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Thread: Integral Domaina and Principal Ideal Domains (PIDs)

  1. #1
    Super Member Bernhard's Avatar
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    Integral Domains and Principal Ideal Domains (PIDs)

    Dummit and Foote, Section 8.2 (Principal Ideal Domains (PIDs) ) - Exercise 4, page 282.

    Let R be an integral domain.

    Prove that if the following two conditions hold then R is a Principal Ideal Domain:

    (i) any two non-zero elements a and b in R have a greatest common divisor which can be written in the form ra + sb for some $\displaystyle r, s \in R $ and

    (ii) if $\displaystyle a_1, a_2, a_3, ... $ are non-zero elements of R such that $\displaystyle a_{i+1} | a_i $ for all i, then there is a positive integer N such that $\displaystyle a_n $ is a unit times $\displaystyle a_N $ for all $\displaystyle n \ge N $
    Last edited by Bernhard; Mar 22nd 2013 at 05:26 PM.
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    Re: Integral Domaina and Principal Ideal Domains (PIDs)

    Suppose $\displaystyle I$ is a (non-empty) ideal in $\displaystyle R$. Given an element $\displaystyle a_1\in R$, we have: either $\displaystyle I=\langle a_1\rangle$ or not. In the former case, we're done, so assume the latter. Choose $\displaystyle b_1\in J- \langle a_1 \rangle$ and set $\displaystyle a_2 = $ gcd$\displaystyle (a_1,b_1)$. Here, we have either $\displaystyle I=\langle a_2\rangle$ or not. In the former case, we're done, so assume the latter... Déjà vu.

    I think you see where this is going. This process terminates because of your second condition.
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    Super Member Bernhard's Avatar
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    Re: Integral Domaina and Principal Ideal Domains (PIDs)

    Thanks Gusbob, most helpful; I can see the overall strategy ... but need a couple of details clarified.

    You write:

    "Suppose I is a (non-empty) ideal in R. Given an element $\displaystyle a_1\in R $, we have: either $\displaystyle I=\langle a_1\rangle $ or not. ... ... "

    Now I follow that statement since it is just simple logic and is true independent of any conditions in the exercise - it says something is true or not!

    But then you write:

    "In the former case, we're done, so assume the latter. Choose $\displaystyle b_1\in I- \langle a_1 \rangle $ and set $\displaystyle a_2 = gcd (a_1,b_1) $. Here, we have either $\displaystyle I=\langle a_2\rangle $ or not. ... ... "

    When you say $\displaystyle I = \langle a_2 \rangle $ or not, I am assuming that there is some link between setting $\displaystyle a_2 $ = gcd $\displaystyle ( a_2 , b_2 ) $ and the fact that $\displaystyle I = \langle a_2 \rangle $ is a possibility.

    What is the link between setting $\displaystyle a_2 $ = gcd $\displaystyle ( a_2 , b_2 ) $ and the fact that $\displaystyle I = \langle a_2 \rangle $ is a possibility

    Can you clarify?

    Would appreciate the help!

    Peter
    Last edited by Bernhard; Mar 23rd 2013 at 05:44 PM.
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    Re: Integral Domaina and Principal Ideal Domains (PIDs)

    Quote Originally Posted by Bernhard View Post
    What is the link between setting $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) $ and the fact that $\displaystyle I = \langle a_2 \rangle $ is a possibility
    $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) $ implies that $\displaystyle a_2=r\cdot a_1$ for some $\displaystyle r\in R$, so $\displaystyle \langle a_1\rangle \subseteq \langle a_2 \rangle$. Now $\displaystyle \langle a_2 \rangle \subseteq I$ by definition, so the question becomes whether $\displaystyle \langle a_2 \rangle$ contains all elements of $\displaystyle I$, formulated as $\displaystyle I=\langle a_2 \rangle$ or not.

    The question of `$\displaystyle I=\langle a_i \rangle$?' is always valid, the motivation being the gcd bit, which guarantees $\displaystyle \langle a_{i-1}\rangle \subseteq \langle a_i \rangle$. What you are doing in this proof is constructing bigger and bigger principal sub-ideals $\displaystyle \langle a_i\rangle$ of $\displaystyle I$, and argue that you must eventually get all of $\displaystyle I$ (otherwise you get a contradiction). Essentially, the reason we want to ask whether $\displaystyle I=\langle a_i \rangle$ at each step is because it is a principal ideal that is bigger (and contains) all previous principal ideals we looked at.

