Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 0 down vote accepted

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!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.