# Three questions about linear algebra

1. Nov 11, 2012

### Artusartos

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?

2. Nov 11, 2012

### HallsofIvy

Staff Emeritus
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.

3. Nov 11, 2012

### Artusartos

Thanks a lot...

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

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

Where $$\eta_i : M_i → M_1 ⊕· · ·⊕M_k$$

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

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

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

$$f(m_1, . . . ,m_k) = f_1(m_1) + · · · + f_k(m_k)$$ ?