I've seen that $\Bbb Q(\sqrt2)=\{a+b\sqrt2:a,b\in\Bbb Q\}$ is the smallest field containing $\sqrt2$. Can this be realized if we consider the splitting field of $x^2-2$?
$\frac{\Bbb Q[x]}{(x^2-2)}=\{q(x)+(x^2-2):q(x)\in\Bbb Q[x]\}$. By the division algorithm, $q(x)=a+bx$ for $a,b\in\Bbb Q$. If we take $[x]:=\sqrt2$, then $[q(x)]=[a+bx]=a+b[x]=a+b\sqrt2$. Hence, $\{a+b\sqrt2:a,b\in\Bbb Q\}$ is the smallest field containing $\sqrt2$.
I would appreciate if anyone could let me know if this logic is correct, or if I am misguided. Thank you.