My attempt $:$
If $\alpha$ is algebraic over a field $F$ then by first isomorphism theorem it is clear that $$F [x] / (f (x)) \simeq F [\alpha]$$ where $f (x)$ is the irreducible polynomial for $\alpha$ over $F$.Now since $F [x]$ is a PID and $f (x)$ is irreducible over $F$ so $(f (x))$ is a maximal ideal and hence $F [x]/ (f (x))$ is a field and hence so is $F [\alpha]$.But then $F [\alpha] [y] = F [\alpha,y]$ is a PID. So again the application of first isomorphism theorem yields $$F [\alpha,y] / (g (y)) \simeq F [\alpha,\beta]$$ where $g (y)$ is the irreducible polynomial for $\beta$ over $F [\alpha]$.But again by similar argunment mentioned above we have $F [\alpha,\beta]$ is a field.
Is my reasoning correct at all? Please check it.
Thank you in advance.