    I hope that made sense, I'm not able to explain this eloquently.
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    Super Member Bernhard's Avatar
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    Re: Integral Domaina and Principal Ideal Domains (PIDs)

    Quote Originally Posted by Gusbob View Post
    $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) $ implies that $\displaystyle a_2=r\cdot a_1$ for some $\displaystyle r\in R$, so $\displaystyle \langle a_1\rangle \subseteq \langle a_2 \rangle$. Now $\displaystyle \langle a_2 \rangle \subseteq I$ by definition, so the question becomes whether $\displaystyle \langle a_2 \rangle$ contains all elements of $\displaystyle I$, formulated as $\displaystyle I=\langle a_2 \rangle$ or not.

    The question of `$\displaystyle I=\langle a_i \rangle$?' is always valid, the motivation being the gcd bit, which guarantees $\displaystyle \langle a_{i-1}\rangle \subseteq \langle a_i \rangle$. What you are doing in this proof is constructing bigger and bigger principal sub-ideals $\displaystyle \langle a_i\rangle$ of $\displaystyle I$, and argue that you must eventually get all of $\displaystyle I$ (otherwise you get a contradiction). Essentially, the reason we want to ask whether $\displaystyle I=\langle a_i \rangle$ at each step is because it is a principal ideal that is bigger (and contains) all previous principal ideals we looked at.

    I hope that made sense, I'm not able to explain this eloquently.
    Thanks again Gusbob, things are getting clearer for me! Just a pedantic point - just to be sure of what is going on:

    You write:

    " ... $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) $ implies that $\displaystyle a_2=r\cdot a_1$ for some $\displaystyle r\in R$ ... ... "

    But I think that $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) $ implies that $\displaystyle a_1 = r\cdot a_2$ since

    $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) \longrightarrow a_2 | a_1 $ and $\displaystyle a_2 | b_1 \longrightarrow a_1 = r_1 \cdot a_2 $ and $\displaystyle b_1 = r_2 \cdot a_2$ for some $\displaystyle r_1 , r_2 \in R$

    Now $\displaystyle (a_1) = \{ r \cdot a_1 | r \in R \} $ and $\displaystyle (a_2) = \{ r \cdot a_2 | r \in R \} $

    So we have $\displaystyle a_1 = r \cdot a_2 \longrightarrow a_1 \in ( a_2 ) \longrightarrow ( a_1 ) \subseteq ( a_2 ) $

    So since you conclude $\displaystyle ( a_1 ) \subseteq ( a_2 ) $ all is well (Sorry if I am being pedantic)

    A more important point:

    In your earlier post on the proof you state at one point:

    "Choose $\displaystyle b_1 \in J - \langle a_1 \rangle $ and set $\displaystyle a_2 = gcd (a_1,b_1) $... ... "

    What motivated you to take $\displaystyle b_1\in J- \langle a_1 \rangle $ . In other words in your proof strategy what made you want to not only exclude $\displaystyle a_1 $ from consideration, but also to exclude all of the principal ideal generated by $\displaystyle a_1 $ ie to exclude all of $\displaystyle ( a_1 ) $


    Peter
    Last edited by Bernhard; Mar 23rd 2013 at 07:50 PM.
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    Re: Integral Domaina and Principal Ideal Domains (PIDs)

    Quote Originally Posted by Bernhard View Post
    Thanks again Gusbob, things are getting clearer for me! Just a pedantic point - just to be sure of what is going on:

    You write:

    " ... $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) $ implies that $\displaystyle a_2=r\cdot a_1$ for some $\displaystyle r\in R$ ... ... "

    But I think that $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) $ implies that $\displaystyle a_1 = r\cdot a_2$ since

    $\displaystyle a_2 $ = gcd $\displaystyle ( a_1, b_1 ) \longrightarrow a_2 | a_1 $ and $\displaystyle a_2 | b_1 \longrightarrow a_1 = r_1 \cdot a_2 $ and $\displaystyle b_1 = r_2 \cdot a_2$ for some $\displaystyle r_1 , r_2 \in R$

    Now $\displaystyle (a_1) = \{ r \cdot a_1 | r \in R \} $ and $\displaystyle (a_2) = \{ r \cdot a_2 | r \in R \} $

