Prime Ideals and Maximal Ideals

Peter

Well-known member
MHB Site Helper
1) Find all prime ideals and all maximal ideals of [TEX] \mathbb{Z}_{12}[/TEX].

2) Find all prime ideals and maximal ideals of $$\displaystyle \mathbb{Z}_2 \ \times \ \mathbb{Z}_4$$.

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Fernando Revilla

Well-known member
MHB Math Helper
Re: Prime Ideals and Maximal Ideals of Z12

Find all prime ideals and all maximal ideals of [TEX] \mathbb{Z}_{12}[/TEX].
According to a well-known theorem, the ideals of $Z_{12}$ are $Z_{12},2Z_{12},3Z_{12},4Z_{12},6Z_{12}$ and $12Z_{12}=\{0\}$. Now, find the corresponding quotient rings. For example, $Z_{12}/2Z_{12}$ (isomorphic to $Z_{6}$) is neither integral domain nor field so, $2Z_{12}$ is neither prime ideal nor maximal ideal. $Z_{12}/4Z_{12}$ (isomorphic to $Z_{3}$) is field so, $4Z_{12}$ is maximal (as a consquenece prime), etc.

Peter

Well-known member
MHB Site Helper
Thanks Fernando, most helpful.

Are you able to give me the statement of the theorem that's you refer to in your post.

Thanks again,

Peter

Fernando Revilla

Well-known member
MHB Math Helper
Are you able to give me the statement of the theorem that's you refer to in your post.
Yes, I quote Proposition 1.1 from Atiyah, MacDonald's Introduction to Commutative Algebra:

Let $A$ be a commutative and unitary ring and $\mathfrak{a}$ an ideal of $A$. Then, there is a one-to-one order-preserving corresponding between the ideals $\mathfrak{b}$ of $A$ which contains $\mathfrak{a}$ and the ideals $\bar{\mathfrak{b}}$ of $A/\mathfrak{a}$, given by $\mathfrak{b}=\phi^{-1}(\bar{\mathfrak{b}})$.

jakncoke

Active member
Ok, here is just another couple of points.

If I is an ideal of a ring R, then I is first and foremost an additive subgroup of the group (R, +).
Since $Z_{12} = <1>$ is a cyclic subgroup under addition. There is exactly one unique subgroup of size p where p is a divisor for 12. So the divisors are 1,2,3,4,6,12

And these additive subgroups are (Subgroup of 1 element)$<1^{12/1}>$,(Subgroup of 2 elements $<1^{12/2}>$, (Subgroup of 3 elements)$<1^{12/3}>$ etc...

All these additive subgroups (Denoted by S) are ideals of $Z_{12}$ for if $r \in Z_{12}$ and $x \in S$ then x*r = r * x = x + x + x + x.. + (r times) $\in S$ (since S is closed under addition).

It is not hard to show that for any cyclic ring R(Ring with a cyclic group under addition), the ideals of R are the subgroups of (R,+), Hence ideals of size p where p divides the order of R

Now for any ideal I of $Z_{12}$, the order of $Z_{12}/I$ is $12/d$ where d is a divisor of 12. So yu have order of $Z_{12}/I$ = 1,12,6,4,3,2. So we know for I to be maximal $Z_{12}/I$ must be a field, or must have order $p^n$ where p is a prime and n is a positive integer. so the ideals where the order of $Z_{12}/I$ = 4,3,2. are maximal ideals. Or maximal ideals are 3$Z_{12}$,4$Z_{12}$ $6Z_{12}$.
Thus these are also prime ideals.

The others, namely, $Z_{12}/2Z_{12} \cong Z_6$, is not a prime ideal because 2*3 = 0 $\in Z_6$, which means it has atleast one zero divisor. and $Z_{12}/12Z_{12} \cong Z_{12}$ $\{e\}$is also not prime since 6*2 = 0 $\in Z_{12}$

(We do not include the ideal $Z_{12}$ in consideration for being a "prime" ideal, akin to how we dont say 1 is a prime number.

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