MHB Prime Ideals and Maximal Ideals

[Math Amateur]

1) Find all prime ideals and all maximal ideals of \mathbb{Z}_{12}.

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

 

[Fernando Revilla]

Re: Prime Ideals and Maximal Ideals of Z12

Find all prime ideals and all maximal ideals of \mathbb{Z}_{12}.

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.

 

[Math Amateur]

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]

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 $\mathfrak{b}=\phi^{-1}(\bar{\mathfrak{b}})$.

 

[jakncoke1]

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 1,2,3,4,6,12And 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 don't say 1 is a prime number.