this is the exercise:
suppose $K|F$ is a field extension , $\alpha,\beta\in K^∗$ , $m,n$ are two integers that $(m,n)=1$ and $α^m,β^n∈F$,prove that $αβ$ is a primitive element of $F(α,β)|F$.
this is my solution:
there are integers $A,B$ that $Am+Bn=1$,now $αβ=(αβ)^{Am+Bn}=α^{Am}β^{Bn}α^{Bn}β^{Am}$,we note $α^{Am}β^{Bn}∈F$, we call it $l$. now,$αβ=lα^{Bn}β^{Am}=l(αβ)^{Am}α^{Bn−Am}$ if we add $αβ=l(αβ)^{Am}α^{Bn−Am}$ to field $F$ ,we would have $α^{Bn}∈F$(because $α^{−Am}∈F$)and then $α^{Am+Bn}=α∈F$,then we have $β∈F$,so $F(α,β)⊂F(αβ)$.trivially we have the inverse inclusion so $F(αβ)=F(α,β)$.
Am i true?
thanks.
This is essentially correct. However, you should give the field $F(\alpha\beta)$ a name; you refer to it in the proof as just $F$, which is not correct. I would rewrite the portion of the proof starting with this sentence as follows:
Now, $\alpha\beta = l\alpha^{Bn}\beta^{Bm} = l(\alpha\beta)^{Am}\alpha^{Bn}\alpha^{-Am}$. Let $L = F(\alpha\beta)$; then $\alpha^{Bn} = (\alpha\beta)^{1-Am}\alpha^{Am}l^{-1}\in L$ since $\alpha^{Am}, l\in F$. Then $\alpha^{Am+Bn} = \alpha\in L$, so that also $\beta\in L$ and thus $F(\alpha,\beta)\subseteq L$. The inverse inclusion is trivial, so that $L = F(\alpha\beta) = F(\alpha,\beta)$.