Results 1 to 4 of 4

Math Help - Principal ideal domain

  1. #1
    Newbie
    Joined
    May 2011
    Posts
    7

    Principal ideal domain

    R is an integral domain.Then prove that if the following two conditions hold then R is a principal ideal domain(PID)
    1)any two non zero elements a, b in R have a greatest common divisor which can be written in the form ra+sb for some r,s belonging to R.
    2)['a_1 denotes a suffix 1]
    If a_1,a_2,a_3,...... are non zero elements of R such that a_(i+1) divides a_i for all i then there is a positive integer N such that a_n is unit times a_N for all n greater than or equal to N.

    How to do the sum?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Re: Principal ideal domain

    Quote Originally Posted by jishnuray View Post
    R is an integral domain.Then prove that if the following two conditions hold then R is a principal ideal domain(PID)
    1)any two non zero elements a, b in R have a greatest common divisor which can be written in the form ra+sb for some r,s belonging to R.
    2)['a_1 denotes a suffix 1]
    If a_1,a_2,a_3,...... are non zero elements of R such that a_(i+1) divides a_i for all i then there is a positive integer N such that a_n is unit times a_N for all n greater than or equal to N.

    How to do the sum?
    let I be any non-zero ideal of R and define S = \{Ra : \ a \in I \}. use condition 2) in your problem to show that S has a maximal element, say Rb. the claim is that I = Rb. suppose our claim is not true and choose a \in I \setminus Rb. then by condition 1) we have Ra+Rb=Rd, where d = \gcd(a,b). clearly d \in I because a,b \in I. thus Rd  \in S. we also have a \in Rd \setminus Rb and so Rb \subset Rd, contradicting the maximality of Rb in S. Q.E.D.
    Last edited by NonCommAlg; June 21st 2011 at 03:56 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: Principal ideal domain

    can this be proved without the use of Zorn's lemma? perhaps i am not understanding the question, but condition two looks like an ascending chain condition on the principal ideals:

    (a_i).

    i was thinking about this, and i ran into the following difficulty:

    given any (non-zero) ideal I, we can form a chain of ideals by taking:

    I_1 = (a_1) where a_1 \in I\setminus{0}.

    I_2 = (a_2) where a_2 is chosen by finding r_1 \in I\setminus I_1 and setting a_2 = gcd(r_1,a_1), so that a_2|a_1,

    I_3 = (a_3) where a_3 is chosen by finding r_2 \in I\setminus I_2 and setting a_3 = gcd(r_2,a_2), so that a_3|a_2,

    and so on, which by (2) terminates with some I_N  =(a_N).

    now, it would be nice if we could conclude that I = I_N, but perhaps for some other choice of the \{a_i\} we would wind up with "a different N". that is, how do we know that we can find an N that makes every element in I\setminus I_N an associate of a_N? (i think this is where Zorn's lemma comes in. if i understand what you wrote, each Ra in your answer, is one of my " I_N"s).

    my apologies if i have stated this badly.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Re: Principal ideal domain

    Quote Originally Posted by Deveno View Post
    can this be proved without the use of Zorn's lemma? perhaps i am not understanding the question, but condition two looks like an ascending chain condition on the principal ideals:

    (a_i).

    i was thinking about this, and i ran into the following difficulty:

    given any (non-zero) ideal I, we can form a chain of ideals by taking:

    I_1 = (a_1) where a_1 \in I\setminus{0}.

    I_2 = (a_2) where a_2 is chosen by finding r_1 \in I\setminus I_1 and setting a_2 = gcd(r_1,a_1), so that a_2|a_1,

    I_3 = (a_3) where a_3 is chosen by finding r_2 \in I\setminus I_2 and setting a_3 = gcd(r_2,a_2), so that a_3|a_2,

    and so on, which by (2) terminates with some I_N  =(a_N).

    now, it would be nice if we could conclude that I = I_N, but perhaps for some other choice of the \{a_i\} we would wind up with "a different N". that is, how do we know that we can find an N that makes every element in I\setminus I_N an associate of a_N? (i think this is where Zorn's lemma comes in. if i understand what you wrote, each Ra in your answer, is one of my " I_N"s).

    my apologies if i have stated this badly.
    yes, you're right. we don't need Zorn's lemma to show that S has a maximal element. i removed it from my proof.
    suppose that S has no maximal element and choose a_1 \in I. since  Ra_1 is not maximal in S, we may choose a_2 \in I such that Ra_1 \subset Ra_2. we continue this way to find an ascending chain of elements of S which does not stop, contradicting condition 2).
    Last edited by NonCommAlg; June 21st 2011 at 03:55 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Principal ideal
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 24th 2010, 12:42 PM
  2. Rings: Principal Ideal domain
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 7th 2010, 05:21 PM
  3. Principal Ideal Domain
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 26th 2010, 12:47 AM
  4. Principal ideal domain question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 30th 2009, 04:22 PM
  5. Principal Ideal Domain
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 23rd 2009, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum