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 .

  • $\begingroup$ For the second quotient ring, note that $(X^2+3, 3)=(X^2, 3)$. That's easier to work with. $\endgroup$ – Robert Shore Mar 21 '19 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.

  • $\begingroup$ So for problem two would that be isomorphic to the integers modulo 3 adjoined with the dual number unit? $\endgroup$ – uhhhhidk Mar 21 '19 at 17:20
  • $\begingroup$ I don't know what the "dual number unit" is. $\endgroup$ – Robert Shore Mar 21 '19 at 17:58
  • $\begingroup$ it's like the imaginary unit but instead of equaling -1 when squared it's equal to 0 $\endgroup$ – uhhhhidk Mar 21 '19 at 18:33
  • $\begingroup$ Then I think you're correct. The result is a ring with 9 elements. $\endgroup$ – Robert Shore Mar 21 '19 at 18:37

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