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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
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