Is Q[x]/I ring-isomorphic to Q[\sqrt{2}]?

[e(ho0n3]

Homework Statement
Prove that Q[x]/\langle x^2 - 2 \rangle is ring-isomorphic to Q[\sqrt{2}] = \{a + b\sqrt{2} \mid a,b \in Q\}.


The attempt at a solution
Denote \langle x^2 - 2 \rangle by I. a_0 + a_1x + \cdots + a_nx^n + I belongs to Q[x]/I. It has n + 1 coefficients which somehow map to a and b. I don't think any injection can do this. I'm stumped. Any hints?

 

[e(ho0n3]

I think I got it: Let f(x) be an element of Q[x]. f(x) may be rewritten as (xx - 2)q(x) + r(x) for some q(x), r(x) in Q[x] with r(x) = 0 or deg r(x) < deg (xx - 2) = 2. Thus, r(x) has the form a + bx, a and b both belonging to Q. Aha!