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Principal ideals of rings without unity

  1. Oct 19, 2011 #1
    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?
  2. jcsd
  3. Oct 19, 2011 #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.
  4. Oct 20, 2011 #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 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]
    Last edited: Oct 20, 2011
  5. Oct 26, 2011 #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.
  6. Oct 26, 2011 #5
    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.
  7. Oct 26, 2011 #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?
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