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

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- Feb 9th 2013, 01:05 PMBernhardPrime and Maximal Ideals
Find all prime ideals and all maximal ideals of $\displaystyle \mathbb{Z}_{12}$.

- Feb 9th 2013, 04:01 PMDevenoRe: Prime and Maximal Ideals
well, first, ask yourself this: what ARE the ideals of Z

_{12}? remember an ideal has to be a subgroup of the additive group. start looking there. - Feb 9th 2013, 10:01 PMBernhardRe: Prime and Maximal Ideals
Definitely tried to take up your challenge but apart from an exhaustive trial of all possible sets of elements of $\displaystyle \mathbb{Z}_{12} $ I did not discover a process (or a general theorem) that would yield all the ideals of $\displaystyle \mathbb{Z}_{12} $

I did notice that Fraleigh (Abstract Algebra) has the following exercise on page 243 [Exercise 3 of Exercieses 26]

"Find all ideals N of $\displaystyle \mathbb{Z}_{12} $"

In the Answers to Odd Numbered Exercises (page 502) Fraleigh gives the answer to this exercise as follows:

<0> = {0}

<1> = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} = $\displaystyle \mathbb{Z}_{12} $

<2> = {0, 2, 4, 6, 8, 10}

<3> = {0, 3, 6, 9}

<4> = {0, 4, 8}

<6> = {0, 6}

These certainly seem to be ideals of $\displaystyle \mathbb{Z}_{12} $, but how did Fraleigh determine these ideals, and further, how do we know they are the only sets which are, in fact, ideals?

Surely this set of ideals is not determined by exhaustive trial of all the subsets of $\displaystyle \mathbb{Z}_{12} $! Why are they the only ideals? Do we have a theorem governing the nature of the ideals of $\displaystyle \mathbb{Z}_{n} $?

I am assuming I apply the requisite tests to the above ideals to determine whether they are (1) prime (2) maximal. Is that corect?

Peter - Feb 10th 2013, 12:00 PMDevenoRe: Prime and Maximal Ideals
Z

_{12}is a CYCLIC group. as such, all its subgroups are cyclic, so it suffices to examine <a> for all elements a.

furthermore, for a cyclic group G, there exists exactly ONE subgroup of order d for each divisor d of the order of G.

Z_{12}is of order 12, and 12 has divisors 1,2,3,4,6 and 12.

for a cyclic group <a>, the order of <a^{k}> is |a|/gcd(k,|a|).

for additive groups, it is common to write a^{k}additively as ka.

Z^{12}has generator 1. looking at every element we see that:

<0> has order 12/gcd(0,12) = 12/12 = 1

<1> has order 12/gcd(1,12) = 12/1 = 12

<2> has order 12/gcd(2,12) = 12/2 = 6

<3> has order 12/gcd(3,12) = 12/3 = 4

<4> has order 12/gcd(4,12) = 12/4 = 3

<6> has order 12/gcd(6,12) = 12/6 = 2

note we have not listed all the elements of Z_{12}, although we have covered every possible divisor. so the elements we have not listed must lead to duplicates of the subgroups listed above. indeed:

<5> = <7> = <11> = <1>, since for all of these numbers k, gcd(k,12) = 1. also:

<10> = <2>, since gcd(10,12) = 2 (we have only ONE subgroup of order 6).

<8> = <4>, because gcd(8,12) = 4.

<9> = <3>, because gcd(9,12) = 3.

**************

this shows that for Z_{n}, all the possible subgroups of (Z_{n},+) are of the form <d>, where d is a divisor of n (it is customary to write <n> as {0}). are these ideals of the RING Z_{n}?

suppose we have an integer d, where d|n, (so n = kd) and we consider the multiples of d mod n. clearly the set {0,d,2d,....,(k-1)d} is a cyclic subgroup of the additive group.

now suppose a is ANY element of Z_{n}. using a somewhat standard abuse of notation, we will identify a with the integer in the range 0 ≤ a ≤ n-1 it is congruent to.

what we now have to do is show that a(td) (mod n) is in <d>, for 0 ≤ t ≤ k-1. however:

a(td) (mod n) = (at)d (mod n) = (at (mod n))(d (mod n)) = ((at (mod n))(d).

