I am working on a proof that shows principal ideal $(2)$ in $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$ is maximal. Let us first say that $\frac{1+\sqrt{5}}{2} = \delta $ in short. The book proves the claim by showing that $\mathbb{Z}[\delta]/(2)$ is a field. First,
$\mathbb{Z}[\delta] \cong \mathbb{Z}[y]/(y^2 -y-1)$ which makes sense since $y^2 -y-1$ is the minimal polynomial of $\delta$ over $\mathbb{Q}$. Then, $\mathbb{Z}[\delta] /(2) \cong (\mathbb{Z}[y]/(y^2 -y-1))/ (2)$ that is still okay. After that, it goes like below and I need help to understand the rest:
$$\mathbb{Z}[\delta] /(2) \cong (\mathbb{Z}[y]/(y^2 -y-1))/ (2) \cong \mathbb{Z}[y]/(2,y^2 -y-1) \cong (\mathbb{Z}/\mathbb{2Z})[y]/(y^2-y-1).$$ - How?
"$y^2-y-1$ has no root in $\mathbb{Z}/\mathbb{2Z}$ so it is irreducible and the quotient is a domain." - Yes.
"Hence, $(\mathbb{Z}/\mathbb{2Z})[y]/(y^2-y-1)$ is isomorphic to the field $\mathbb{F_4}$ with four elements " -How? Is the quotient finite?
"Thus, $(2)$ is a maximal ideal."
I appreciate any help, thanks.
1) A finite domain is a field.
2) $\mathbf Z/2\mathbf Z[y]/(p(y))$, where p(y) is polynomial of degree $d$ is a $\mathbf Z/2\mathbf Z$-vector space of dimension $d$, hence it has $2^d$ elements.
Some more details to justify the equivalences above:
If $I, J$ are ideals of a ring $R$, the ideal generated by $J$ in $R/I$ is $\;J\cdot R/I=(I+J)/I$, so $$(R/I)\big/J\cdot R/I=(R/I)\Big/\big((I+J)/I\big)\simeq R/(I+J).$$
For the last isomorphism, you can tensor the short exact sequence $$0\longrightarrow (y^2-y-1)\longrightarrow \mathbf Z[y]\longrightarrow \mathbf Z[y]/(Y^2-y-1)\longrightarrow0$$ by $\mathbf Z/2\mathbf Z$, and use that for polynomial rings $$R[y]\otimes_R R/I\simeq R[y]/IR[y]\simeq R/I[y].$$