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Three questions about linear algebra

  1. Nov 11, 2012 #1
    There is a theorem in our textbook that says:

    Every abelian group G is a Z-module. Moreover, the Z-module structure
    on G is unique: for n ∈ Z and g ∈ G, ng is the n-th power of g in the group structure of G. (Thus, if n > 0, ng = g + · · · + g, the sum of n copies of g.) Finally, every group homomorphism between abelian groups is a Z-module homomorphism.

    My questions:

    a) "Finally, every group homomorphism between abelian groups is a Z-module homomorphism."...what's the difference between saying that something is a Z-module homomorphism or just any homomorphism?

    b) So, in order to show that any abelian group is a Z-module, we show:

    (m + n)a = ma + na
    m(a + b) = ma + mb
    (mn)a = m(na)
    1a = a

    For some reason, I can't see where the "abelian" property is being used when showing that an abelian group is a Z-module.

    c) Let f : A → B be a ring homomorphism. Then B is a left A-module
    via a · b = f(a)b.

    So when they say a · b = f(a)b, are they assuming that f(a)b is in B?

    Thanks in advance
     
  2. jcsd
  3. Nov 11, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    A "group homorphism" preserves the properties of the group: f(a+ b)= f(a)+ f(b) and f(-a)= -f(a). Saying that it is also a "Z-module homomorphism" means it also preserves the Z-module property: the properties shown in (b), below.

    Look at m(a+ b)= ma+ mb. With m= 2, say, that says (a+ b)+ (a+ b)= (a+ a)+ (b+ b). That requires commutativity.

    I wouldn't use the word "assume". We are told that "f: A→B". That means that, for every a in A, f(a) is in B.

     
  4. Nov 11, 2012 #3
    Thanks a lot...

    I have one more question if you don't mind...

    Let [tex]f_i : M_i → N[/tex] be A-module homomorphisms for 1 ≤ i ≤ k. Then
    there is a unique A-module homomorphism [tex]f : M_1 ⊕· · ·⊕M_k → N[/tex] such that [tex]f ◦ \eta_i = f_i[/tex] for 1 ≤ i ≤ n. Explicitly,
    [tex] f(m_1, . . . ,m_k) = f_1(m_1) + · · · + f_k(m_k)[/tex]

    Where [tex] \eta_i : M_i → M_1 ⊕· · ·⊕M_k[/tex]

    I understand what the theorem says...but then it says that (the theorem) explicitely means:
    [tex] f(m_1, . . . ,m_k) = f_1(m_1) + · · · + f_k(m_k)[/tex]

    I have trouble connecting this to what it said in the theorem. In other words, the theorem says [tex]f ◦ \eta_i = f_i[/tex]. So where exactly is the composite in [tex] f(m_1, . . . ,m_k) = f_1(m_1) + · · · + f_k(m_k)[/tex]?

    In other words, how is the theorem "explicitely" being shown in

    [tex] f(m_1, . . . ,m_k) = f_1(m_1) + · · · + f_k(m_k)[/tex] ?
     
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