Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

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).

Does this help?  Write back if you'd like to talk about this 
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

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/