Relationship between $\Bbb Q(\sqrt2)$ and the splitting field of $x^2-2$

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.

• 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 $\mathbb Q$. – awllower Mar 28 at 5:51
• Looks fine to me. – Dbchatto67 Mar 28 at 5:52