I am looking for an elementary demonstration of this:
Suppose $K$ a field, $\mathrm{char}(K)=0$ and $K(\alpha) \supseteq K$ and $K(\beta) \supseteq K$ field extensions. Denote $n=[K(\alpha):K]$ and $m=[K(\beta):K]$ and suppose $\gcd(m,n)=1$. Then $K(\alpha,\beta)=K(\alpha + \beta)$.
To let everyone know what my level is: I am an undergraduate math student and my interest sparked when my professor showed us the demonstration of what I have written above in the specific case $K = \mathbb{Q}$, which only works thanks the specific structure of $\mathbb{C}$. I also found a paper that demonstrates it in fields with $\text{char}(K) \ne 0$, but involves subjects that are not in my course. link
I've tried to find mn distinct homomorphisms from $K(\alpha+\beta)\to E$ where $E$ is an algebraic closure of $K(\alpha + \beta)$. I've tried using the $mn$ homomorphisms from $K(\alpha,\beta)$, which has degree $mn$ over $K$, and I can say that my problem is equivalent of proof that: call $\alpha_i$ and $\beta_j$ for $i\in 1,\ldots,n$ and $j\in 1,\ldots,m$ the conjugates of $\alpha$ and $\beta$, the other roots of minimum polynomial. Thus $\alpha+\beta$ has degree $mn$ over $K$ $\iff$ "$\alpha+\beta=\alpha_i+\beta_j\iff \alpha=\alpha_i\wedge \beta=\beta_j$". My problem is that I don't know how prove that.
Thank you in advance
Edit: The specific structure of $\mathbb{C}$ is that, if you consider only $(\mathbb{C},+)$ you can orderer it in the following way:
$z = a + bi; w = c +di$ ($a,b,c,d \in \mathbb{R}$)
$z < w \iff a < c \vee a = c \land b < d$
In that way we have a maximum $\alpha_{i} + \beta_{j} = \max(\alpha_{i}) + \max(\beta_{j})$. Now we have that: $$\sigma(\max(\alpha_{i}) + \max(\beta_{j}))= \sigma(\max(\alpha_{i})) + \sigma(\max(\beta_{j})) = \max(\alpha_{i}) + \max(\beta_{j})$$ If and only if $\sigma(\max(\alpha_{i})) = \max(\alpha_{i}) \land \sigma(\max(\beta_{j})) = \max(\beta_{j}).$ This lets us prove that there exist $mn$ distinct homomorphisms. Since $\alpha + \beta$ and $\alpha_{i} + \beta_{j}$ are conjugate, we can conclude.