when we reduce at (mod n), we obtain an integer between 0 and n-1, say b. next we can reduce b (mod k), to obtain cd, where c is between 0 and k-1, showing that a(td) is indeed in <d>.

let me illustrate this with n = 12, and d = 3. so <d> = <3> = {0,3,6,9}. our "k" here is 4. suppose we take a = 5, and t = 2.

so we need to show that 5(2*3) is in <3>. first we see that 5(2*3) (mod 12) = (4*2)3 (mod 12) = 10*3 (mod 12).

now 10*3 = (2*4 + 2)*3 = (2*4)3 + 2*3 = 2(4*3) + 2*3 = 2*12 + 6 = 0 + 6 = 6 (mod 12) (this is why we can reduce at (mod k), instead of at (mod n), because kd = n = 0 (mod n)).

or: the "long way"- 5(2*3) = 5*6 = 30 = 6 + 24 = 6 (mod 12). that's part of the beauty of "modular arithmetic", you can reduce (mod 12) before evaluating sums and products, or after, you get the same answer either way.

this shows that for the ring Z_{n}, the subgroup <d> for any divisor d, is also an ideal, and that these are the only POSSIBLE ideals. in other words, the ideals of Z_{n}are all principal.

(unfortunately, Z_{n}isn't a PID, since it will have zero-divisors if n is not prime, so it isn't in general, an integral domain).

this simple sub-structure of the ring Z_{n}lets us see something very clearly: an ideal of Z_{n}, (d), is maximal if and only if d is prime. for example, in Z_{12}:

4 is not prime, it has the divisor 2. this means that (4) is contained in (2), which is easy to see. thus the maximal ideals of Z_{12}are (2) and (3).

it's pretty easy to see that (2) is maximal, if we add any odd integer k , and consider the ideal (k,2), we clearly get ALL odd elements of Z_{12}.

perhaps it's not so obvious (3) is maximal. but gcd(2,3) = 1, right? this means that, for example, we have 3*3 - 4*2 = 1, so that 1 is in (2,3), so the ideal (2,3) = Z_{12}.

in fact, if we choose any k NOT in (3), we will have gcd(k,3) = 1, since all the multiples of 3 (and so the elements that have 3 as a factor) are already in (3).

the same reasoning holds for any prime p where p|n. this is the reason why ideals such that I+J = R are called co-prime ideals, we are generalizing this basic example.

so that takes care of the maximal ideals. what about the prime ideals? for Z_{n}i claim:

the maximal ideals ARE the prime ideals. consider (p), where p is a prime dividing n. this is a maximal ideal, as we have seen (for if (p) is contained in (a), for some a|n in Z_{n}, then a|p, so either a = p, so that (a) = (p), or a = 1, so that (a) = Z_{n}).

now if ab is in (p), we have ab = kp (mod n). that is ab = kp + tn. suppose that a is not in (p). this means that a = sp + u, where u is between 1 and p-1, and s is between 0 and n/p.

let's write b = xp + y, where x is between 0 and n/p, and y is between 0 and p-1.

so ab = (sp + u)(xp + y) = (sxp + sy + ux)p + uy = kp + tn. hence:

tn + uy = (sxp + sy + ux - kx)p <--this is an equation of integers, not congruence classes.

if we take both sides mod p (remember, p divides n), we get:

uy = 0 (mod p), that is, p divides uy. by our choice of u, p does NOT divide u, so p MUST divide y. since y < p, the only choice left is y = 0, that is b = xp, so that b is in (p). hence (p) is a prime ideal.

i admit this is somewhat long-winded, but i want to make it clear how the structure of the ordinary integers is "partially preserved" in Z_{n}.

so the maximal ideals in Z_{n}are prime. let's go the other way: suppose (d) is a prime ideal of Z_{n}. if d is composite, we have d = ab, for 1 ≤ a,b < d.

so then ab = d is in (d), but neither a nor b can be in (d), since d is the smallest (non-zero) element of (d) = {0,d,2d,...,(n/d - 1)d}. so if (d) is prime, d must be prime, and as we have seen, (p) is maximal for a prime p dividing n.

this shows the prime ideals are also the ideals (p) for a prime p|n, which are maximal.

if you feel up to it, find the maximal ideals of Z_{60}now.