Consider the $\mathbb{C}$-vector space $M=\mathbb{C}^n$. I am trying to prove that as an $R$-module, $M$ is finitely generated iff $R=\mathbb{C}[x]$.
Suppose first that $R=\mathbb{C}[x]$. Since $M$ is a $\mathbb{C}$-module, we know that the V.S. basis $B=\{m_1, ..., m_n\}$ would also generate $M$. This means there are numbers $c_1,\dots, c_{n}\in \mathbb{C}$ such that given any $m\in M$, we can write $$m=c_1m_1+\cdots+c_{n}m_{n}. $$ However, the $c_{i}$'s are also constant polynomials in $\mathbb{C}[x]$ so we get that $B$ is a finite generating set for the $R$-module $M$.
I don't know how to prove the converse and would appreciate some help.
Edit: This exercise arose from the following notes I am reading:
As D_S noted, the converse is false. Let $M=\mathbb{C}[X]$, the free $\mathbb{C}[X]$-module of rank $1$. Then no module of the form in the example is isomorphic to $M$. To see this, let $N=\mathbb{C}^n$ be any $\mathbb{C}[X]$-module with respect to some linear transformation $T:\mathbb{C}^n\to\mathbb{C}^n$. Then there is some polynomial on which $T$ is $0$ (such as the characteristic polynomial). Say $a_0+a_1T+\cdots+a_mT^m=0$. Then $a_0+a_1X+\cdots+a_mX^m$ is an element of $\mathbb{C}[X]$ which annihilates $N$ since for any $v\in N$, we have $$(a_0+a_1X+\cdots+ a_mX^m)\cdot v=a_0v+a_1T(v)+\cdots+a_mT^m(v)=0$$ and therefore $N$ is not a free module. Consequently, $N\not\cong M$.