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.

  • $\begingroup$ I think the smallest field containing $\mathbb Q$ and $\pm\sqrt2$ is by definition the splitting field of $x^2-2\in\Bbb Q[x]$ over $\mathbb Q$. $\endgroup$ – awllower Mar 28 at 5:51
  • $\begingroup$ Looks fine to me. $\endgroup$ – Dbchatto67 Mar 28 at 5:52

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