Question

Prove that if N is a submodule of a finitely generated module M over a ring R .Then M/N is also finitely generated R module.

Answer 1

Suppose $M$ is generated by {a₁,a2,...,a_n}.

Let J = <ai + N | 1 ≤ i ≤ n>.

Clearly, J is finitely generated.

We show that M/N = J.

Suppose $y \in J$ . Then y = a_k + N where a_k $\in$ M for some $k \in$ {1,2,...,n}.

So $y \in M/N$ . Hence $J \subset M/N$

Conversely, suppose $z \in M/N.$ Then $z=x+N$ where $x\in M$ .

Since $M$ is finitely generated, $\exist$ r_i$\in R$ such that $x=\Sigma$ r_ia_i.

Consider $z=x+N \\=(\Sigma r_i a_i ) + N \\=\Sigma(r_i a_i + N) = \Sigma r_i (a_i + N).$

Thus $J$$=M/N.$