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### Ideals of Z

Date: 07/30/2003 at 00:39:36
From: Vivien
Subject: Ideals, Rings

What are the ideals of Z?

How to find all of the subgroups of (Z, +)?

I learned from some answered questions in Dr. Math that "Ideals of Z
must be of the form mZ for some integer m." Why is that true?

Date: 07/30/2003 at 02:16:07
From: Doctor Jacques
Subject: Re: Ideals, Rings

Hi Vivien,

It is indeed true that all ideals of Z are of the form mZ for some
integer m (we say that all ideals are principal), and this is a very
important result.

Let J be an ideal of Z. This means that:

(a)  for all x, y in J, (x +/- y) is in J
(b)  for all x in J and n in Z, nx is in J

If J is the 0 ideal, {0}, then J = 0Z and we are done.

If this is not the case, J contains at least one non-zero element a.
We may assume that a is positive, since J must contain both a and
-a = (-1)a.

Consider the set J+ of positive elements of J. This is a non-empty
set of positive integers, and therefore it contains a smallest
element. Let d be the smallest element of J+.

We must show that J = dZ, i.e.

1) every element of dZ (every multiple of d) is in J
2) every element of J is in dZ, i.e. is divisible by d.

(1) is easy : as d is an element of J by construction, every element
nd is in J, by the definition of an ideal (property (b)).

For (2), let us take an element x of J. We must show that x is
divisible by d.

We may divide x by d:

x = qd + r

where q is the quotient, and r is the remainder, and we have:

0 <= r < d

We may also write that as:

r = x - qd

Now, J contains d and x, and therefore also contains:

qd           (property (b))
x - qd = r   (property (a))

and so J contains r. Now, if r > 0, r is a positive element of J,
smaller than d, and this is a contradiction, because we chose d as
the smallest positive element of J.

We may conclude that r = 0, i.e. x is a multiple of d. As x is any
element of J, this proves (2).

some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/

Date: 07/30/2003 at 05:54:00
From: Vivien
Subject: Thank you (Ideals, Rings)

Hi, thanks a lot! It really helps me. Now my concepts are
much clearer.

Date: 08/05/2003 at 22:17:57
From: Vivien
Subject: Ideals, rings

1) Find the prime ideals of Z
2) What are the maximal ideals of Z?

Date: 08/06/2003 at 03:18:01
From: Doctor Jacques
Subject: Re: Ideals, rings

Hi Vivien,

To answer those questions, we must use the definitions of prime and
maximal ideals, and the fact that all ideals of Z are principal, i.e.
of the form mZ for some integer m (this was one of your previous
questions).

J is a prime ideal if, whenever xy is in J, either x or y is in J,
i.e.:

xy in J --> x in J or y in J     [1]

As J = mZ, this means:

m | xy --> m | x or m | y        [2]

(where | means "divides")

Consider first the case m <> 0. If m is prime, property [2] is true,
and mZ is a prime ideal. Conversely, if m is not prime, say m = pq, m
divides pq, but does not divide p or q, so mZ is not a prime ideal.

If m = 0, mZ is the zero ideal {0}. property [1] means:

xy = 0 --> x = 0 or y = 0        [3]

and this is also true (it is the "zero factor" theorem, which means
that Z does not contain zero divisors).

In conclusion, the prime ideals of Z are {0} and the ideals mZ, where
m is prime.

An ideal J is maximal if J is a proper ideal (not the whole ring) and
is not strictly contained in another proper ideal, i.e. if there are
no proper ideals between J and Z.

Note that, if J and K are ideals of Z, we have J = mZ and K = nZ. J
is contained in K iff n | m (do you see why?), and the inclusion is
strict if n is a proper factor of m. J is therefore a maximal ideal
if m nas no proper factor, i.e. if m is prime. Note that {0} is not a
maximal ideal.

It can be shown that, in any ring, every maximal ideal is prime. The
converse is not always true (for example, the zero ideal is prime but
not always maximal - this is a special case - but there are also
other more meaningful examples in some other rings).

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
Associated Topics:
College Modern Algebra

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