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I Prime and Maximal Ideals in PIDs .. Rotman, AMA Theorem 5.12

  1. Sep 1, 2016 #1
    I am reading Joseph J. Rotman's book: Advanced Modern Algebra (AMA) and I am currently focused on Section 5.1 Prime Ideals and Maximal Ideals ...

    I need some help with understanding the proof of Theorem 5.12 ... ...


    Theorem 5.12 reads as follows:


    ?temp_hash=61cd73afb237213fbd511810f1e3a61d.png



    In the above text Rotman writes the following:


    " ... ... If ##(p) \subseteq J = (a)##, then ##a|p##. Hence either ##a## and ##p## are associates, in which case ##(a) = (p)##, or ##a## is a unit, in which case ##J = (a) = R##. ... ... ... "


    My question is as follows:


    Rotman argues, (as I interpret his argument), that ##a|p## implies that either ##a## and ##p## are associates ... or ... ##a## is a unit ...


    Can someone please explain (slowly and clearly smile.png ) why this is the case ... ... ?


    Hope someone can help ... ...

    Peter
     

    Attached Files:

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  3. Sep 1, 2016 #2
    I have been reflecting on my question and believe the answer is something like the following:


    Firstly ... we are given that ##p## is irreducible ...

    Now ... ##p## irreducible

    ##\Longrightarrow p## is non-zero and ##p## not a unit ... and ... where ##p## equals a product,

    say, ##p = ra## ... then one of ##a## and ##r## is a unit ...


    Now, ##a|p \Longrightarrow p = ra## for some ##r \in R##

    So then we have that:

    ##p## irreducible and ##p = ra \Longrightarrow## one of ##a## and ##r## is a unit ...

    If ##r## is a unit then ##a## and ##p## are associates ... ...

    ... otherwise ##a## is a unit ...



    Can someone confirm that this is correct ... or alternatively point out shortcomings and errors in the analysis ...

    Peter
     
  4. Sep 1, 2016 #3

    andrewkirk

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    We have ##a|p## so ##p=ba## for some ##b##. Since ##p\in I=(p)## and ##I## is a prime ideal, it must be the case that either ##a## or ##b## is in ##I##.

    If ##a\in I## we have ##p|a## which, together with the earlier-observed ##a|p## makes ##p,a## associates.

    If ##a\not\in I## then ##b\in I## in which case ##b=pc## for some ##c##. So ##p=pca##. We write this as
    $$0=p-pca=p(1-ca)$$
    and since ##p\neq 0## and a PID is an Integral Domain, we must then have ##1-ca=0##, whence ##ca=1## so ##a|1##, making ##a## a unit.
     
  5. Sep 1, 2016 #4

    andrewkirk

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    That reasoning looks sound to me. It's a different route from mine, as yours explicitly uses the irreducibility of ##p## whereas mine uses the primeness of the ideal ##I##. I suspect the connection between the two lies in the proof of Proposition 5.6. But either one will suffice.
     
  6. Sep 1, 2016 #5
    Thanks Andrew ... your posts were REALLY helpful ...

    Appreciate your help on this issue ...

    Peter
     
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