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Is $\mathbb{Q}[x]/(x^2-x-1)$ isomorphic to $\mathbb{Q}[x]/(x^2-5)?$

My guess is yes. I am trying to find an isomorphism between the two. Universal Property of Quotient certainly helps. I am thinking of something like homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[x]/(x^2-5)$, but getting stuck...

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To get started, recall that if $f$ is irreducible over a field $F$, then $F[x]/\langle f(x) \rangle \cong F[\alpha]$, where $\alpha$ is any root of $f$. This fact can be derived by considering the evaluation homomorphism $\operatorname{ev}_{\alpha}:F[x] \rightarrow F[a]$ defined by $f(x) \mapsto f(\alpha)$ and applying the isomorphism theorem.

In this case, we have $\displaystyle \mathbb{Q}[x]/ \langle x^2 - x - 1 \rangle \cong \mathbb{Q} \Big[\frac{1}{2} + \frac{\sqrt{5}}{2}\Big]$ and $\mathbb{Q}[x]/\langle x^2 - 5 \rangle \cong \mathbb{Q}[\sqrt{5}]$. It is easy to show that $\mathbb{Q} \Big[\frac{1}{2} + \frac{\sqrt{5}}{2}\Big] = \mathbb{Q}[\sqrt{5}]$ by showing that each is a subset of the other.

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Hint: Consider the map $\mathbb Q[x] \rightarrow \mathbb Q[x]/(x^2-5)$, $x\mapsto \frac12 x - \frac12$ and show that it induces your isomorphism.

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Hint:(Sometimes it is easier/convenient to prove a general theorem than a specific one.) For two irreducible quadratic polynomials $f(x), g(x)\in \mathbf{Q}[x]$ the quotient rings are isomorphic as fields iff the ratio of their discriminants is the square of a non-zero rational number.

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Yes, they are isomorphic. In fact, by checking the zeroes of $X^2-X-1$ we find $$\frac{\mathbb{Q}[X]}{(X^2-X-1)}\cong \mathbb{Q}(\frac{\sqrt{5}-1}{2}) = \mathbb{Q}(\sqrt{5}) \cong \frac{\mathbb{Q}[X]}{(X^2-5)}.$$ For an explicit isomorphism, $\overline{X} \mapsto \overline{\frac{X-1}{2}}$ will do the job (which is exactly the above isomorphism). You talk about an isomorphism between $\mathbb{Q}[X]$ and $\frac{\mathbb{Q}[X]}{(X^2-5)}$. However, I think you mean homomorphism, because these two rings can never be isomorphic($\mathbb{Q}[X]$ does not contain an element $y$ for which $y^2=5$).

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  • $\begingroup$ Thanks for the typo correction. $\endgroup$ – Yang Ethan Jul 15 '16 at 2:57

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