Results 1 to 2 of 2

Thread: Irreducible and gcd.

  1. #1
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722

    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 $\displaystyle \neq$ 1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.

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

    This look right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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 $\displaystyle \neq$ 1, in which case p|a, a contradiction. Hence if b = 1, g.c.d (a,p) = 1.

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

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

    i.e. $\displaystyle p \mid d.$ we also have $\displaystyle d \mid a.$ hence $\displaystyle p \mid a.$ contradiction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Apr 10th 2011, 10:41 AM
  2. Irreducible in F_p[x] ?
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Dec 29th 2010, 04:00 AM
  3. Replies: 2
    Last Post: Sep 10th 2010, 08:05 AM
  4. irreducible
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 11th 2008, 06:27 AM
  5. 3 irreducible in Z[sq.rt.-5]
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Nov 6th 2008, 07:26 PM

Search Tags


/mathhelpforum @mathhelpforum