# One contradiction between maximal ideals and prime ideals.

• Dec 23rd 2012, 06:06 PM
Rita
One contradiction between maximal ideals and prime ideals.
First of all, according to the textbook that I use, there are some theorems about maximal and prime ideals.
(a) If R is a commutative ring such that R2=R, then every maximal ideal M in R is prime.
(b) In a commutative ring R with identity maximal ideals always exist.
(c) In a commutative ring R with identity 1R, which is not equal to 0, an ideal P is prime iff the quotient ring R/P is an integral domain.

It seems all right, but the contradiction happens when I think about Zm, the integers modulo m with m not prime.
This is my reasoning:
First, Zm is a commutative ring with identity 1R, which is not equal to 0. And since the identity exists in this ring, Zm2=Zm, so maximal ideals exist and they are prime.
On the other hand, since m is not prime, Zm has zero divisors, say xy=m, 1<x,y<m, then for all ideals I in Zm, Zm/I has zero divisors because (x+I)(y+I)=xy+I=I where x+I and y+I are not equal to I. This means Zm/I is not an integral domain for all ideals I in Zm. By (c), all ideals are not prime. But what about the maximal ideals? They do exist and they're prime.

Can anyone tell me what's wrong with my thought?
• Dec 23rd 2012, 09:16 PM
Deveno
Re: One contradiction between maximal ideals and prime ideals.
let's look at Z12.

since this is a cyclic ring, every ideal is of the form (a), for some a. so here are the ideals of Z12:

(0) = {0}
(1) = Z12, not maximal, not prime (by definition the entire ring is excluded from being a prime ideal)
(2) = {0,2,4,6,8,10} <---this is a maximal ideal.
(3) = {0,3,6,9} <---this is a maximal ideal.
(4) = {0,4,8} <--this is NOT a maximal ideal, as it is contained in (2).
(5) = (1) = Z12(since 5*5 = 1 (mod 12))
(6) = {0,6} <---not maximal, contained in (3) and (2)
(7) = (1) (verify!)
(8) = {0,8,4} = (4) <--- not maximal
(9) = {0,9,6,3} = (3) <---maximal
(10) = {0,10,8,6,4,2} = (2) <---maximal
(11) = (1) (obviously since 11 = -1).

it should be clear that Z12/(2) is isomorphic to Z2, which is an integral domain (a field, in fact).

also Z12/(3) is isomorphic to Z3, also an integral domain.

so let's look at what happens for Z12/(4) (note (4) is neither maximal, nor prime: we have 4 = 2*2 in (4), but 2 is not in (4)).

here are the cosets (equivalence classes):

(4) = 0 + (4)
1 + (4)
2 + (4)
3 + (4)

now (2 + (4))(2 + (4)) = 2*2 + (4) = 4 + (4) = (4) = 0 + (4), so 2 + (4) is a zero-divisor in Z12/(4).

this tells us (by (c)) that (4) must not be a prime ideal of Z12, and thus (a) tells us (4) cannot be maximal (but we knew that already, right?).

*********

the flaw in your reasoning is in assuming Zm/I must contain 0-divisors for any I. yes if xy = m, then (x + I)(y + I) = xy + I = I. but with m = 12, we have the following possibilities:

x = 1, y = 0 <--then y + I = I
x = 2, y = 6 note that we have x in I for I = (2), and y in I for I = (3), which are the only maximal ideals of Z12.
x = 3, y = 4 now x is in I for I = (3), and y is in I for I = (2).

we never have a situation where neither x NOR y is in a maximal ideal.