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Math Help - Irreducible and gcd.

  1. #1
    Super Member Deadstar's Avatar
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    Irreducible and gcd.

    R is an integral domain.
    Assume that greatest common divisors exist in R. Let p be an irreducible element of R and let a be an element of R that is not divisible by p. Show that 1 is a greatest common divisor of a and p.

    Thoughts...
    p is irreducible, hence if p|ab then either a=1 or b=1.
    If b = 1 then either a = 1, in which case g.c.d (a,p) = 1, or a \neq 1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.

    So assume b \neq 1, hence a = 1. Hence g.c.d (a,p) = 1.

    This look right?
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    R is an integral domain.
    Assume that greatest common divisors exist in R. Let p be an irreducible element of R and let a be an element of R that is not divisible by p. Show that 1 is a greatest common divisor of a and p.

    Thoughts...
    p is irreducible, hence if p|ab then either a=1 or b=1. incorrect! p is irreducible not prime!
    If b = 1 then either a = 1, in which case g.c.d (a,p) = 1, or a \neq 1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.

    So assume b \neq 1, hence a = 1. Hence g.c.d (a,p) = 1.

    This look right?
    let d=\gcd(a,p). then d \mid p. so p=kd and thus either k or d is a unit, because p is irreducible. if d is a unit, then \gcd(a,p)=1. if k is a unit, then ks=1, for some s and hence d=ps,

    i.e. p \mid d. we also have d \mid a. hence p \mid a. contradiction.
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