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I wonder, if the ring without unity $2\mathbb{Z}$, considered as a modul over itself, is a free modul.

For a ring with unity, which is not the nullring the answer is clearly yes, because one can write every element r as r $\cdot$ 1, so {1} is a linearly independed spanning set. Oh, I think I don't have to exclude the nullring.

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  • $\begingroup$ Hint: Can there be two independent elements? Can any element generate it? $\endgroup$ – Tobias Kildetoft Oct 28 '15 at 10:11
  • $\begingroup$ Consider the element $2$. Note that $6 = 2*2 + 2$. Can you show that any other element can be generated in a similar manner? $\endgroup$ – Guy Paterson-Jones Oct 28 '15 at 10:20
  • $\begingroup$ Check that the definition of free module is satisfied. The basis of $2\Bbb{Z}$ is $\{ 2 \}$: check that it is an independent generating set. $\endgroup$ – Crostul Oct 28 '15 at 10:28
  • $\begingroup$ en.wikipedia.org/wiki/Free_module . So under point 1. in the wiki articel it is said ...with coefficients in R. But if I write 6=2*2+2 it means for me 6=2*2+1*2 and 1 is no element of 2Z. The same is to write 6 as 3*2 because 3 is no element of 2Z. And this exactly is my problem here. If I consider 2Z only as abelian group or Z-Modul 2 is clearly a generator or linearly independed generating set in the second case. Are the definitions inconsistent here or is 2Z no free modul as questioned? – $\endgroup$ – i-have-no-clue Oct 28 '15 at 11:46
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For a ring $R$ without a unit, we can make it into a ring with unit by forming the ring $\mathbb{Z}\ltimes R$ with underlying abelian group $\mathbb{Z}\oplus R$ and multiplication $(n,r)(n',r') = (nn',nr'+n'r' + rr')$.

Note that a module over a ring without a unit satisfies one less axiom than a module defined over a ring with a unit (i.e $1x = x$ does not make sense for modules over a ring without a unit). There is an isomorphism of categories between $R$-modules and $(\mathbb{Z}\ltimes R)$-modules. An $R$-module $M$ can be made into a $(\mathbb{Z}\ltimes R)$-module by defining $(n,r)m = nm+rm$, and a $(\mathbb{Z}\ltimes R)$-module can be made into an $R$-module by defining $rm = (0,r)m$. Since we know that for rings with unit the free module with a single generator is the ring we conclude the free $R$-module (where $R$ is a ring without unit) with a single generator is $\mathbb{Z}\ltimes R$. In particular $2\mathbb{Z}$ is not a free $2\mathbb{Z}$-module, while $\mathbb{Z} \ltimes 2\mathbb{Z}$ is.

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  • $\begingroup$ So, am I right that 2Z possesses no generating subset at all? $\endgroup$ – i-have-no-clue Oct 28 '15 at 17:29
  • $\begingroup$ @i-have-no-clue Let us agree on the definition of a generating set first. A subset $S$ of a module $M$ is a generating set when for each module $M'$, if $S \subseteq M' \subseteq M$, then $M'=M$. If we agree, then $\{2\}$ is a generating set for $2\mathbb{Z}$. What is false is that it is freely generated, in particular although it might seem strange the fact that $2\cdot 2 = 2+2$ does not follow from the axioms and hence would not be true if $2\mathbb{Z}$ were free (more precisely $\mathbb{Z}\ltimes 2\mathbb{Z}$ is generated by $\{(1,0)\}$ and $2 (1,0)= (0,2) \neq (2,0) = (1,0) + (1,0)$) . $\endgroup$ – Nex Oct 28 '15 at 18:47

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