Find all of the prime ideals of $\mathbb{Z}_3 \times \mathbb{Z}_4$

Question

My assignment was to find all of the prime ideals of $\mathbb{Z}_3 \times \mathbb{Z}_4$ so I first found all of the ideals which are:

$\mathbb{Z}_3 \times {0}$, ${0} \times \mathbb{Z}_4$, $\mathbb{Z}_3 \times \mathbb{Z}_2$,${0} \times \mathbb{Z}_2$

I then just modded the whole group by each of the ideals (ignoring the trivial cases). Which told me that

${0} \times \mathbb{Z}_4$, and $\mathbb{Z}_3 \times \mathbb{Z}_2$ where both maximal and prime.

Is there an easier way to do this by just looking at the structure of the ring? It feels like the way I approached this question was much slower than it should have been.

Use the Chinese remainder theorem and look for the prime ideals $I$ of the (principal) ring $A=\mathbb Z/ 12\mathbb Z.$ ($|A/I|$ must be a prime factor of $12$.) — Anne Bauval, Oct 20 '22 at 22:58
You may find this interesting : Find all prime ideals and maximal ideals of Z/12Z and Describe all prime and maximal ideals of Zn. — Anne Bauval, Oct 20 '22 at 23:13

suckling pig · Accepted Answer · 2022-10-21T04:05:09.043

2

First of all, when you're talking about ideals, you've got rings, not groups.

Second, in $\Bbb Z_3×\Bbb Z_4\cong \Bbb Z_{12}$, the prime ideals are the prime ideals in $\Bbb Z$ containing $(12)$, by the correspondence theorem.

So we get just $(12)\subset(2),(3)\subset \Bbb Z$, which correspond to $(\bar2),(\bar3)\subset \Bbb Z_{12}$. These are $\{(2,2),(1,0),(0,2),(2,0),(1,2),(0,0)\}, 0×\Bbb Z_4$ (after isomorphism).

So, note: $\Bbb Z_3×\Bbb Z_2$ is wrong. It contains $(1,1)$, so any ideal containing it is the whole ring.

You can also tell because the respective quotients, $\Bbb Z_2,\Bbb Z_3$ are integral domains.

edited Oct 21 '22 at 04:05

answered Oct 21 '22 at 00:53

suckling pig

1

The whole ring is a trivial ideal of any ring. – user317176 Oct 21 '22 at 03:21
@DougM yes, but by a prime ideal is meant a proper ideal. Also there was a typo. I meant $\Bbb Z_3×\Bbb Z_2$, as the OP wrote. – suckling pig Oct 21 '22 at 04:08

score 1 · Answer 2 · answered May 06 '25 at 18:17

Without referring to correspondence theorem and ideals of $\Bbb Z$.

Any ideal is an additive subgroup. So all proper ideals of $Z_{12}$ are $\langle2\rangle$, $\langle3\rangle$, $\langle4\rangle$, $\langle6\rangle$ as you a little bit wrong listed. Now, $$(6)=6(\Bbb Z_3\times\Bbb Z_4)=0\times2\Bbb Z_4$$ $$(4)=4(\Bbb Z_3\times\Bbb Z_4)=\Bbb Z_3\times0$$ are not prime ideals (e.g. $(2,2)\notin (6), (0,3)\notin(6)$ but $(2,2).(0,3)=(0,2)\in(6)$) and $$(2)=2(\Bbb Z_3\times\Bbb Z_4)=\Bbb Z_3\times2\Bbb Z_4$$ $$(3)=3(\Bbb Z_3\times\Bbb Z_4)=0\times\Bbb Z_4$$ are prime ideals, again by definition of a prime ideal.

Find all of the prime ideals of $\mathbb{Z}_3 \times \mathbb{Z}_4$

2 Answers2