1. ## Group theory question

Hi,

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

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

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

Thanks!

2. ## Re: Group theory question

Hi Ant,

Take a look at $\displaystyle H_{1}=2\mathbb{Z}, H_{2}=3\mathbb{Z}$ and $\displaystyle H_{3}=5\mathbb{Z}$ as subgroups of $\displaystyle (\mathbb{Z}, +)$ 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!

3. ## Re: Group theory question

Yes of course. $\displaystyle 2 \in H_{1}$, $\displaystyle 3 \in H_{2}$ and $\displaystyle 2+3=5 \in H_{3}$ yet clearly these subgroups are not equal. Thanks! No wonder I couldn't prove it!

4. ## 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 $\displaystyle (x) \subset J_{1}$, $\displaystyle (y) \subset J_{2}$, $\displaystyle (x+y) \subset J_{3}$. For $\displaystyle J_{1}, J_{2}, J_{3}$ maximal ideals.

Our goal is to prove that $\displaystyle J_{1} = J_{2} = J_{3}$ 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...

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

Good luck!

6. ## Re: Group theory question

Thanks! I'll try that and see if I can come up with anything

7. ## Re: Group theory question

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 $\displaystyle R$ contains two distinct maximal ideals $\displaystyle M_{1}$ and $\displaystyle M_{2}$. Without loss of generality take $\displaystyle x\in M_{1}-M_{2}.$ Since $\displaystyle x\in M_{1}$ and $\displaystyle M_{1}\neq R$, $\displaystyle x$ is a non-unit. Now take a look at the ideal $\displaystyle (x)+M_{2}$ 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.

8. ## Re: Group theory question

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!

9. ## 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).

10. ## 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 $\displaystyle J$ of all non units in $\displaystyle R$.

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

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

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

(Maximality) Recall that $\displaystyle J$ contain all non units. This means that if we want to find an ideal of $\displaystyle R$ which is larger than $\displaystyle J$ we must include some non unit of $\displaystyle R$. But the inclusion of a non unit will immediately give us all of $\displaystyle R$. So $\displaystyle J$ is maximal.

11. ## 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).

12. ## Re: Group theory question

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