# Quotient rings examples/intuition

I'm confused with some ideas about quotient rings and wanted to check if my intuition was flawed in any ways:

1: $$\mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5})\cong \mathbb{Z}/2\mathbb{Z}$$ The way I approached this was by using the homomorphism $$f(a+b\sqrt{-5})=a+5b\: \; mod(2)$$ and showing that the kernel included all multiples of 2 in the ring, as well as for $$(1+\sqrt{-5})$$. Is this a proper approach to proving the isomorphism? I'm sorry if this is blatantly obvious, I felt like it was a little too easy and am skeptical if whether this is a valid way of showing the isomorphism

2: $$\mathbb{Z}[X]/(X^{2}+3,3)\cong (\mathbb{Z}/3\mathbb{Z})[X]/(X^2)$$ For this one, I found this answer somewhere and I am confused because I ended up with something else. Like the previous example, I set up a homomorphism with a proper kernel to arrive at the isomorphism: $$f(a+bx)=a+b\sqrt{-3}\; mod(3)$$. This clearly has the ideal of 3 (unless I'm mistaken) and $$(x^{2}+3)$$ as its kernel, so would the above also be isomorphic to $$(\mathbb{Z}/3\mathbb{Z})[\sqrt{-3}]$$? Apart from this, how would the original isomorphism be proved?

I don't know what I'm not understanding/doing wrong. I apologize for the naivety, I'm trying to teach myself algebra using only online resources and it hasn't been very easy. Thanks for any help .

• For the second quotient ring, note that $(X^2+3, 3)=(X^2, 3)$. That's easier to work with. – Robert Shore Mar 21 at 2:49

The intuition for quotient rings is that the generators of the ideal give you elements that are zero in the quotient ring. So in your first example, we know that $$2=0$$ and $$1+\sqrt{-5}=0 \Rightarrow \sqrt{-5}=-1\equiv 1 \pmod{2}$$. That gives us a pretty good idea how we should try to define our homomorphism. Define $$\phi(a+b\sqrt{-5})=a+b~(\mod 2).$$
Show that $$\phi$$ is a ring homomorphism (not an isomorphism) from $$\Bbb Z[\sqrt{-5}]$$ to $$\Bbb Z/2\Bbb Z$$ and that $$\ker(\phi)=(2, 1+\sqrt{-5})$$ and you've proved your isomorphism, because for any ring homomorphism $$\phi:R \to S, R/\ker(\phi) \cong \operatorname{image}(\phi)$$.
Similarly, for problem $$2$$, you know $$3=0$$ and $$X^2+3=X^2=0$$. So for any polynomial $$p(X)=a_0+a_1X+X^2q(X) \in \Bbb Z[X]$$, define $$\phi(p)=a_0 (\mod 3)+ a_1 (\mod 3)X$$. (You're ignoring the terms of higher degree in $$p$$ because they are all multiplied by $$X^2$$, which will be $$0$$ in the quotient ring.) As above, show that this really is a ring homomorphism and show that it has the correct kernel, and you're done.