Results 1 to 3 of 3

Math Help - Irreducible but not prime

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    15

    Irreducible but not prime

    I need an example of a ring that is irreducible but not prime.

    there is this example of such a ring in the book saying in Z[-(root(3)] where
    4= 2*2
    but also 4= (1+root(-3))*(1-root(-3))
    okay I get it that and I know 1+root(-3) doesn't divide 2 hence is is not prime. but also if it was reducible then we should have units
    since the definition of irreducible is that a=b*c then b or c is unit. what is the unit of this ring !!!!?

    (ps: if one day I understand everything in abstract algebra, surely I will rewrite the book)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by hamidr View Post
    I need an example of a ring that is irreducible but not prime.

    there is this example of such a ring in the book saying in Z[-(root(3)] where
    4= 2*2
    but also 4= (1+root(-3))*(1-root(-3))
    okay I get it that and I know 1+root(-3) doesn't divide 2 hence is is not prime. but also if it was reducible then we should have units
    since the definition of irreducible is that a=b*c then b or c is unit. what is the unit of this ring !!!!?
    You mean an irreducible element but not a prime element in the domain?

    You can find such elements in a non-UFD. In UFD, every irreducible element is a prime element though.

    Now, take some non-UFD examples.

    Consider D = F[x^2, xy, y^2], where F is a field. Then, (x^2)(y^2) = (xy)(xy) (Note: the parenthesis here is not denoted for an ideal ). We see that (xy) | (x^2)(y^2), but it does not imply either (xy) |(x^2) or (xy) | (y^2). Thus xy is not a prime element. However, xy is an irreducible element in D. For example, if F = \mathbb{Q}, xy = (3xy) \times (1/3), then you see that 1/3 is a unit here and xy is not factorized (x)(y) in this domain.

    As per your example, 1+\sqrt{-3} is not a prime element, but it is an irreducible element. You cannot factorize 1+\sqrt{-3} = ab unless a or b is a unit in \mathbb{Z}[\sqrt{-3}].

    (ps: if one day I understand everything in abstract algebra, surely I will rewrite the book)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2010
    Posts
    78
    Your example (4=2*2=(1+sqrt(-3))(1-sqrt(-3))) works, but the reason is not correct. You should show that neither of them differ by a unit, and everything in the factorization (2,1+sqrt(-3),1-sqrt(-3)) is irreducible.

    The unit in Z[sqrt(-3)] is just 1 and -1, since you can define a positive-integer-valued multiplicative "norm" on non-zero elements, and it's easy to show that a unit must have a norm 1, while the only elements in Z[sqrt(-3)]* having norm 1 are 1 and -1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: May 2nd 2011, 07:24 PM
  2. A simple problem about prime and irreducible
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 17th 2009, 12:10 AM
  3. Replies: 3
    Last Post: April 13th 2009, 12:00 PM
  4. Show that a prime is not irreducible in Z[i]
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 21st 2008, 01:12 PM
  5. Prime and Irreducible in Z[i]
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 17th 2008, 09:09 PM

Search Tags


/mathhelpforum @mathhelpforum