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Thread: Principal ideal domain

  1. #1
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    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?
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  2. #2
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    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 $\displaystyle I$ be any non-zero ideal of $\displaystyle R$ and define $\displaystyle S = \{Ra : \ a \in I \}$. use condition 2) in your problem to show that $\displaystyle S$ has a maximal element, say $\displaystyle Rb$. the claim is that $\displaystyle I = Rb$. suppose our claim is not true and choose $\displaystyle a \in I \setminus Rb$. then by condition 1) we have $\displaystyle Ra+Rb=Rd$, where $\displaystyle d = \gcd(a,b)$. clearly $\displaystyle d \in I$ because $\displaystyle a,b \in I$. thus $\displaystyle Rd \in S$. we also have $\displaystyle a \in Rd \setminus Rb$ and so $\displaystyle Rb \subset Rd$, contradicting the maximality of $\displaystyle Rb$ in $\displaystyle S$. Q.E.D.
    Last edited by NonCommAlg; Jun 21st 2011 at 03:56 PM.
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  3. #3
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    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:

    $\displaystyle (a_i)$.

    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:

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

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

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

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

    now, it would be nice if we could conclude that $\displaystyle I = I_N$, but perhaps for some other choice of the $\displaystyle \{a_i\}$ 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 $\displaystyle I\setminus I_N$ an associate of $\displaystyle a_N$? (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 "$\displaystyle I_N$"s).

    my apologies if i have stated this badly.
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  4. #4
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    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:

    $\displaystyle (a_i)$.

    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:

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

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

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

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

    now, it would be nice if we could conclude that $\displaystyle I = I_N$, but perhaps for some other choice of the $\displaystyle \{a_i\}$ 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 $\displaystyle I\setminus I_N$ an associate of $\displaystyle a_N$? (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 "$\displaystyle I_N$"s).

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