1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homomorphism on rings

  1. Dec 12, 2007 #1
    can someone please explain what these mappings really means? like what is being mapped and mapped to..?? i get confused by the direct sum & product that gets mapped..

    Z [tex]\oplus[/tex] Z ->Z
    Z -> ZxZ
  2. jcsd
  3. Dec 12, 2007 #2
    the first is a mapping from the direct sum Z circle(+) Z to Z, the second is a mapping from Z to the direct product Z x Z, i'm not sure where the confusion is, does this help?
  4. Dec 12, 2007 #3
    thanks! can u tell me if i'm understanding correctly?

    so for Z [tex]\oplus[/tex] Z -> Z (say Z is the integer field),
    can i have some maps like this (4 maps): map((a,b)) = [tex]\pm a \pm b [/tex]
    i mean as long as the value on the right side stays in Z?

    and for Z [tex]\rightarrow[/tex]ZxZ, we are mapping from an elt to an ordered pair like
    map(a)= (ca,ca) where c in Z?
  5. Dec 12, 2007 #4
    Z is not a field

    so you mean can you have the following maps
    f:Z circle(+) Z ->Z given by f(a, b) = a + b
    f:Z circle(+) Z ->Z given by f(a, b) = a - b
    f:Z circle(+) Z ->Z given by f(a, b) = -a + b
    f:Z circle(+) Z ->Z given by f(a, b) = -a - b
    they are all well defined as is the bottom one in your post

    That doesn't mean they are ring homomorphisms though(looking at the 3rd one)
    f(a, b)f(c, d) = (a - b)(c - d) = ac - bc -ac + bd
    f((a, b)(c, d)) = f(ac, bd) = ac - bd
    Last edited: Dec 12, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Homomorphism on rings
  1. Ring homomorphism (Replies: 4)