Ideals of ZDate: 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/ |
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