Let $R$ be a unital ring, let $F$ be a free left $R$-module with basis $B$, and let $K \subset F$ be an $R$-submodule. Let
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\begin{align*}
q: F \to F/K, \quad x \mapsto x + K
\end{align*}
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be the canonical quotient homomorphism. Then the set $q(B) = \{q(b) : b \in B\}$ generates $F/K$ as a left $R$-module, and $\ker q = K$. Equivalently, if the relations among the generators $q(B)$ are understood as the elements $x \in F$ whose image in $F/K$ is zero, then the module of relations is exactly $K$.
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Conversely, for every left $R$-module $M$, there exist a set $S$, a free left $R$-module $P$ with basis indexed by $S$, and an $R$-submodule $N \subset P$ such that $M \cong P/N$.