Results 1 to 3 of 3

Math Help - idempotents

  1. #1
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301

    idempotents

    What are three idempotents of the quotient ring  \mathbb{Q}[x]/(x^4+x^2) ?

    So an element is of the form  a+x^2+x^4 where  a \in \mathbb{Q}[x] . So  (a+x^2+x^4)^2 = a+x^2+x^4 .


    So one idempotent would be:  0 . The others would be just guess and check?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Sampras View Post
    What are three idempotents of the quotient ring  \mathbb{Q}[x]/(x^4+x^2) ?

    So an element is of the form  a+x^2+x^4 where  a \in \mathbb{Q}[x] . So  (a+x^2+x^4)^2 = a+x^2+x^4 .


    So one idempotent would be:  0 . The others would be just guess and check?

    Besides 0 I'd check 1 (!!) and x^2+1

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Sampras View Post
    What are three idempotents of the quotient ring  \mathbb{Q}[x]/(x^4+x^2) ?

    So an element is of the form  a+x^2+x^4 where  a \in \mathbb{Q}[x] . So  (a+x^2+x^4)^2 = a+x^2+x^4 .


    So one idempotent would be:  0 . The others would be just guess and check?
    use Chinese Remainder Theorem for rings:

    since x^2 and x^2+1 are coprime, we have \frac{\mathbb{Q}[x]}{<x^4+x^2>} \cong \frac{\mathbb{Q}[x]}{<x^2>} \times \frac{\mathbb{Q}[x]}{<x^2+1>}. obviously for any two rings R,S, an element (a,b) \in R \times S is an idempotent iff a and b are idempotents.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Idempotents and Projective R-Modules
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 16th 2013, 08:48 AM
  2. [SOLVED] finitely many idempotents
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 3rd 2010, 08:04 PM
  3. idempotents
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 1st 2010, 11:55 PM

Search Tags


/mathhelpforum @mathhelpforum