The issue I could have coming up with one might be a misunderstanding of some of the terms used.

Is it true that a generating set works the same way as a generating element but for a higher dimension $\Bbb Z$-module?

Is it true $\Bbb Z_2$ would not have a generating set?

Is it true a free $\Bbb Z$-module is simple a $\Bbb Z$-module that has a basis?

What are examples of $\Bbb Z$-modules without a basis?

In order to solve this specific problem would I pick a $\Bbb Z$-module I know can be an $\Bbb F$-module like $\Bbb Z_7 $-module and then try to find a generating set that does not contain a basis?



Yes, "generating set" is more or less a generalisation of "generating element". In the special case of $\Bbb Z$-modules, a module with a generating element is a cyclic group, while modules that do not have a generating element are non-cyclic, abelian groups. They still have generating sets, though.

More generally, if $R$ is a commutative, unital ring, and $M$ is an $R$-module, then we call $\{m_1, m_2,\cdots\}$ (it might be finite, or it might be infinite) a generating set iff any element $m\in M$ may be written as a summ the following way: $$ m = r_1m_1 + r_2m_2 + \cdots $$ This sum must, of course, only have a finite number of terms to even make sense in the context of abstract algebra (we have no notion of "convergence" here). If there is a generating set of $M$ with only one element, then that element is a generating element. Also note that every module is its own generating set, since $m = 1\cdot m$.

The set $\{m_1, m_2,\cdots\}$ is called a basis iff, in addition to being a generating set, the above sum is unique for each $m$. Another way of thinking about it is that the $m_i$ are linearly independent over $R$.

That being said, let's look at a few of your questions: You asked whether $\Bbb Z_2$ didn't have a generating set as a $\Bbb Z$-module. As per the above, it does. It even has a generating element, namely $\bar 1$. It does not, however, have a basis, because $\bar 1\in \Bbb Z_2$ can be written both as $1\cdot\bar 1$ and as $3\cdot\bar 1$.

Yes, a free $R$-module is, by definition, an $R$-module that has a basis. This goes for $\Bbb Z$ as well as any other unital, commutative ring.

Easy examples of $\Bbb Z$-modules that do not have a basis are the ones that have torsion. In other words, elements of finite order. If we have a $\Bbb Z$-module $M$ and a non-zero element $m\in M$ such that $m + \cdots + m = 0$, then there are no unique decompositions of elements in $M$, no matter what generating set we choose, because we can freely add a certain number of $m$'s (which also has a decomposition) and get back to the same element we had before.


The $\mathbb Z$-module $\mathbb Z$ is generated by the set $\{2,3\}$ but this set does not contain a basis, since the only 2 bases are $\{1\}$ and $\{-1\}$.

Any finite abelian group $A$ is a $\mathbb Z$-module without a basis: $0=|A|\cdot x$ holds for any $x \in A$. This shows that there are no linear independent elements at all in $A$.


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