# 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

## 1 Answer

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.

• So for problem two would that be isomorphic to the integers modulo 3 adjoined with the dual number unit? – uhhhhidk Mar 21 at 17:20
• I don't know what the "dual number unit" is. – Robert Shore Mar 21 at 17:58
• it's like the imaginary unit but instead of equaling -1 when squared it's equal to 0 – uhhhhidk Mar 21 at 18:33
• Then I think you're correct. The result is a ring with 9 elements. – Robert Shore Mar 21 at 18:37