I'm trying to solve this problem from my abstract algebra course:
Find all prime and maximal ideals of $\mathbb{Z}_{36}=\mathbb{Z}/36\mathbb{Z}$.
I've seen other posts about this topic, but I still don't get it since they have orders so much lower that $36$. I think in this case the prime and the maximal ideals are the same, and I guess that's because $\mathbb{Z}_{36}$ is a principal ideal domain (if this were the case, how can I also prove this? In a general case $\mathbb{Z}_n$ assuming every $\mathbb{Z_n}$ verifies that it is a principal ideal domain).
After some research I found out that the maximal ideals of $\mathbb{Z}_{36}$ are $k\mathbb{Z}/36\mathbb{Z}$ for $k\in\{1,\dots,35\}$, but these are so much posibilities to check to not exist some easier way to see it, like checking only the coprimes to $36$ (or the non-coprimes, I'm not sure about what of them should I check, that's why I'm asking). How can I act in these kind of problems?
Thanks in advance, any help will be appreciated.
Best Answer
$\Bbb Z_{36}$ is not a principal ideal domain, however it is a principal ideal ring. In general for a commutative ring $R$ with ideal $I$ the ideals of $R/I$ correspond bijectively to the ideals of $R$ containing $I$. This correspondence preserves prime/maximal/... ideals. So in this case the ideals in $\Bbb Z$ containing $36\Bbb Z$ are exactly $$\Bbb Z,2\Bbb Z,3\Bbb Z,4\Bbb Z,6\Bbb Z,9\Bbb Z,12\Bbb Z,18\Bbb Z,36\Bbb Z$$ (All divisors of $36$) By quotienting out $36\Bbb Z$ we get the corresponding ideals of $\Bbb Z/36\Bbb Z$. I will let you figure out which of these are maximal.