    So we have $\displaystyle a_1 = r \cdot a_2 \longrightarrow a_1 \in ( a_2 ) \longrightarrow ( a_1 ) \subseteq ( a_2 ) $

    So since you conclude $\displaystyle ( a_1 ) \subseteq ( a_2 ) $ all is well (Sorry if I am being pedantic)

    A more important point:

    In your earlier post on the proof you state at one point:

    "Choose $\displaystyle b_1 \in J - \langle a_1 \rangle $ and set $\displaystyle a_2 = gcd (a_1,b_1) $... ... "

    What motivated you to take $\displaystyle b_1\in J- \langle a_1 \rangle $ . In other words in your proof strategy what made you want to not only exclude $\displaystyle a_1 $ from consideration, but also to exclude all of the principal ideal generated by $\displaystyle a_1 $ ie to exclude all of $\displaystyle ( a_1 ) $


    Peter
    I have been reflecting on my own question which was concerned with selecting a new element from which to generate a new principal ideal when faced with the failure of $\displaystyle a_1 $ to generate a principal ideal = I.

    Gusbob's strategy was to choose an element from $\displaystyle I - ( a_1) $ and I wished to know why he excluded elements in $\displaystyle ( a_1 ) \ne a_1 $ from consideration.

    I now think that the reason is as follows:

    If we choose $\displaystyle b_1 \in (a_1) $ where $\displaystyle b_1 \ne a_1 $ then

    $\displaystyle b_1 \in ( a_1 ) \longrightarrow (b_1) \subseteq (a_1) $

    and so we are restricted to a subset of $\displaystyle a_1 $ - BUT, we need a larger principal ideal domain and not a smaller one.

    Can someone confirm that my thinking is correct.

    FURTHER EDIT

    Sorry to backtrack from my idea above - why when, say I fails to equal $\displaystyle ( a_1) $ do we assume that we need a larger ideal - all we are trying to show is that any ideal we examine is principal - why if $\displaystyle I = ( a_1 ) $ do we assume that we need to examine a larger ideal??




    Peter
    Last edited by Bernhard; Mar 23rd 2013 at 09:09 PM.
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    Re: Integral Domaina and Principal Ideal Domains (PIDs)

    Quote Originally Posted by Bernhard View Post
    Sorry to backtrack from my idea above - why when, say I fails to equal $\displaystyle ( a_1) $ do we assume that we need a larger ideal - all we are trying to show is that any ideal we examine is principal - why if $\displaystyle I = ( a_1 ) $ do we assume that we need to examine a larger ideal??
    You're right about my point about gcd, the $\displaystyle a_1$ and $\displaystyle a_2$ should have been switched. I'll assume the quote above reads $\displaystyle I \not= ( a_1 ) $. The reason is that $\displaystyle \langle a_1\rangle \subseteq I$ always. If $\displaystyle I=\langle a \rangle$ is principal, then $\displaystyle a_1 = r\cdot a $ for some $\displaystyle r$, which means $\displaystyle \langle a_1\rangle \subseteq \langle a\rangle =I$. As such, it makes sense to take bigger ideals.
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    Super Member Bernhard's Avatar
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    Re: Integral Domaina and Principal Ideal Domains (PIDs)

    Hi Gusbob, I am still struggling with this - originally you say the following:

    "Suppose I is a (non-empty) ideal in R. Given an element $\displaystyle a_1\in R $, we have: either $\displaystyle I=\langle a_1\rangle or not $."

    Then you conclude that if $\displaystyle I \ne ( a_1 ) $ then $\displaystyle (a_1) \subseteq I $

    Why exactly does it follow: it seems to me that from $\displaystyle I \ne (a_1) $ all we can conclude is that either I is a subset of $\displaystyle (a_1) $ , I is a superset of $\displaystyle (a_1) $ or I and $\displaystyle (a_1) $ are disjoint.

    How do you know that $\displaystyle (a_1) \subseteq I $?

    Peter

    PS Sorry if I am being slow here!
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    Re: Integral Domaina and Principal Ideal Domains (PIDs)

    Quote Originally Posted by Bernhard View Post
    How do you know that $\displaystyle (a_1) \subseteq I $?
    Much apologies. I chose $\displaystyle a_1 \in R$. It should have been $\displaystyle a_1 \in I$. Sorry for all the confusion.
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