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

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SUMMARY

The discussion confirms that the quotient ring Q[x]/⟨x² - 2⟩ is indeed ring-isomorphic to Q[√2]. The proof involves rewriting any polynomial f(x) in Q[x] as f(x) = (x² - 2)q(x) + r(x), where r(x) is a remainder polynomial of degree less than 2, taking the form a + bx with a, b in Q. This establishes a clear mapping between elements of Q[x]/⟨x² - 2⟩ and elements of Q[√2], demonstrating the isomorphism definitively.

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


The attempt at a solution
Denote [tex]\langle x^2 - 2 \rangle[/tex] by I. [tex]a_0 + a_1x + \cdots + a_nx^n + I[/tex] 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?
 
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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!
 

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