\alpha,\beta)$ can have at most $mn$ distinct elements i.e. {$\alpha^r.\beta^s:0\le r\le m-1,0\le s\le n-1 $}).
But we see $[F(\alpha,\beta):F]$ is a common multiple of $m$ and $n$. Since, they are relatively prime $lcm(m,n) = mn$. Thus, $[F(\alpha,\beta):F] = mn$.
You are correct that $n$ and $m$ are the degrees of the irreducible polynomial.
To see the theorem, notice that $\{\alpha^r\beta^s\}$ forms a basis of $F(\alpha,\beta)$ if and only if $F(\alpha)\cap F(\beta)=F$. Thus $[F(\alpha, \beta):F]=nm$ if the intersection contains only the ground field.
Let $\gamma\in F(\alpha)\cap F(\beta)$. Then $F(\gamma)\subseteq F(\alpha)\cap F(\beta)$, and therefore $\deg(\gamma)\mid \deg(\alpha),\deg(\beta)$. Since $n$ and $m$ are coprime, this implies $\deg(\gamma)=1$, so $\gamma\in F$.