Results 1 to 8 of 8

Math Help - Finding all homomorphisms

  1. #1
    Junior Member
    Joined
    May 2011
    Posts
    59

    Finding all homomorphisms

    My goal was to find all homomorphisms \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}, and \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}, and \mathbb{Q} \to \mathbb{Q}. Here's what I have...

    For \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} I believe they are (a,b) \mapsto 0, (a,b) \mapsto (a,b), (a,b) \mapsto (b,a), (a,b) \mapsto (a,0), and (a,b) \mapsto (0,b).

    For \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} I have the identity, the 0-map, (a,b) \mapsto a, and (a,b) \mapsto b.

    And for \mathbb{Q} \to \mathbb{Q} I just have the 0-map and the identity.

    Did I get them all? And when there are more 'creative' ones like (a,b) \mapsto (b,a), how do we know when we've found them all?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    what about (a,b) ---> (ma,nb) and (a,b)--->(nb,ma)?

    and for Z⊕Z -->Z, what about (a,b) --> ma+nb?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2011
    Posts
    59
    Neither of the first two work unless n, m= 0 or 1 because upon finding \phi\[(a,b)\]\phi\[(c,b)\], there are n^2 and m^2 terms that do not appear with \phi\[(a,b)(c,d)\], which just has n and m.

    The third one doesn't work since \phi\[(a,b)(c,d)\]=mac+nbd and \phi\[(a,b)\]\phi\[(c,b)\]=m^2ac+mand+nbmc+n^2bd

    The second does give (a,b) \mapsto (b,0) and (a,b) \mapsto (0,a).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,391
    Thanks
    757
    perhaps you should indicate what kind of structure the homomorphisms are to preserve?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Agreed. Ring homomorphisms?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2011
    Posts
    59
    Yes, ring homomorphisms. I suppose that would be good to know.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by AlexP View Post
    My goal was to find all homomorphisms \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}, and \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}, and \mathbb{Q} \to \mathbb{Q}. Here's what I have...

    For \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} I believe they are (a,b) \mapsto 0, (a,b) \mapsto (a,b), (a,b) \mapsto (b,a), (a,b) \mapsto (a,0), and (a,b) \mapsto (0,b). (a,b) → (a,a) ? (a,b) → (b,b) ?

    For \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} I have the identity, the 0-map, (a,b) \mapsto a, and (a,b) \mapsto b.

    And for \mathbb{Q} \to \mathbb{Q} I just have the 0-map and the identity.

    Did I get them all? And when there are more 'creative' ones like (a,b) \mapsto (b,a), how do we know when we've found them all?
    A ring homomorphism has to take idempotents to idempotents. In the ring \mathbb{Z} \oplus \mathbb{Z}, the only idempotents are (0,0), (1,0), (0,1) and (1,1). Also, every element of \mathbb{Z} \oplus \mathbb{Z} is a linear combination of (1,0) and (0,1), so a homomorphism is specified by what it does to those two elements. That severely restricts the possible choices for a homomorphism.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    May 2011
    Posts
    59
    Thank you, that does clear things up. Ring homomorphisms are more rigid than I was thinking they are...I saw something like (a,b) \mapsto (b,a) as being somewhat random and didn't know how it fit in with the properties of homomorphisms, but now I see what the basis for it is. Thanks to everyone else too.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 6th 2011, 11:39 AM
  2. Finding all the homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 5th 2010, 08:30 PM
  3. Homomorphisms to and onto
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 23rd 2009, 09:21 AM
  4. homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 11th 2008, 04:30 PM
  5. Homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: April 23rd 2007, 11:53 AM

Search Tags


/mathhelpforum @mathhelpforum