I can't help but notice that they look exactly the same. For example:
$\mathbb{Q}(\sqrt{3})$ = $\lbrace p + q\sqrt{3}:p,q \in \mathbb{Q}\rbrace$
That seems pretty much exactly an ideal. Only the sqrt takes the ideals place. The operations of this ring seem pretty similar as well
In my comment above I said that the similarity is superficial, but perhaps this is not entirely true. Certainly the multiples of $\sqrt 3$ do not form an ideal, but they do form an additive subgroup and a subspace as a $\mathbb Q$ vector space. Hence they form a normal subgroup and one could take the quotient to obtain (a group isomorphic to) $\mathbb Q$. While this is a homomorphism of vector spaces/abelian groups, it is not a ring homomorphism.