# Proof for the equivalence between a quotient ring and a polynomial ring

Can someone give me an idea, how to prove that $\mathbb{Q}[x]/I=\mathbb{Q}[\hat{x}]$, where $I$ is the ideal generated by $-1+x+x^{2}$ and $\hat{x}$ is the equivalence of $x$ in $\mathbb{Q}[x]/I$ ?

I know some facts about $\mathbb{Q}[x]/I$ (it is a field, since the polynomial is irreducible etc.), but my problem is that we did so much theory in class the I literally can't see the forest for the trees. I tried applying the proof from theorem 3 from page 512 from Dummit \& Foote's {}Abstract Algebra'', but somehow I couldn't get it to work, because my hunch to solve this problem was to use the evaluation homomorphism, but plugging an equivalance class of polynomials into a polynomial itself is just extremely confusing.

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A naive way of answering the question: The element $\widehat x$ satisfies $-1+\widehat x+\widehat x^2=0$ and nothing else. --- It seems to me the present question is very similar to that question of yours. – Pierre-Yves Gaillard Jan 8 '12 at 10:49

Let $E \hookrightarrow F$ be a field extension, fix $x_0 \in F$. Consider in $E[x]$ the collection $V$ of polynomials such that evaluating at $x_0$ yields 0, this is easily seen to be an ideal (if it's nonempty). In particular, $V = (p)$, by considering polynomials in $V$ of the smallest degree.
As you say, we have $ev: E[x] \twoheadrightarrow E[x_0]$, and quotienting out by the kernel we have by the first isomorphism theorem $E[x]/(p) \simeq E[x_0]$.
So for your question, you just have to note that $-1 + x + x^2$ is irreducible, and hence is the $p$ (up to fiddling with multiplication by units) as above. Done!