$\mathbb{Q}(\sqrt{d}) \cong \mathbb{Q}[t]/(t^2 - d)$ Let $d$ be an integer that is not the square of an integer, and defined the subfield of $\mathbb{C}$ as
$\mathbb{Q}( \sqrt{d} ) = \{ a + b \sqrt{d} \ | \ a,b \in \mathbb{Q} \}$.
I need to prove that 
$\mathbb{Q}(\sqrt{d}) \cong \mathbb{Q}[t]/(t^2 - d)$
I tried to prove it the following way. I defined a ring homomorphism $\phi: \mathbb{Q}[t] \to \mathbb{Q}( \sqrt{d} )$ by setting $\phi(f(t)) = f( \sqrt{d} )$. Now all that if left there to prove is that $\ker \psi = (t^2 - d)$
Obviously, $(t^2 - d) \subseteq \ker \psi$. 
I also tried to use the division with remained in $\mathbb{C}[t]$: for a $f(t) \in \mathbb{Q}[t]$, $f( \sqrt{d} ) = 0 \Leftrightarrow f(t) \in (t - \sqrt{d}) \subseteq \mathbb{C}[t]$, so in $\mathbb{C}[t]$ we have
$(t^2 - d) \subseteq \ker \psi \subseteq (t - \sqrt{d})$

Consider $f:Q[t]\rightarrow Q(\sqrt d)$ defined by $f(t)=\sqrt d$, $(t^2-d)\subset Ker f$ and $f$ is surjective, it induces a morphism $\bar f:Q[t]/(t^2-d)\rightarrow Q(\sqrt d)$, since $d$ is not a square, $(t^2-d)$ is a maximal ideal and $Q[t]/(t^2-d)$ is a field, this implies that $Ker \bar f=0$ or $Q[t]/(t^2-d)$, it is $0$ since $\bar f$ is surjective.
