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Math Help - Prime elements and maximal ideals in a PID

  1. #1
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    Prime elements and maximal ideals in a PID

    Here is the problem I have:

    Let R be a PID and p \in R. Prove that the ideal \langle p \rangle is maximal in R if and only if p is a prime element in R. (Recall p is a prime in R if p is not a unit and if p|ab then p|a or p|b)

    I know that prime elements generate prime ideals (and I think I can prove that if I have to) and the only trick I had up my sleeve was that prime ideal -> ID and finite ID's are fields -> maximal, but it's not finite, so I am having difficulties constructing a proof...
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  2. #2
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    Hi

    One way is easy:

    (p)\ \text{maximal ideal}\Leftrightarrow R/(p)\ \text{field}\Rightarrow R/(p)\ \text{integral domain}\Leftrightarrow (p)\ \text{prime ideal}

    Now assume (p) prime. Take x,y\in R/(p) i.e. classes of elements which are not in (p). Since R/(p) is an integral domain, xy=0\Rightarrow x=0\ \text{or}\ y=0.
    So imagine there is a r\in R such that (p) \subseteq (r)\subsetneq R.
    Is there a divisibility relation between r and p? Then think of what that gives you in R/(p). Using the fact that R is a PID, conclude!
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  3. #3
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    This is very much like what I managed to come up with. I finally broke down and showed that prime elements iff prime ideals, and then went that route.

    The only tricky part was getting the two different divisibility cases for prime->maximal. One was obvious containment, thus maximal, and the other showed that the multiple was a unit, and thus containment.

    Thanks a bunch!
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