TITLE: Ideals and quotient rings.
QUESTION [5 upvotes]: Let $R=\{a+bi \space | \space a,b\in \mathbb{Z} \} $
Let I be the ideal $\langle 3+5i \rangle $
In the quotient ring $ R/I $
(a) Without finding zero divisors explain why $R/I$ cannot be an integral domain.
(b) Explain why $I$ is not prime.
In the first part of the question I found that the order of $(1+I)$ is $34$.
I was hoping that this would help me answer these two questions but I can't figure it out. I can know that they are equivalent statements as $I$ is prime iff $R/I$ is an integral domain. my problem is I can't seem to figure out how write out what an element in the quotient ring looks like. If I could show there was an element of a different order than $34$ I would be done.
Edit: i think we can use the fact that $(17+I)+(17+I) = (34+I)$ doesn't this imply that the order of $(17+I) =2$ ? But by theorem the only possible order was already proved to be 34 so R/I is not a FID?

REPLY [2 votes]: Another way: 
$\mathbb Z[i]$ has Krull dimension $1$. Therefore, a quotient by a nonzero prime ideal of $\mathbb Z[i]$ would leave you with a domain of Krull dimension $0$, which is a field. But clearly $2$ does not have any inverse in this ring, so it can't be a field.