Quotienting out polynomial rings by polynomial-generated ideals

Question

I'm studying in Lorenzini's Arithmetic Geometry, and it has been a while since I've taken a rigorous algebra course. I'm trying to understand a certain step in his proof that $\mathbb{Z}[\sqrt{5}]$ does not have the property of a unique factorization of ideals.

"We claim that the ideal $I$ generated by $2$ and $(1-\sqrt{5})$ in $\mathbb{Z}[\sqrt{5}]$ is maximal.

Indeed, $\mathbb{Z}[\sqrt{5}]/I \cong (\mathbb{Z}/2\mathbb{Z})[y]/(1-y, y^{2} - 5) = (\mathbb{Z}/2\mathbb{Z})[y]/(1-y) \cong \mathbb{Z}/2\mathbb{Z}$."

I understand that because the result is a field, then $I$ is maximal.

What I'm not understanding, however, is the first isomorphism and the second equality. I think I understand the last isomorphism (is it because $1-y$ is not irreducible over $\mathbb{Z}/2\mathbb{Z}$?). Additionally, I'm not certain where the $1-y$ and $y^2 + 5$ really come from (why are we quotienting out those things?), other than $y^2-5 = 0$ has a root $\sqrt{5}$.

Answer 1

I think that Lorenzini’s proof is unnecessarily complicated, although it may be good for showing a general method. Why not bypass it entirely and show that $\Bbb Z[\sqrt5\,]/I$ is isomorphic to $\Bbb F_2$ directly? Consider any element of the factor ring, represented at first by a number $m+n\sqrt5$. If either $m$ or $n$ is even, we can drop it away and assume that the class is represented by $\epsilon_1+\epsilon_2\sqrt5$, where the epsilons may be either $0$ or $1$. But if both are $1$, it’s congruent mod $1-\sqrt5$ to zero. So there are only two classes, $0$ and $1$.

Answer 2

For the first isomorphism, you can "quotient by $2$ first" so that $$\mathbb{Z}[\sqrt{5}]/I \cong \mathbb{F}_2[\sqrt{5}]/(1-\sqrt{5}) \cong \mathbb{F}_2[y]/(1-y,y^2-5).$$ To check these, you can explicitly construct the natural maps $\mathbb{Z}[\sqrt{5}] \to \mathbb{F}_2[\sqrt{5}]/(1-\sqrt{5})$ and $\mathbb{F}_2[y] \to \mathbb{F}_2[\sqrt{5}]/(1-\sqrt{5})$ and compute the kernels.

For the second equality, note that in the polynomial ring over $\mathbb{F}_2$, you have the equality $$y^2-5 = y^2-1 = (y-1)^2 = (1-y)^2,$$ so you can simplify $(1-y,y^2-5) = (1-y)$.

Answer 3

You have that $\mathbb Z[\sqrt{5}]\simeq \mathbb Z[y]/(y^2-5)$ and one isomorphism is the one sending $\sqrt{5}\mapsto y$. Therefore $\mathbb Z[\sqrt{5}]/I\simeq \mathbb Z[y]/(2,1-y,y^2-5)\simeq \mathbb Z/2\mathbb Z[y]/(1-y,y^2-5)$. Now note that $(1-y,y^2-5)=(1-y,y^2-1)=(1-y,(y-1)(y+1))=(1-y)$ as ideals of $\mathbb Z/2\mathbb Z[y]$, and the claim follows.