# Isomorphisms between fields

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...)

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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
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.
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