Hey,Need some help figuring out what a maximal ideal is: the question is to list all maximal ideals in Z_8 and Z_10. The maximal ideal is similiar to the normal subgroup, but I would think there would only be one per ring.Thanks for the help--
4/16/2006 1:09:56 PM
I think you're right. As I recall the maximal ideal is the biggest one not equal to the ring itself, lets see Wolfram has a nice defintion and some comments relevant to your question
4/16/2006 7:21:06 PM
I'll give you a couple of hints. First, no, there is not just one maximal ideal per ring; there can be arbitrarily many distinct maximal ideals.Second, if R is a ring and I is a maximal ideal, the quotient ring R/I is a field. That limits the possible cardinalities of R/I quite a bit (in the finite case, the cardinality must be the power of a prime).
4/16/2006 7:21:58 PM
mathman, I didn't have time to respond to you yesterday when I made my previous post.Anyway, first, your notation is a bit confusing. It's very nonstandard to refer to, for example, Z_7 as an ideal (maximal or not) of Z_8.(In Z_8, I'll use 0,1,2,...,7 to denote the respective equivalence classes)The only sensible way to interpret it, of course, is as the ideal generated by 7, generally denoted by (7).However, 7 is a unit: 7*7=1. If an ideal contains a unit, the ideal is the entire ring itself, and cannot be maximal.
4/17/2006 5:37:07 PM
from how I understand it, which is a pretty weak understanding, the order of an ideal has to devide the order of the ring, which I believe has already been said in different words. But with this, it's also apparent that in Z_8, <7>, which is relatively prime (and thus generates Z_8), cannot be a maximal ideal, since as Cabbage said, it's the entire ring.I think if the ideals have the same order, say, |<2>| = |<6>| in Z_8, then you just say that only <2> would be maximal, but I'd really like somebody else's opinion on this as well.Note: I'm responding more to make myself think through this again rather than for virga's sake.
4/18/2006 12:40:31 AM
4/18/2006 4:03:23 AM
yeah.. that they weren't distinct was more what I meant. Thanks for clarifying.
4/18/2006 8:28:45 AM
Actually, <6> is not an ideal of Z_8 because six does not divide 8 ---When finding the maximal ideals I've learned that you list out all the of the divisors of the n, and then find the orders of those elements, and see what is contained in what -- like drawing the lattice. What is at the top are the maximal ideals.omg 1600th post[Edited on April 18, 2006 at 10:46 AM. Reason : 1600]
4/18/2006 10:37:05 AM
4/18/2006 2:39:54 PM
4/18/2006 2:53:26 PM
Just to make one more thing clear (hopefully it's already clear, but I'm not sure):If I was grading answers to the problem, "Find all maximal ideals of Z_8" and saw the following answers:A. (2) is the only maximal idealB. (6) is the only maximal idealC. (0,2,4,6) is the only maximal idealD. (2) and (6) are the only maximal ideals(Of course, I would expect there to be some jusitification for the answers, but let's only look at the final answers for now).A, B, and C are all entirely correct, they simply expressed the same ideal in different manners. There are a few other natural ways to express it as well, such as (2,6) (the ideal generated by the set {2,6}). (2) and (6) seem to me the most elegant ways to describe the ideal (because of their conciseness, and I see no particular reason to prefer one over the other), but all are correct.D would bother me. It implies to me that the student believes (2) and (6) are distinct ideals, when in fact they are the same.
4/18/2006 6:02:22 PM
Hey Cabbage, thanks for your comments. It's been a while since I've gotten my hands dirty with the modular arithmatic examples. I see my error, wolfram was talking about maximal ideals of the integers, but we are interested in maximal ideal(s) of Z mod 8. To summarize,(1)= Z_8(2)=(6)={2,4,6,0} (generating with 2)(3)=(5)=(7)={3,6,1,4,7,2,5} (generating with 3)(4)={4,0}So Z_8 has additive subgroups of order 2 and 4 (which is 2^2=prime^2). These are also subrings since they are closed under multiplication mod 8 (both left and right). Then to make them an ideal we need that they are closed under (left/right) multiplication by the whole ring Z_8,{0,4}"times"{0,1,2,3,4,5,6,7}={0,4,8,12,16,20,24,28}={0,4} mod 8.the same goes for (2) it also closes back on (2) under multiplication by the whole Ring Z_8. So (2) and (4) are not just additive subgroups, they are ideals. By explicit computation we can see they are the only "proper" ideals of Z_8, hence (2)=(6) is the maximal ideal of Z_8.Of course, my presentation here is one rooted in abject ignorance to the many nice labor saving theorems that you have to play with. Memories of the Euler-phi function dance in my head, but the dance is blurred by the passing years I'm afraid...
4/18/2006 6:09:24 PM
No problem, mathman, glad to help. Everything's perfect, except for one nitpick (which looks more like a typo or omission rather than a real mistake):
4/18/2006 6:21:27 PM
cabbage got it[Edited on April 18, 2006 at 6:27 PM. Reason : .]
4/18/2006 6:26:38 PM
4/18/2006 6:34:36 PM