# Group theory question

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• Nov 29th 2012, 03:07 PM
Ant
Group theory question
Hi,

Let $H_{1}, H_{2}, H_{3}$ be subgroups of a group $G$ under addition. Moreover suppose $x\in H_{1}, y\in H_{2}, x+y \in H_{3}$

I'm wondering if it then follows that $H_{1} = H_{2} = H_{3}$

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

Thanks!
• Nov 29th 2012, 03:25 PM
GJA
Re: Group theory question
Hi Ant,

Take a look at $H_{1}=2\mathbb{Z}, H_{2}=3\mathbb{Z}$ and $H_{3}=5\mathbb{Z}$ as subgroups of $(\mathbb{Z}, +)$ 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 PM
Ant
Re: Group theory question
Yes of course. $2 \in H_{1}$, $3 \in H_{2}$ and $2+3=5 \in H_{3}$ yet clearly these subgroups are not equal. Thanks! No wonder I couldn't prove it!
• Nov 29th 2012, 03:45 PM
Ant
Re: 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 $(x) \subset J_{1}$, $(y) \subset J_{2}$, $(x+y) \subset J_{3}$. For $J_{1}, J_{2}, J_{3}$ maximal ideals.

Our goal is to prove that $J_{1} = J_{2} = J_{3}$ 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 PM
GJA
Re: 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 $R$ contains two distinct maximal ideals $M_{1}$ and $M_{2}$. Without loss of generality take $x\in M_{1}-M_{2}.$ Since $x\in M_{1}$ and $M_{1}\neq R$, $x$ is a non-unit. Now take a look at the ideal $(x)+M_{2}$ and see if you can use the assumption to get a contradiction.

Good luck!
• Nov 29th 2012, 04:22 PM
Ant
Re: Group theory question
Thanks! I'll try that and see if I can come up with anything
• Nov 29th 2012, 08:20 PM
Deveno
Re: Group theory question
Quote:

Originally Posted by GJA
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 $R$ contains two distinct maximal ideals $M_{1}$ and $M_{2}$. Without loss of generality take $x\in M_{1}-M_{2}.$ Since $x\in M_{1}$ and $M_{1}\neq R$, $x$ is a non-unit. Now take a look at the ideal $(x)+M_{2}$ and see if you can use the assumption to get a contradiction.

Good luck!

oh i like that! (x) + M2 is an ideal containing M2, and so we have two choices:

a)(x) + M2 = R
b)(x) + M2 = M2.

b) is out of the question since x is in (x) + M2 (as the element 1x + 0) and by supposition, x is not in M2.

the key to ruling out a) is that M2 is proper, and thus doesn't contain any units, and neither does (x). but certainly 1 is in R.
• Nov 30th 2012, 12:51 AM
Ant
Re: Group theory question
Quote:

Originally Posted by Deveno
oh i like that! (x) + M2 is an ideal containing M2, and so we have two choices:

a)(x) + M2 = R
b)(x) + M2 = M2.

b) is out of the question since x is in (x) + M2 (as the element 1x + 0) and by supposition, x is not in M2.

the key to ruling out a) is that M2 is proper, and thus doesn't contain any units, and neither does (x). but certainly 1 is in R.

This seems to work perfectly. However, as far as I can see, at no point in this argument do we use the fact that if x,y are non units them so is their sum, x+y. This concerns me!
• Nov 30th 2012, 04:47 AM
Deveno
Re: 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-1r)x = 1, contradicting the fact that x is not a unit (and we know x is not a unit, because x is in M1, and M1 ≠ 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 M2 is ALSO not a unit.

now if (x) + M2 = R, then:

rx + m = 1, for some r in R, and some m in M2.

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 AM
Ant
Re: Group theory question
Ah okay, thanks. For some reason I was thinking that (x) + M2 was the union of (x) and M2. 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 $J$ of all non units in $R$.

Claim 1: $J$ is an ideal of $R$.
Proof: It's clear that $0$ is in $R$. Let $x$ be a non unit, assume $-x$ is a unit. So there exists $u$ s.t. $-xu = 1$ then $x(-u) = 1$ so $x$ is a unit. So $-x$ must be a non unit. Closure follows by assumption. so $J$ 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, $x$ be a non unit. Assume $ux$ is a unit. So there exists $w$ s.t $uxw=wux =1 = (wu)x$ So $wu$ is inverse of $x$ and thus $x$ is a unit. Contradiction proves $ux$ is non unit). So $J$ is an ideal.

Claim 2: $J$ is unique maximal.
Proof: $J \ne R$ because $R$ contains $1$. So we must still prove unique maximality.

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

(Maximality) Recall that $J$ contain all non units. This means that if we want to find an ideal of $R$ which is larger than $J$ we must include some non unit of $R$. But the inclusion of a non unit will immediately give us all of $R$. So $J$ is maximal.
• Nov 30th 2012, 06:37 AM
Deveno
Re: 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 AM
Ant
Re: Group theory question
Quote:

Originally Posted by Deveno
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).

Yes, that's a bit more succinct.

Apparently they're called "local rings"
Local ring - Wikipedia, the free encyclopedia