Principal ideals of rings without unity

In summary, the book and lecturer have different definitions of an ideal. The book says that an ideal is a ring with a unity, and the lecturer says that an ideal is a ring with no unity. The ring without unity example the lecturer gave is when you set r=1 and n=1, and a is in the set. But of course, if a is in the set, then a+a=2a is also in the set.
  • #1
Wingeer
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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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  • #2
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.
 
  • #3
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 can't 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]
 
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  • #4
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.
 
  • #5
Wingeer said:
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.
 
  • #6
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?
 

1. What are principal ideals of rings without unity?

Principal ideals of rings without unity are subsets of the ring that are generated by a single element. In other words, they are sets of elements that can be obtained by multiplying a given element by all the other elements in the ring.

2. How do principal ideals differ from regular ideals?

Unlike regular ideals, principal ideals are generated by a single element and therefore have a simpler structure. However, not all ideals are principal, and some rings may have both principal and non-principal ideals.

3. Can a ring without unity have a principal ideal?

Yes, a ring without unity can have principal ideals. In fact, every ideal in a ring without unity is a principal ideal, since there is no unity element to generate non-principal ideals.

4. What is the significance of principal ideals in ring theory?

Principal ideals are important in ring theory because they allow us to study the structure of rings without unity in a simpler way. They also have applications in algebraic number theory and algebraic geometry.

5. Are there any connections between principal ideals and prime elements in rings without unity?

Yes, there is a close connection between principal ideals and prime elements in rings without unity. In fact, every principal ideal is a prime ideal, and every prime element generates a principal ideal in a commutative ring without unity.

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