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Commutative Ring, Maximal Ideal

Date: 12/08/2003 at 02:05:15
From: Johny
Subject: maximal ideal

I've been asked to prove that in a comutative ring any ideal is
contained in some maximal ideal.

But I'm not sure why there should be any maximal ideal.


Date: 12/09/2003 at 03:01:34
From: Doctor Jacques
Subject: Re: maximal ideal

Hi Johny,

Note first that we need an additional condition: the ring must contain 
an element 1, i.e., it must be a "unit ring".  The "official"
definition of a ring does not require the existence of an element 1, 
although many authors do understand "ring" as meaning "unit ring".

The proof makes use of Zorn's lemma:

  MathWorld: Zorn's Lemma
    http://mathworld.wolfram.com/ZornsLemma.html 
  
  Proving Mathematical Induction is Correct
    http://mathforum.org/library/drmath/view/55696.html 

Let R be a commutative unit ring, and J a proper ideal of R.  We 
consider the set X of proper ideals of R containing J.

The idea is to order that set by inclusion, and to show that the 
resulting order statisfies the hypotheses of Zorn's lemma, i.e., that 
it is inductive.  Zorn's lemma then ensures the existence of a maximal 
element in X.  In this case, this means a maximal (proper) ideal of R 
containing J.

We must prove that, if Y is a totally ordered subset of X, Y has an 
upper bound in X.

"Y is totally ordered" means that, if L and K are two elements of Y 
(proper ideals of R), then one of them is contained in the other. 
(This does not necessarily mean that Y is a sequence:

  K1 < K2 < ...

since Y could be uncountable.)

Let M be the union of the ideals in Y, i.e. the set of elements that 
belong to at least one of the ideals of Y.  M is obviously an upper 
bound for Y: any element of Y is contained in M.  We must show that M 
belongs to X, i.e., that M is a proper ideal of R.

We show first that M is an ideal. We must prove:

  (1) M is not empty.

  (2) If x and y are in M, x - y is in M.

  (3) If x is in M and r is in R, then rx is in M.

(1) is obvious: M contains J as a subset.

To prove (2), assume that x and y are elements of M.  By the 
definition of M, this means that x and y belong to two ideals of Y, 
say x in K1 and y in K2.  Now, by hypothesis, one of these ideals is 
contained in the other.  If, for example, K1 is contained in K2, then 
K2 contains both x and y, and therefore also contains x - y, since K2 
is an ideal.  As K2 is a subset of M, M also contains x - y.

To prove (3), note that x belongs to some ideal K in Y, and therefore 
rx belongs to the same ideal K, and, as K is a subset of M, rx belongs 
to M.

To prove that M is a proper ideal, note that an ideal is proper if 
and only if it does not contain 1 (this is where we need the existence 
of 1).

If M were not proper (M = R), then M would contain 1, and, by the 
definition of M, one of the ideals of Y would contain 1, i.e., it 
would not be a proper ideal.  This contradicts the definition of X as 
the set of proper ideals containing J.

To summarize, we have shown that every totally ordered subset of X 
has an upper bound in X.  By Zorn's lemma, X contains a maximal 
element, and, in this case, this means a maximal ideal containing J.

To show that the existence of 1 (more accurately, a multiplicative 
identity) is required, consider the ring Q' defined on Q by taking 
addition as usual, but defining multiplication by x#y = 0 for all
x, y.  It is easy to check that this does satisfy the definition of a 
ring (except for the existence of 1, of course).  Now, the ideals of 
Q' are simply the additive subgroups of Q, and (Q,+) does not contain 
a maximal subgroup.

More generally, the theorem is true whenever R is finitely generated 
as a module over itself (if R is a unit ring, it is generated by 1).

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: 12/10/2003 at 01:35:04
From: Johny
Subject: maximal ideal

Hi Dr. Jacques,

Thank you very much!  It helped a lot.  However, it seems like I'm 
still missing something.  Doesn't this method prove that there is 
actually the ideal not identical with R that contains all other 
ideals?  If we take X as the set of all ideals and order them by set 
inclusion, do the conditions of the Kuratowski-Zorn lemma still hold? 
Certainly, any chain would still have an upper bound, doesn't it?  I 
would think there's something wrong, but can't think of what it is. 

Johny


Date: 12/10/2003 at 10:14:46
From: Doctor Jacques
Subject: Re: maximal ideal

Hi again, Johny,

First let me congratulate you--the right reaction to a dubious proof 
is to ask, "Does this allow me to prove something that I know is not 
true?" 

There are several subtle aspects to your question.

You may consider all the proper ideals of R, ordered by inclusion.  In 
this case, Zorn's lemma proves that there is a maximal proper ideal--
this is simply a weaker result.  In fact, we proved that there is a 
maximal ideal containing J, where J is any proper ideal.

Note that Zorn's lemma does not say that there is a greatest element, 
i.e., an ideal that contains all the other ideals.  It merely asserts 
that there is a proper ideal not strictly contained in any other 
proper ideal.

Generally, in terms of a partial order relation <=, we have to 
distinguish two things:

  (1)  A maximal element is an element m such that, if x <= m, 
       then m = x.

  (2)  A greatest element is an element M such that, for all x, 
       x <= M.

A greatest element is always a maximal element, but the converse is 
not true.  Zorn's lemma only asserts the existence of a maximal 
element when the set is inductively ordered.

For example, if you consider the set of integers > 1, and the relation 
a <= b iff b divides a, then any prime number is a maximal element (no 
other element divides it), but not a greatest element (it does not 
divide all the numbers under consideration).  In fact, this translates 
directly into proper ideals of Z.

Another point to be kept in mind is that the union of all proper 
ideals is generally not an ideal.  If J and K are two ideals (or, more 
generally two subgroups) such that neither of them contains the other, 
we can find j in J \ K and k in K \ J.  Now, j + k does not belong to
J (since then k would be in J) not to K (since then j would be in K), 
and therefore j + k does not belong to J (union) K.  J (union) K is 
not closed under addition, so it is not an ideal.

We could also, instead of the union of ideals, consider the operation 
defined as <J,K> = L, where L is the smallest ideal containing J and 
K.  By definition, L is indeed an ideal, but, if J and K are proper 
ideals, it does not follow that L is a proper ideal.  For example, in 
Z, <2> and <3> are proper ideals, but <2,3> = Z is not.  The fact that 
neither <2> nor <3> contains 1 does not imply that <2,3> does not 
contain 1, since <2,3> is not simply the union of the other two 
ideals.  The argument used in the proof only works for totally ordered 
sets of ideals.

I hope this clarifies things a little.  Please feel free to write back 
if you require further help.

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


Date: 12/10/2003 at 11:47:42
From: Johny
Subject: Thank you (maximal ideal)

Thank you very much, Dr. Jacques!  I think now I understand it.  Thank
you again!

Johny
Associated Topics:
College Modern Algebra

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