How to find all the maximal ideals of $\mathbb Z_n?$

Question

How to find all the maximal ideals of $\mathbb Z_n?$

I think $(0)$ is the only maximal ideal of $\mathbb Z_n$ for if $a$ is a non-unit in a maximal ideal of $\mathbb Z_n$ then $(a,n)=1\implies\exists~u,v\in\mathbb Z$ such that $au+nv=1\implies au=1~(\equiv\mod n)\implies 1\in$ the maximal ideal !

Am I right?

One suggestion is to characterize all ideals first, and see which ones are maximal. — Sungjin Kim, Sep 05 '16 at 18:36

score 34 · Answer 1 · edited Jan 03 '22 at 10:07

34

The ideals of $\mathbb{Z}_n$ are, first of all, additive subgroups of $\mathbb{Z}_n$. These we know to all have the form $\langle d\rangle$, where $d$ divides $n$. But, as we know, the set $\langle d\rangle$ is the ideal generated by $d$. So we have just proven that the ideals in $\mathbb{Z}_n$ are precisely the sets of the form $\langle d\rangle$ where $d$ divides $n$. Since we are interested in maximal ideals, and this concept is defined in terms of containment of ideals in one another, we now need to determine when we can have $\langle d_1\rangle\subset \langle d_2\rangle$. This is the case if and only if $d_1 \in \langle d_2\rangle$.

Here is the main result that you are seeking for: An ideal $I$ in $\mathbb{Z}_n$ is maximal if and only if $I = \langle p \rangle$ where $p$ is a prime dividing $n$.

edited Jan 03 '22 at 10:07

user26857

52,094

answered Apr 19 '13 at 04:25

Srijan

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actually $p$ should be irreducible right? The fact that it is prime is a consequence of the fact that $\Bbb{Z}_n$ is a UFD – Bman72 Jan 06 '15 at 10:15
This answer is quite similar to this one. But this one is explained in a more detailed way. It can be helpful for someone http://ramanujan.math.trinity.edu/rdaileda/teach/m4363s07/HW3_soln.pdf – Byag Oct 03 '19 at 15:54

Alex Becker · Answer 2 · 2013-04-19T04:37:09.993

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Hint: The maximal ideals in $\mathbb Z_n$ are precisely the ideals in $\mathbb Z$ properly containing $(n)$ which are maximal w.r.t. this condition.

edited Apr 19 '13 at 04:37

answered Apr 19 '13 at 04:21

Alex Becker

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You used correspondence theorem for maximal ideals, right ? – Pritam Roy Aug 14 '22 at 14:06

score 6 · Answer 3 · answered Apr 19 '13 at 04:02

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I'm afraid not - in fact, $a\in \mathbb{Z}_n$ is a non-unit if and only if $(a,n)>1$.

answered Apr 19 '13 at 04:02

Zev Chonoles

129,973

score 2 · Answer 4 · answered Sep 05 '16 at 18:36

The ring $\mathbb Z_p$ is a principal ideal domain: its ideals are $\{0\}$ and $p^k \mathbb Z_p$ ($k \in \mathbb N$).

Proof. Let $I \neq \{0\}$ be a nonzero ideal of $\mathbb Z_p$ and $0 \neq a \in I$ an element of minimal order, say $v(a) = k$. Write $a = p^k u$ with $p$-adic unit $u$. Hence $p^k = u^{-1} a \in I$ and $(p^k) \subseteq I$. Conversely, for any $b \in I$ let $w(b) \ge k$ and write $$b = p^w u' = p^k p^{w-k} u' \in p^k \mathbb Z_p.$$ This shows that $I \subseteq p^k\mathbb Z_p$. $\mathbb Z_p$ is a local ring, its maximal ideal is $p\mathbb Z_p$.

Unless you meant $\mathbb Z/n\mathbb Z$ and not the ring of $p$-adic numbers...

score 1 · Answer 5 · answered Apr 19 '13 at 04:03

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Try an example like $\mathbb Z_6$!

answered Apr 19 '13 at 04:03

Ted Shifrin

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$\mathbb{Z}_6! = \mathbb{Z}_6 \times \mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_2 \times \mathbb{Z}_1$ – Alex Provost Apr 19 '13 at 05:06
@AlexProvost Nice....! – Sungjin Kim Sep 05 '16 at 18:28

score 0 · Answer 6 · answered Dec 05 '23 at 12:36

$\varphi: \mathbb{Z}\to \mathbb{Z}_n , Ker(\varphi)=n\mathbb{Z}$, we have 1-1 injection between the (prime or maximal)ideal $I$ of $\mathbb{Z}$ containing $Ker(\varphi)$ and the (prime or maximal)ideal $\varphi(I)$ of $\mathbb{Z}_n$

Then:

The prime ideals of $\mathbb{Z}_n$: $\varphi((p_1)),...,\varphi((p_k)), \varphi((0))$, where $n=p_1^{e_1}...p_k^{e_k}$
The maximal ideals of $\mathbb{Z}_n$: $\varphi((p_1)),...,\varphi((p_k))$

How to find all the maximal ideals of $\mathbb Z_n?$

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