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Math Help - 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  r, s \in R and

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

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

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

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

    Can you clarify?

    Would appreciate the help!

    Peter
    Last edited by Bernhard; March 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  a_2 = gcd  ( a_1, b_1 ) and the fact that  I = \langle a_2 \rangle is a possibility
     a_2 = gcd  ( a_1, b_1 ) implies that a_2=r\cdot a_1 for some r\in R, so \langle a_1\rangle \subseteq \langle a_2 \rangle. Now \langle a_2 \rangle \subseteq I by definition, so the question becomes whether \langle a_2 \rangle contains all elements of I, formulated as I=\langle a_2 \rangle or not.

    The question of ` I=\langle a_i \rangle?' is always valid, the motivation being the gcd bit, which guarantees \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 \langle a_i\rangle of I, and argue that you must eventually get all of I (otherwise you get a contradiction). Essentially, the reason we want to ask whether 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
     a_2 = gcd  ( a_1, b_1 ) implies that a_2=r\cdot a_1 for some r\in R, so \langle a_1\rangle \subseteq \langle a_2 \rangle. Now \langle a_2 \rangle \subseteq I by definition, so the question becomes whether \langle a_2 \rangle contains all elements of I, formulated as I=\langle a_2 \rangle or not.

    The question of ` I=\langle a_i \rangle?' is always valid, the motivation being the gcd bit, which guarantees \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 \langle a_i\rangle of I, and argue that you must eventually get all of I (otherwise you get a contradiction). Essentially, the reason we want to ask whether 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:

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

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

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

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

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

    So since you conclude  ( 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  b_1 \in J - \langle a_1 \rangle and set  a_2 = gcd (a_1,b_1) ... ... "

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


    Peter
    Last edited by Bernhard; March 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:

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

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

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

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

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

    So since you conclude  ( 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  b_1 \in J - \langle a_1 \rangle and set  a_2 = gcd (a_1,b_1) ... ... "

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

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

    I now think that the reason is as follows:

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

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

    and so we are restricted to a subset of  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  ( 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  I = ( a_1 ) do we assume that we need to examine a larger ideal??




    Peter
    Last edited by Bernhard; March 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  ( 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  I = ( a_1 ) do we assume that we need to examine a larger ideal??
    You're right about my point about gcd, the a_1 and a_2 should have been switched. I'll assume the quote above reads  I \not= ( a_1 ) . The reason is that \langle a_1\rangle \subseteq I always. If I=\langle a \rangle is principal, then a_1 = r\cdot a for some r, which means \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  a_1\in R , we have: either  I=\langle a_1\rangle or not ."

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

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

    How do you know that  (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  (a_1) \subseteq I ?
    Much apologies. I chose a_1 \in R. It should have been a_1 \in I. Sorry for all the confusion.
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