Let $M$ be a finitely generated module over a commutative ring with unity. Then there eixsts a set of generators $\{m_1,…,m_n\}$ with $n$ minimal.
Proof) Since $M$ is finitely generated, there is $i$ such that $\{l_1,…,l_i\}$ generates $M$. Since for all $j\leq i$, there are $j$ elements generating $M$ or not, there exists a minimal such $n$.
I was wondering if the above is a correct argument.
Best Answer
Your argument is not correct.
For example, consider the abelian group $G=\mathbb{Z}_6=\{0,1,2,3,4,5\}$, regarded as a $\mathbb{Z}$-module. Note that the set $A=\{3,4\}$ generates $G$, and no proper subset of $A$ generates $G$. But $A$ is not a generating set with the least cardinality, since for example, the set $B=\{1\}$ is also a generating set.
Instead, just use well-ordering . . .
Suppose $M$ is a finitely generated module.
Let $S$ be the set of positive integers $n$ such that $M$ can be generated by $n$ elements. By hypothesis, $S$ is nonempty, hence by the Well-Ordering Principle, $S$ has a least element.