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Rings, ideals, prime and maximal

  1. Sep 26, 2008 #1
    I just turned in this homework and I want to know if I got it right. The proof is pretty simple, but I think I might be using a theorem in the wrong way.

    1. The problem statement, all variables and given/known data

    [tex]\{P_{i} : i \in \Lambda \}[/tex] is a family of prime ideals in a ring, R. Prove that [tex]R \setminus \{ \cup_{i \in \Lambda} P_{i} \}[/tex] is a multiplicative system.


    2. Relevant theorems

    1. Let R be a ring. An element [tex]u \in R[/tex] is called a unit if it has an inverse in R.
    2. Let R be a ring. The union of all maximal ideals of R is the set of non-units in R.
    3. Any Maximal ideal is also a prime ideal .

    3. The attempt at a solution
    Any maximal ideal is also a prime ideal. Hence,
    [tex]\cup_{M max} M \subseteq \cup_{i \in \Lambda} P_{i}[/tex].
    Then every element in
    [tex]R \setminus \{ \cup_{i \in \Lambda} P_{i} \}[/tex]
    is a unit. (We have removed all non-units and, possibly, some units since there may be prime ideals that are not maximal.) The set of units is closed under multiplication because, if a and b are units, then c=ab has an inverse [tex]c^{-1}=b^{-1}a^{-1}[/tex].
    [tex]1 \in R \setminus \{ \cup_{i \in \Lambda} P_{i} \}[/tex].
    1 is not in any of the prime ideals, hence,
    [tex]R \setminus \{ \cup_{i \in \Lambda} P_{i} \}[/tex]
    is a multiplicative system.



    So would this work? It bothers me that when we remove [tex] \cup_{i \in \Lambda} P_{i} [/tex] from R we are not just taking out the maximal ideals, but also the prime ideals that may not be maximal. Could this lead to some non-units remaining in the set?
     
  2. jcsd
  3. Sep 26, 2008 #2

    morphism

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    What is a multiplicative system? A multiplicative group?
     
  4. Sep 27, 2008 #3
    A multiplicative system is a set closed under multiplication and containing the multiplicative identity.
     
  5. Sep 27, 2008 #4
  6. Oct 1, 2008 #5

    morphism

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    Then I don't think your proof is valid as it stands.
     
  7. Oct 1, 2008 #6
    Because of the issue I mentioned or something else.
     
  8. Oct 3, 2008 #7

    morphism

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    Because of the issue you mentioned.
     
  9. Mar 24, 2009 #8
    This doesn't require any theorems, really. Only the definition of a prime ideal. Just take an arbitrary product of elements in the set, ie. elements that are NOT in any of these prime ideals. Hence if the set were not multiplicatively closed, this product would be in the union of the prime ideals and hence contained in some prime ideal (by definition of a set theoretic union) and hence, by definition of prime ideal, some element would be in the prime ideal, contradicting the element NOT being in any of the prime ideals.
     
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