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Principal ideals of rings without unity

by Wingeer
Tags: ideals, principal, rings, unity
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Wingeer
#1
Oct19-11, 12:25 PM
P: 79
Both my book and lecturer have in the definition a ring omitted the requirement of a unity.
I was reading in my book about ideals, more specifically principal ideals. I stumbled over a formula that differed by whether or not the ring had a unity. As an example I state the two for principal left ideals for a ring R:
[tex](a)_l = \{ar+na | r \in R, n \in \mathbf{Z} \}[/tex]
[tex](a)_l = \{ar | r \in R,\}[/tex]
Why is the extra term omitted if the ring does not have a unity? I bet the explanation is easy answer, but despite how hard I am looking at it, I cannot figure it out.
I also looked at an example. I took 2Z which has no unity and looked at 4Z which is a principal left (or right) ideal generated by 4. The formula then dictates that:
[tex](4)_l = \{ 4r + 4n | r \in 2 \mathbf{Z}, n \in \mathbf{Z} \}[/tex]
But then why not just say that:
[tex](4)_l = \{ 4n | n \in \mathbf{Z} \} = 4 \mathbf{Z}[/tex]

Am I doing something horrendously wrong here?
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micromass
#2
Oct19-11, 01:12 PM
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I'm not used to working with rings without a unity. But I think I have the answer.

They key is that you of course want that [itex]a\in (a)[/itex]. But if we set

[tex]\{ar~\vert~r\in R\}[/tex]

then there is nothing which forces a to be in that set. In a ring with unity, we could simply set r=1 and it follows immediately that a is in the set. But without unity, we simply do not know.

That's why we set

[tex]\{ar+na~\vert~r\in R,n\in \mathbb{Z}\}[/tex]

taking r=0 and n=1 gives us that a is in the set. So in this case we do have [itex]a\in (a)[/itex]. But of course, if we have a in there, then we also need to have a+a=2a in there. So that's what the n is for.
Wingeer
#3
Oct20-11, 07:04 AM
P: 79
Aha! That makes perfect sense.
I guess the same explanation goes for the principal two sided ideal:
[tex] (a) = \{ \sum_{i} r_i a s_i + ra + as + na | r,s,r_i,s_i \in R, n \in \mathbf{Z} /} [/tex]
Where the sum is finite. If R had a unity, then first off we could set r_i and s_i = 1 which would imply that a is in the ideal. Moreover, without a unity we can't guarantee that either ra or as exists since we cant set any r_i or s_i equal to 1. But they have to exist by the definition of a principal two sided ideal.
If R has a unity, then we would have:
[tex] (a) = \{ \sum_{i} r_i a s_i | r_i,s_i \in R /} [/tex]

Wingeer
#4
Oct26-11, 11:46 AM
P: 79
Principal ideals of rings without unity

Another question somewhat related:
I just read in my book that "An ideal A in a ring R is maximal if and only if the pair X,A, for all ideals X not a subset of A, is comaximal".
What does it mean for two ideals to be comaximal? That X+A=R. Is this just taking every element in X, and every element in A and putting them together in one bigger rng? How can there exist several ideals X that satisfy this? Does this also mean that if a ring R has a maximal ideal A then you have a family of ideals that "spans" the whole ring?
And yeah, R does not necessarily have a unity.
micromass
#5
Oct26-11, 12:02 PM
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Quote Quote by Wingeer View Post
Another question somewhat related:
I just read in my book that "An ideal A in a ring R is maximal if and only if the pair X,A, for all ideals X not a subset of A, is comaximal".
What does it mean for two ideals to be comaximal? That X+A=R. Is this just taking every element in X, and every element in A and putting them together in one bigger rng? How can there exist several ideals X that satisfy this? Does this also mean that if a ring R has a maximal ideal A then you have a family of ideals that "spans" the whole ring?
And yeah, R does not necessarily have a unity.
I think you need an example. In the ring [itex]\mathbb{Z}[/itex]. If a and b satisfy gcd(a,b)=1, then [itex]a\mathbb{Z}[/itex] and [itex]b\mathbb{Z}[/itex] are comaximal.

So for example, [itex]4\mathbb{Z}[/itex] is comaximal with [itex]15\mathbb{Z}[/itex] and [itex]3247\mathbb{Z}[/itex]. So you can see that there are multiple ideal with which an ideal can be comaximal with.

Note that the maximal ideals of [itex]\mathbb{Z}[/itex] are exactly of the form [itex]p\mathbb{Z}[/itex] with p prime.
Wingeer
#6
Oct26-11, 12:04 PM
P: 79
I actually deduced that result earlier today. Should have thought of it when regarding multiple ideals comaximal to another one. Thanks. :-)
What, however, if the ring has no unity?


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