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Math Help - Prime elements in integral domains

  1. #1
    Super Member Bernhard's Avatar
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    Prime elements in integral domains

    In Dummit and Foote, Section 8.3 on Unique Factorization Domains, Proposition 10 reads as follows:

    Proposition 10: In an integral domain a prime element is always irreducible.

    The proof reads as follows:

    ================================================== =========

    Suppose (p) is a non-zero prime ideal and p = ab.

    Then  ab = p \in (p) , so by definition of prime ideal, one of a or b, say a, is in (p).

    Thus a = pr for some r.

    This implies p = ab = prb and so rb = 1 and b is a unit.

    This shows that p is irreducible.

    ================================================== ============

    My question is as follows: Where in this proof do D&F use the fact that p is in an integral domain??? (It almost reads as if this applies for any ring)

    Peter
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    Re: Prime elements in integral domains

    p = ab = prb \implies rb=1
    You use the fact that an integral domain has a cancellation law.
    Last edited by Gusbob; March 14th 2013 at 12:57 AM.
    Thanks from Bernhard
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    Super Member Bernhard's Avatar
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    Re: Prime elements in integral domains

    Thanks

    I was just imagining that they were pre-multiplying both sides by the inverse of p, but of course that assumes p has an inverse, and we do not necessarily have a field - only an integral domain.

    Presumably the cancellation law can operate without p having an inverse???

    Peter
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    Re: Prime elements in integral domains

    Quote Originally Posted by Bernhard View Post
    Presumably the cancellation law can operate without p having an inverse???
    Yes. Consider the integral domain \mathbb{Z}. If 2a=2b, then obviously a=b. But 2 does not have an inverse in \mathbb{Z}.
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