How to prove this isomorphism of a quotient ring

Question

I'm trying to understand more about ring theory and the concept of ideals has been confusing.

I'm trying to understand why this is true: $\mathbb Z[\sqrt{-5}]/(1+\sqrt{-5})\simeq\mathbb Z/6\mathbb Z$, but I don't know where to start.

If someone could show me how to prove this or link me somewhere I'd appreciate it. Thanks.

Answer 1

Consider the map $f:\Bbb Z[\sqrt{-5}]\to\Bbb Z/6\Bbb Z$ given by $f(1)=1, f(\sqrt{-5})=5$. Show that this indeed does give a homomorphism.

It's easy to see that $(1+\sqrt{-5})$ is contained in the kernel. It's a little more work to show that it contains all of the kernel, but it's not too bad. It will become relevant that $6\in (1+\sqrt{-5})$.

Answer 2

Evaluate this quotient by considering it as two successive quotients in two different ways: $$\frac{\mathbb{Z}[X]}{(5+X^2, 1+X)}$$

One order gives you $\mathbb{Z}[\sqrt{-5}]/(1+\sqrt{-5})$, one order gives you $\mathbb{Z}/(5+(-1)^2) = \mathbb{Z}/(6)$.

Answer 3

Recall that $\mathbb{Z[\sqrt{-5}]}$ is the set of all elements of the form $a+b\sqrt{-5}$, where $a,\,b\in\mathbb{Z}$. Now, in order to get the isomorphism, as a quotient ring is involved, try to apply the first isomorphism theorem.

This means that if we define a homomorphism from $\mathbb{Z[\sqrt{-5}]}$ to $\mathbb{Z}_6$ whose kernel is $(1+\sqrt{-5})$, we are done.

An obvious choice is to put $f(a+b\sqrt{-5})\equiv a+5b\pmod 6$ so that $f(a+b\sqrt{-5})=af(1)+bf(\sqrt{-5})$, which is uniquely determined from the values of $f(1)$ and $f(\sqrt{-5})$; observe that $f(1)\equiv 1\pmod 6$ and $f(\sqrt{-5})\equiv -1\pmod 6$.

Since $f(1+\sqrt{-5})=0$, we have $(1+\sqrt{-5})\subseteq\mathrm{Ker}(f)$.

Then, it suffices to prove that $\mathrm{Ker}(f)\subseteq (1+\sqrt{-5})$, i.e., for any $a+b\sqrt{-5}\in \mathrm{Ker}(f)$ there exists $c+d\sqrt{-5}$ s.t. $a+b\sqrt{-5}=(1+\sqrt{-5})(c+d\sqrt{-5})$. But this means that $a+b\sqrt{-5}=(c-5d)+\sqrt{-5}(c+d)$ and solving this system of linear equations you find out that any element of the kernel factors through $1+\sqrt{-5}$.

Answer 4

Consider the unique ring homomorphism $\chi\colon\mathbb{Z}\to\mathbb{Z}[\sqrt{-5}]/(1+\sqrt{-5})$. If $I=(1+\sqrt{-5})$, for simplicity, then $$ -\sqrt{-5}+I=1+I $$ and therefore the homomorphism $\chi$ is surjective. We just need to compute its kernel. For $z\in\mathbb{Z}$, we have $z\in I$ if and only if $$ z=(a+b\sqrt{-5})(1+\sqrt{-5})=a-5b+(a+b)\sqrt{-5} $$ for some $a,b\in\mathbb{Z}$. This implies $b=-a$, so $z=6a\in 6\mathbb{Z}$. Conversely $6=(1-\sqrt{-5})(1+\sqrt{-5})\in I$, so $6\in\ker\chi$. Therefore $\ker\chi=6\mathbb{Z}$.