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Commutative Ring

  1. Sep 21, 2010 #1
    Suppose we let R be a commutative ring with identity, and let I be any ideal of R. And we define the RADICAL of I to be the set N(I) = {[tex]r \in R[/tex]: [tex]r^n \in I[/tex] for some positive integer n}.

    I need the proof that:

    1. If I contains a unit of R, then show that I = R.
    2. N(N(I))=N(I)

    An integral domain is a commutative, unital ring that contains no zero-divisors. So I'm guesing R must be an ID. If R is a ring with identity 1R, then [tex]a \in R[/tex] is a unit if ab = ba = 1R for some b in R, and b is the inverse of a. Now if we suppose I contains a, how do we show that I=R?

    For the second proof we can assume N(I) is an ideal of R.

    Any help or suggestions are appreciated.
     
  2. jcsd
  3. Sep 21, 2010 #2
    1) What is the definition of an ideal. What are the properties defining the unit? Can you deduce that [tex]1_R[/tex] is in [tex]I[/tex]?
     
  4. Sep 21, 2010 #3

    Hurkyl

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    You don't have to assume there are no zero-divisors for these equalities to be true, and the problem you stated certainly didn't list it as a hypothesis.


    What have you done on the problem? In my mind, they practically solve themselves just from writing down what it means for two sets to be equal in terms of what objects are members -- i.e. A=B is equivalent to "x in A implies x in B and x in B implies x in A" .


    So, I imagine your problem is either a lack of confidence, or you are having difficulty writing out definitions -- either way, we can't help effectively unless you show us what you have been able to do on the problem (even if you think it is trivial or pointless), so that it is clear where you have the mental block.
     
  5. Sep 21, 2010 #4
    I is a (two-sided) ideal of R if [tex]rI := \{ rx: x \in I \} \subseteq I[/tex] and [tex]Ir := \{ xr : x \in I \} \subseteq I[/tex] for all [tex]r \in R[/tex]. I is a proper ideal if [tex]I \neq R[/tex].

    We have [tex]a \in R[/tex] is a unit of R, and we let [tex]x \in I[/tex]. Since I is an ideal of R:

    [tex]a.x=x.a=1_R \subseteq I[/tex]

    I guess we can conclude that 1R is in I. And does this show that I = R?

    Hurkyl, it's not a lack of confidence... I just don't know how to tackle the problem and that's why I've posted it here as a last resort. :smile:
     
  6. Sep 21, 2010 #5

    Office_Shredder

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    You're on the right track but you haven't quite wrapped it up. Why is [tex] a\cdot x=1_R[/tex]? Just because you have a unit in R, doesn't mean that its inverse is in I. But you don't just know that there's a unit in R, you know more than that
     
  7. Sep 22, 2010 #6
    How do I need to show that the inverse of a unit is in R? :confused:
     
  8. Sep 22, 2010 #7

    Office_Shredder

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    You said let [tex]a\in R[/tex] be a unit. Then you said there exists [tex]x\in I[/tex] such that [tex]a\cdot x=1_R[/tex]. This never uses the fact that you know there is a unit in I; rather, you just assumed that a given unit in R has its inverse in I (it's true that one with this property exists, but it might not be the a that you picked)


    The argument you want should be flipped: Let [tex]a\in I[/tex] be a unit, then there is [tex] x\in R[/tex]
     
  9. Sep 23, 2010 #8
    Okay, here's what I did now:

    Let [tex]a \in I[/tex] be a unit of R, and let [tex]r \in R[/tex].

    I is an ideal of R, [tex]I \triangleleft R[/tex], if

    [tex]rI=\{ ra: a \in I \} \subseteq I[/tex]
    [tex]Ir=\{ ar: a \in I \} \subseteq I[/tex]

    [tex]\forall r \in R[/tex]. So

    [tex]r.a=a.r =1_R \subseteq I[/tex]

    [tex]r=a^{-1}[/tex]

    This is true since a is a unit of R and [tex]r \in R[/tex].

    Is this correct now?
     
