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 find all $t \in \mathbb{R}$ such that $\mathbb{Q} [t]$ is isomorphic to $\mathbb{Q} [x] / (x^2+x-1)$ ?

(I only know that $\mathbb{Q} [\alpha]$ , where $\alpha$ is the equivalence class of $x$ in $\mathbb{Q} [x] / (x^2+x-1)$, is isomorphic to this same structure; but for plugging numbers in $\mathbb{Q} [\ ]$, I don't have any idea...)

share|improve this question
4  
First, can you see why $\mathbb{R}$ has a unique subfield isomorphic to $K = \mathbb{Q}[x]/(x^2 + x - 1)$? So $t$ must lie in this subfield. Second, any element of $K$ generates some subfield of $K$. What are the subfields of $K$? –  Qiaochu Yuan Jan 9 '12 at 23:16

1 Answer 1

Think about it this way. If you pick some real number $r$ then the evaluation map $\mathbb{Q}[x]\to\mathbb{R}$ given by $p\mapsto p(r)$ defines a ring homomorphism onto $\mathbb{Q}[r]$. Now, since $\mathbb{Q}$ is a field it's easy to see that the kernel of this map is the ideal generated minimal polynomial of $r$ over $\mathbb{Q}$, call that $m_r$. Then, the FIT gives $\mathbb{Q}[x]/(m_r)\cong\mathbb{Q}[r]$. Thus, you just need to find which real number has $x^2+x-1$ as its minimal $\mathbb{Q}$-polynomial. This shouldn't be hard.

share|improve this answer
7  
Note that the OP seeks all $\:t\in \mathbb R\: $ such that $\:\mathbb Q[t]\:$ is isomorphic to $\:\mathbb Q[x]/(x^2+x-1)\:.$ –  Bill Dubuque Jan 9 '12 at 23:14

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.