Results 1 to 4 of 4

Math Help - Ring/Field help.

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    20

    Ring/Field help.

    From what I understand, fields, as far as numbers Zeta are concerned, are zeta in some prime number. I can't seem to figure out why.

    Also, an inverse of the product for any number in the ring is another number or itself that can be multiplied with it and yield 1. In a field, all the numbers have an inverse, but in zeta 6 (for example), which is not a field the only numbers that have an inverse are 1 and 5, in this case being themselves. Not sure I got everything down. What are these numbers called, btw? I am studying abroad in Spain, and in Spanish they are "unidades" but I don't know what they are in English.

    Thanks.
    Last edited by HeadOnAPike; November 2nd 2009 at 03:48 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by HeadOnAPike View Post
    From what I understand, fields, as far as numbers Zeta are concerned, are zeta in some prime number. I can't seem to figure out why.

    Also, an inverse of the product for any number in the ring is another number or itself that can be multiplied with it and yield 1. In a field, all the numbers have an inverse, but in zeta 6 (for example), which is not a field the only numbers that have an inverse are 1 and 5, in this case being themselves. Not sure I got everything down. What are these numbers called, btw? I am studying abroad in Spain, and in Spanish they are "unidades" but I don't know what they are in English.

    Thanks.

    Unidades, en castellano = units, in english. And indeed: the only two units in \mathbb{Z}_6 are the residue classes 1 and 5.

    In a field every NON-ZERO elements has an inverse.

    Of all the rings \mathbb{Z}_n of residues classes modulo n, the only ones that are fields are those with n a prime number, since then for any non-zero element m of it, which has an integer representative 0\leq m<p , as (m,p) = 1, we can write by the Euclidean algorithm am+bp=1 \,,\,\,a\,,\,b\in \mathbb{Z}\Longrightarrow am=1-bp, and thus a (mod p) is the mult. inverse of p.

    You better get a good algebra book and try to read this there since otherwise it'll be pretty hard for you to grab all this stuff.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    20
    It's not required of me to understand why the only fields are of prime numbers in zeta though, so I think I am alright. But am I right about units of a non-field being multiplicate inverses?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by HeadOnAPike View Post
    It's not required of me to understand why the only fields are of prime numbers in zeta though, so I think I am alright. But am I right about units of a non-field being multiplicate inverses?

    Thanks.

    You mean "...units of a non field HAVING multiplicative inverses"? Yes, this is right: a unit in a ring is an element u s.t. uv=1 for some element v in the ring.

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ring, field, Galois-Field, Vector Space
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 15th 2012, 03:25 PM
  2. ring and field
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 5th 2011, 06:00 AM
  3. Quotient ring / field
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 12th 2010, 12:29 PM
  4. ring+conditon=field
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 3rd 2009, 06:58 PM
  5. ring -> field
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 3rd 2009, 06:38 PM

Search Tags


/mathhelpforum @mathhelpforum