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$\mathbb{Z_6}$ is considered a $\mathbb{Z_6}$ free module since it has a basis $\{1\}$ also known as a generator. My question is, don't free modules need a basis over a field? $\mathbb{Z_6}$ is not a field.

My other question is how do we know that a vector space over a field can form a basis?

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For a module the scalars are elements of a ring, which is not necessarily a field.

Every vector spaces can be shown to have a basis using Zorns Lemma. The idea is to use Zorns Lemma to show that a vector space must have a maximal set of linearly independent vectors. Since the set is maximal it must be a basis.

A module may not have a basis. For example consider $Z_{n}$ as $Z$-module.

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If $R$ is any ring (with identity), a (left) $R$-module is free if there exists a set $I$ and an $R$-module isomorphism $M\cong\bigoplus_{i\in I}R$ (the RHS is a direct sum of copies of the left $R$-module $R$ indexed by $I$). It is equivalent that there exists a set $\{m_i\in M:i\in I\}$ such that every element $x$ of $M$ admits an expression $x=\sum_{i\in I}r_ix_i$ with unique $r_i$ which are non-zero for only finitely many $i\in I$. Such a set is called an $R$-basis. So an $R$-module $M$ is free if and only if it admits a basis. It is clear that $\{1\}$ is a basis for the left $R$-module $R$ (just check the conditions in the definition). But a generator is not the same thing as a basis. For example, the element $1+6\mathbf{Z}$ in $\mathbf{Z}/6\mathbf{Z}$ is a $\mathbf{Z}$-module generator, but it is not a $\mathbf{Z}$-module basis. It is however a $(\mathbf{Z}/6\mathbf{Z})$-module basis.

For the $R$-module $\bigoplus_{i\in I}R$, there is a standard basis $\{e_i:i\in I\}$ where $e_i$ is the ``vector" whose $j$-th component for $j\neq i$ is $0$ and whose $i$-th component is $1$. If $M$ is a general $R$-module admitting a basis as above indexed by $I$, we get an isomorphism of $R$-modules $\bigoplus_{i\in I}R\cong M$ by sending $e_i$ to $m_i$.

Using Zorn's lemma as indicated in jojo's answer, one can prove that every module over a field is free, i.e., admits a basis. This can be generalized (using basically the same argument) to modules over division rings. At least over commutative rings $R$, every $R$-module is free if and only if $R$ is a field.

So, to answer your first question, the notion of a free module over a ring $R$ has nothing to do with fields. It just so happens that every module over a field has a basis, and therefore is free.

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