  10. Sep 23, 2010 #9
    Once you have

    [tex]1_R\in I[/tex]

    (You should write [tex]1_R\in I[/tex] and not [tex]1_R\subseteq I[/tex] - while [tex]I[/tex] and [tex]aI[/tex] are subsets, [tex]1_R[/tex] is an element)

    is it true that [tex]\forall a\in R,\, a=1_R a=a1_R\in I[/tex]?

    And if so, can you deduce that [tex]I=R[/tex]?
     
    Last edited: Sep 23, 2010
  11. Sep 24, 2010 #10
    Why do we need to know if that is true or not? We just needed to show that for [tex]r \in R[/tex] and [tex]a \in I[/tex], ar and ra both are in I. That's what I've tried to show in my previous post making use of the fact that [tex]a \in I[/tex] is also a unit of R.

    Can I deduce that I=R from my previous work? :confused:
     
  12. Sep 24, 2010 #11
    It is good to be aware of the fact for any ideal [tex]I\subset R[/tex], we have [tex]I=R[/tex] if and only if [tex]1_R\in I[/tex].
     
  13. Sep 25, 2010 #12
    Ok, thanks for reminding me. But I have already shown that 1R is in I, didn't I? So we can conclude that I=R?

    P.S. I think it's true that [tex]\forall a\in R,\, a=1_R a=a1_R\in I[/tex], because 1R is the unity (identity under multipication).
     
    Last edited: Sep 25, 2010
  14. Sep 25, 2010 #13
    Right. The logic is this: a is a unit, a is in the ideal. It follows that 1 is in the ideal. It follows that I=R.
     
  15. Sep 25, 2010 #14
    Thank you very much for your help. Could you also give me some clues on how to get started on the second question?
     
  16. Sep 26, 2010 #15

    Office_Shredder

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    To show N(N(I))=N(I), like most set equalities you need to do two things
    1) Show [tex]N(N(I))\subset N(I)[/tex]
    2) Show [tex] N(I)\subset N(N(I))[/tex]

    Which means write down what it means for an element of your ring to belong to one set, and show that implies the element belongs to the other set as well. Number 2 is probably the easier one to start with, why don't you try it out and see where you get?
     
  17. Sep 26, 2010 #16

    Hurkyl

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    Maybe it would help if you didn't drop the "there exists"? Also, maybe use different letters for dummy variables that appear in different places? (Since its existential, you can always pick for one the same value as another if that is what makes things work)
     
  18. Sep 27, 2010 #17
    For [tex]N(N(I)) \subset N(I)[/tex] I start like this: Let [tex]x \in N(N(I))[/tex] then we know that since [tex]N(I)=\{ r \in R: r^n \in I \}[/tex], then we will have [tex]N(N(I))= \{ r^* \in R: (r^n)^m=r^{n.m} \in I \}[/tex]. Again, how does this imply that r* must belong to the other set N(I)? Can we just say since r* from N(N(I)) belongs to I, it must also belong to N(I), then x also belongs to N(I)?
     
  19. Sep 27, 2010 #18

    Office_Shredder

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    If r is in N(N(I)), you just said that there exist some numbers n,m such that rnm is in I. Compare this to what your definition of N(I) is
     
  20. Sep 28, 2010 #19
    Thank you very much, I see that. This is exactly what the definition of N(I) is, but the problem is that I don't know how to exactly write it down/explain it in words...

    Similarly for [tex]N(I) \subseteq N(N(I))[/tex], we let [tex]r \in N(I)[/tex], therefore r is in R such that [tex]r^n \in I[/tex]. This fits in with the definition of N(N(I)):

    [tex]N(N(I))= \{ r \in R: (r^n)^m=r^{n.m} \in I \}[/tex] (is this even correct?)

    Because if say m is 1 then rn in I and so r belongs to N(N(I)). But is there a better way of explaining this?
     
  21. Sep 28, 2010 #20
    I think it may be less confusing for you to first prove (or, better, just notice once for all) that for any [tex]K[/tex]

    [tex]K \subseteq N(K)[/tex]. Then specialize: in particular for [tex]K=...[/tex].
     
    Last edited: Sep 28, 2010
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