Problem: $\Bbb{Q}(\sqrt{2})$ is a field extension of $\Bbb{Q}$.
Solution: Let $F=\Bbb{Q}$ be a field of rational numbers and take $p(x)=x^2-2$, irreducible over $\Bbb{Q}$ By Eisenstein's Criterion.
Define a map $\psi:F\to {F[x]}/{\left<x^2-2\right>}$ by $\psi(a)=a+\left<x^2-2\right>$, for $a\in F$.
This map makes sense. because $\left<x^2-2\right>$ is irreducible over $F$. Thus $\left<x^2-2\right>$ is a maximal ideal. Therefore ,${F[x]}/{\left<x^2-2\right>}$ is a field. Clearly $\psi$ is injective ring homomorphism. Therefore, ${F[x]}/{\left<x^2-2\right>}$ is a field extension of $F$. Of course I know that $\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$.
Is it enough to show that for extension?
I am writing this answer because it answers also two to three questions. It would be helpful for others who looks answers like$\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$.
Let $F=\Bbb{Q}$ be a field of rational numbers and take $p(x)=x^2-2$, irreducible over $\Bbb{Q}$ By Eisenstein's Criterion.
Define a map $\psi:F\to {F[x]}/{\left<x^2-2\right>}$ by $\psi(a)=a+\left<x^2-2\right>$, for $a\in F$.
This map makes sense because $\left<x^2-2\right>$ is irreducible over $F$. Thus $\left<x^2-2\right>$ is a maximal ideal. Therefore ,${F[x]}/{\left<x^2-2\right>}$ is a field. Clearly $\psi$ is injective ring homomorphism. If we'll show that ${F[x]}/{\left<x^2-2\right>}\cong F$. Then $\Bbb{Q}(\sqrt{2})$ is extension of $\Bbb{Q}$.
So first we will show that $\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$.
Define a map $\psi:\Bbb{Q}\sqrt{2}\to \Bbb{Q}[x]/<x^2-2>$ by $\psi(a+b\sqrt{2})=[a+bx]$ for any $a,b\in\Bbb{Q}$.
$\psi$ is homomorphism:
$\psi((a+b\sqrt{2})+(c+d\sqrt{2}))=\psi((a+c)+(b+d)\sqrt{2}))=[(a+c)+(b+d)x]=[a+bx]+[c+dx]=\psi(a+b\sqrt{2})+\psi(c+d\sqrt{2}$.
$\psi((a+b\sqrt{2})(c+d\sqrt{2})=\psi((ac+2bd)+(ad+bc)\sqrt{2}))=[(ac+2bd)+(ad+bc)x]$. and, $\psi((a+b\sqrt{2})\psi(c+d\sqrt{2})=[a+bx][c+dx]=[ac+adx+bcx+bdx^2]=[ac+(ad+bc)x+bdx^2]=[(ac+2bd)+(ad+bc)x]$.
Therefore, $\psi$ is homomorphism.
$\psi$ is injective:
Let, $\psi(a + b\sqrt{2}) = \psi(c + d\sqrt{2})$, then $[a + bx] = [c + dx]$. Since there is a unique polynomial of degree $1$ or less for each congruence class $\operatorname{mod} p(x)$. $\Rightarrow$ $a + bx = c + dx$ $\Rightarrow$ $ a = c$ and $b = d$. Hence $\psi$ is injective.
$\psi$ is surjective:
Let $y\in \Bbb{Q}[x]/<x^2-2>$ then $y=[l+mx]$ for some $l,m\in \Bbb{Q}$. Therefore $\psi(l+m\sqrt{2})=y=[l+mx]$.
$\Rightarrow$ $\psi$ is surjective. And hence $\psi$ is isomorphism. Thus $\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$. Thus $\Bbb{Q}(\sqrt{2})$ is extension of $\Bbb{Q}$.