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!
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...
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!
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.
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).
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.
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