Hi,

Let be subgroups of a group under addition. Moreover suppose

I'm wondering if it then follows that

Could anyone offer any help as to whether or not this is true?

Thanks!

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- Nov 29th 2012, 03:07 PMAntGroup theory question
Hi,

Let be subgroups of a group under addition. Moreover suppose

I'm wondering if it then follows that

Could anyone offer any help as to whether or not this is true?

Thanks! - Nov 29th 2012, 03:25 PMGJARe: Group theory question
Hi Ant,

Take a look at and as subgroups of to see if you can come up with a counterexample.

If this is too cryptic let me know and I'll try to provide more details. Good luck! - Nov 29th 2012, 03:36 PMAntRe: Group theory question
Yes of course. , and yet clearly these subgroups are not equal. Thanks! No wonder I couldn't prove it!

- Nov 29th 2012, 03:45 PMAntRe: Group theory question
The problem I'm working is actually:

Let R be a commutative ring with unity. Prove that if the sum of two non units is a non unit then R has a unique maximal ideal.

My working so far:

Let x,y be non zero non units. So the ideals they generate are proper subgroups of R. Furthermore the ideal that x+y generates is also proper. We also know that every proper ideal is contained in a maximal ideal.

so , , . For maximal ideals.

Our goal is to prove that i.e. that There is unique maximal ideal.

The only thing I can think of to do at the moment, is use the closure of ideals to show that the intersection of (x) and (y) will contain xy = yx. but I'm not sure how, if at all, that helps me... - Nov 29th 2012, 04:12 PMGJARe: Group theory question
Seems like a fun problem! I think looking at the ideal generated by a non unit is a good idea. Here's my two cents (for what it's worth):

By way of contradiction suppose contains two distinct maximal ideals and . Without loss of generality take Since and , is a non-unit. Now take a look at the ideal and see if you can use the assumption to get a contradiction.

Good luck! - Nov 29th 2012, 04:22 PMAntRe: Group theory question
Thanks! I'll try that and see if I can come up with anything

- Nov 29th 2012, 08:20 PMDevenoRe: Group theory question
oh i like that! (x) + M

_{2}is an ideal containing M_{2}, and so we have two choices:

a)(x) + M_{2}= R

b)(x) + M_{2}= M_{2}.

b) is out of the question since x is in (x) + M_{2}(as the element 1x + 0) and by supposition, x is not in M_{2}.

the key to ruling out a) is that M_{2}is proper, and thus doesn't contain any units, and neither does (x). but certainly 1 is in R. - Nov 30th 2012, 12:51 AMAntRe: Group theory question
- Nov 30th 2012, 04:47 AMDevenoRe: Group theory question
sure we do. take any element of (x) (which is to say rx for some r in R). this cannot be a unit, for if so, we have, say rx = u, then we have:

(u^{-1}r)x = 1, contradicting the fact that x is not a unit (and we know x is not a unit, because x is in M_{1}, and M_{1}≠ R

-this is using the fact that if an ideal of a commutative ring with unity contains a unit, it contains 1, and thus it is the entire ring).

by the same reasoning, any element of M_{2}is ALSO not a unit.

now if (x) + M_{2}= R, then:

rx + m = 1, for some r in R, and some m in M_{2}.

so we have:

non-unit + non-unit = unit, contradicting what we are given as a condition on R. thus any two maximal ideals of R cannot be distinct (the assumption that allowed us to assume x existed). - Nov 30th 2012, 06:20 AMAntRe: Group theory question
Ah okay, thanks. For some reason I was thinking that (x) + M

_{2}was the union of (x) and M_{2}. Which is why I thought we didn't need to use the closure under + of non units. BTW I've since realized that in fact considering the union isn't helpful as it may no even be an ideal.

If anyone is interested, here's another proof (which I believe is also correct!):

Consider the set of all non units in .

Claim 1: is an ideal of .

Proof: It's clear that is in . Let be a non unit, assume is a unit. So there exists s.t. then so is a unit. So must be a non unit. Closure follows by assumption. so forms an abelian group under . The product of two non units is clearly non unit, and so is the product of a unit with a non unit. (let u be a unit, be a non unit. Assume is a unit. So there exists s.t So is inverse of and thus is a unit. Contradiction proves is non unit). So is an ideal.

Claim 2: is unique maximal.

Proof: because contains . So we must still prove unique maximality.

(Uniqueness) Consider an arbitrary proper ideal of , . is proper and therefore cannot contain any units. As is the set of all units, we have that .

(Maximality) Recall that contain all non units. This means that if we want to find an ideal of which is larger than we must include some non unit of . But the inclusion of a non unit will immediately give us all of . So is maximal. - Nov 30th 2012, 06:37 AMDevenoRe: Group theory question
for claim 2 i would word it like so:

let I be a maximal ideal of R. since I is a maximal ideal it is proper, and therefore contains no units. since J contains all non-units, I is contained in J, hence I = J (by the maximality of I, since J ≠ R).

i would be curious to see what kind of ring R might have to be, since the integers don't qualify: -2 and 3 are not units, but -2+3 is. the ring Q[x] also doesn't appear to work:

neither x nor 1-x are units, but their sum is. the only examples of such rings that spring to mind are fields (which have boring maximal ideals: {0}), but there might be others (i haven't thought about it too much). - Nov 30th 2012, 06:49 AMAntRe: Group theory question
Yes, that's a bit more succinct.

Apparently they're called "local rings"

Local ring - Wikipedia, the free encyclopedia