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Having trouble working with modules

  1. Mar 30, 2012 #1
    Hi guys,

    Basically I'm playing around with modules at the moment, and I can't work out why we cant have the group of integers as an F-module (F a field), where the left action is the identity.

    i.e F x Z ----> Z

    where we have f.z = z

    f in F, z in Z

    If this were possible, then Z would be a vector space wouldn't it, this is probably a stupid question but would be grateful if somebody could point out where I'm going wrong, i've been trying to work it out for hours.

    Thanks!

    C
     
  2. jcsd
  3. Mar 30, 2012 #2

    micromass

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    Does the axiom

    [tex](f+f^\prime)z=fz+f^\prime z[/tex]

    Still hold??

    What about 0z=0 ??
     
  4. Mar 30, 2012 #3
    Ahhh, so simple!

    Thank you so much, I think i'll noodle around with them a bit more so I can understand them better.

    Really appreciate it.

    C
     
  5. Mar 30, 2012 #4
    Another question for you :),

    I've been playing around with these and had a look at a bit of representation theory.

    I was looking at group algebra's, where G is a finite group, and \mathbb{C} is our field.


    I was trying to find a C[G] module, V, that has a left action is the basis of C[G] (i.e. group elements of G) on V as the identity map,

    i.e. g.v=v where g \in{G}


    I wasn't sure if the axiom (g+h).v = g.v + h.v held, but I think its to do with the fact that g+h isn't a basis element and thus

    (g+h).v \neq v

    and that the axiom holds trivially as our left action is a group homomorphism

    i.e. (g+h).v = g.v + h.v by definition

    am I right in saying this, otherwise I can't work out how we get the identity rep for group algebra's under the correspondence theorem in representation theory.

    Thanks in advance

    Sorry if this is unclear, just say if you can't work out what i'm trying to say

    C
     
  6. Mar 30, 2012 #5

    micromass

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    Yes, that seems right indeed!!
     
  7. Mar 30, 2012 #6
    Awesome, cheers Micromass!

    C
     
  8. Mar 30, 2012 #7


    Ah wait no,

    What I said that (g+h) isnt a basis element is wrong. It is an element of G and thus a basis element. Therefore we have

    (g+h).v = g.v +h.v

    implies v=v+v

    So I'm still stuck as to how we can get left action to be the identity.

    thought we almost had it!

    C
     
    Last edited: Mar 30, 2012
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