unitary modules over Z are simply abelian groups

Question

I have just started with modules and this is not the first time that i have come across something like this. What does it actually mean when you say "this does not add any new structure to the set"? As we add the trivial multiplication operation in a group(a.b=0),although it becomes a ring, they say it does not add structure to it and so is of no importance to us. Same is the case with trivial module structure and also the title of my question. So, what should one understand when someoone says that unitary modules over Z is abelian group. I hope i'm not sounding silly.

Answer 1

It should be clear that, in general, a unitary $R$-module is just an abelian group $A$ with "extra structure", namely a unity-preserving ring-homomorphism $R \to \text{End}(A)$, where the latter ring is the ring of group homomorphisms $A \to A$ with "pointwise" addition and functional composition as the multiplication.

This homomorphism is often couched as an $R$-action, or "scalar multiplication" satisfying certain axioms.

So we can turn any unital $\Bbb Z$-module into an abelian group "naturally", by just forgetting the $\Bbb Z$-action.

On the other hand, suppose we start with an abelian group, $A$. If we wish to impose a (unital) $\Bbb Z$-action upon $A$, we must first have, for any $a \in A$, $1\cdot a = a$. That is, $1 \in \Bbb Z$ must map to the identity endomorphism of $A$.

Now, since we have a ring-homomorphism $\sigma:\Bbb Z \to \text{End}(A)$, we must have:

$\sigma(k+m) = \sigma(k) + \sigma(m)$

In particular, we must have that $2$ maps to the endomorphism that sends $a \mapsto a+a$.

One can then show by induction, that for all $a \in A, n \in \Bbb Z^+$, $n \cdot a = \sum\limits_{i=1}^n a = a + a +\cdots + a$.

Our ring-homomorphism must send $0_{\Bbb Z} \to z$, where $z(a) = 0_A$, for all $a \in A$; that is $0 \cdot a = 0_A$.

It then follows from $-n + n = 0$ in the integers (since we have a ring-homomorphism), that for $n > 0$, $(-n)\cdot a = -(n \cdot a)$, for all $a \in A$.

So the fact that we insist on a unital $\Bbb Z$-module, means that such a structure can (at most) occur in only one way with $A$. It is straight-forward to verify that this $\Bbb Z$-action does indeed give a $\Bbb Z$-module structure to $A$. It is also easy to see that these processes are "inverses" of each other, and when we "forget" the $\Bbb Z$-module structure of $A$, and then restore it, we get the "same" (at least isomorphic to) $\Bbb Z$-module we started with.

If we drop the unital requirement, we lose the uniqueness of the $\Bbb Z$-action (for example, we could have the trivial action, as you point out).

Answer 2

This is just a variation on a theme (i.e. David's answer), but a different perspective might be useful for some people.

To begin with, we view $\mathbb Z$ simply as a cyclic group on one generator. As such it is the free (abelian) group on one element, so group homomorphisms $\mathbb Z \to A$ from $\mathbb Z$ to an (abelian) group $A$ corresponds precisely to the elements of $A$.

All of which is to say that $(\mathbb Z,+)$ is an incredibly basic object, but notice that it also means that there is a natural identification of $\mathrm{End}(\mathbb Z)$ with $\mathbb Z$, and $\mathrm{End}(\mathbb Z)$ is naturally a ring (under pointwise addition and composition) so that the identification immediately equips $\mathbb Z$ with the structure of a ring (if one is paying attention to left and right actions etc. it is actually natural to consider $\mathbb Z$ as the opposite ring to $\mathrm{End}(\mathbb Z)$, but since $\mathbb Z$ is commutative the point isn't too important here).

Moreover, if $A$ is an abelian group, then this $A$ is naturally a module for this ring because $\mathrm{End}(\mathbb Z)$ acts on $\mathrm{Hom}(\mathbb Z,A)$ by precomposing (so here we see one reason why we might want to take $\mathbb Z = \mathrm{End}(\mathbb Z)^{\text{op}}$). $$ \mathrm{End}(\mathbb Z) \times \mathrm{Hom}(\mathbb Z,A) \to \mathrm{Hom}(\mathbb Z,A); \quad (\phi_n,\theta_a)\mapsto \theta_a \circ \phi_n $$ But then using the freeness of $\mathbb Z$ this exactly says that $A$ is a $\mathbb Z$-module, and so we recover the fact that any abelian group is canonically a $\mathbb Z$-module.

(In the same vein, you can deduce that any ring $R$ is a $\mathbb Z$-algebra, that is, there is a canonical homomorphism of rings $c\colon \mathbb Z \to R$, etc..)