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We know that any Abelian group is a $\mathbb{Z}$-module. Is the converse true?

2 Answers2

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Yes, there is a bijective correspondence between $\mathbb Z$-modules and abelian groups.

From module to group, just forget the scalar multiplication; the module laws directly require that the module's addition constitutes an abelian group.

From group to $\mathbb Z$-module, declare $n\cdot a$ to be $\underbrace{a+a+\cdots+a}_n$ and so forth.

hmakholm left over Monica
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    @aboozar: What are you trying to do there? It is (usually) explicitly part of the module axioms that the addition must be commutative, not something that needs a separate proof. – hmakholm left over Monica Feb 19 '15 at 15:08
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    @aboozar: Every module everywhere, over anything, has an abelian addition. That's part of the definition of "module". (Your reasoning shows that this part of the definition could have been omitted -- even for $\mathbb Z_2$ -- but it usually isn't). – hmakholm left over Monica Feb 19 '15 at 15:18
  • @HenningMakholm Commutativity of module addition comes for free if rings have a unity and modules are unitary. – egreg Feb 19 '15 at 16:42
  • @egreg: Um, yes? – hmakholm left over Monica Feb 19 '15 at 16:52
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    @HenningMakholm Yes! $(1+1)(x+y)=(1+1)x+(1+1)y=x+x+y+y$; on the other hand, $(1+1)(x+y)=1(x+y)+1(x+y)=x+y+x+y$. Subtract $x$ from the left and $y$ from the right and you get $x+y=y+x$. – egreg Feb 19 '15 at 16:54
  • @egreg: Again: Yes? That's the argument from aboozar's first comment. I'm not sure what your point is here. – hmakholm left over Monica Feb 19 '15 at 16:55
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    @HenningMakholm Of course. In my definition of vector space, for instance, I don't assume commutativity of addition. Showing commutativity is a good exercise in applying the axioms. – egreg Feb 19 '15 at 16:57
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    This answer is much better than the accepted one. The equivalence of Z modules and abelian groups comes down to the very elementary fact that when looking at 'multiplication' by an integer, that expression can be expressed as 'addition', so the multiplication operation isn't actually adding anything new. – Jonathan Hebert Jul 25 '18 at 15:08
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Let $A$ be a commutative ring with unit. To give yourself an $A$-module $M$ is equivalent to giving yourself an abelian group $M$ plus an action of $A$ on $M$, that is, a ring morphism $\varphi : A \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$ where $\textrm{Hom}_{\textrm{Ab.}}(N',N)$ is the set of morphism of abelian groups from $N'$ to $N$ if $N,N'$ are abelian groups.

Note that $\textrm{Hom}_{\textrm{Ab.}}(M,M)$, also noted $\textrm{End}_{\textrm{Ab.}}(M)$, is a ring with unit, but is not necessarily commutative. Now as you know, for any ring $R$ with unit, you have a unique ring morphism $\mathbf{Z}\rightarrow R$. So that in particular if $M$ is an abelian group, you have a unique ring morphism $\varphi : \mathbf{Z} \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$, that is, a unique structure of $\mathbf{Z}$-module on $M$. Inversely, having one structure of $\mathbf{Z}$-module on $M$ does not give very much, that is, not more that the already existing unique ring morphism $\varphi : \mathbf{Z} \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$.

In a pedantic way : the forgetful functor from the category of $\mathbf{Z}$-modules to the category of abelian groups is an equivalence of category (and indeed worth mentionning, even an isomorphism of categories).

Olórin
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  • @ Robert Green: thank you for your complete answer.I have another question in category theory: why the cokernel of zero-morphism from $0$ to $A$ is $A$ whit identity morphism? –  Feb 19 '15 at 17:20
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    Because it's the "smaller" map you can compose the zero-morphism to get the zero morphism, property of the cokernel. – Olórin Feb 19 '15 at 17:22
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    note that the forgetful functor from $\mathbb{Z}$-modules to abelian groups is not just an equivalence of categories but actually an isomorphism of categories! Isomorphisms of categories are extremely rare "in nature". This means that abelian groups and $\mathbb Z$-modules are really "the same thing" in the strongest possible sense. – Tashi Walde Feb 24 '18 at 16:10