# How to solve this problem about faithful module.

Let $R$ be a non-zero commutative ring with identity and $M$ a unital $R$-module. The $R$-module $M$ is called faithful if $rM=0$ for $r\in R$ implies $r=0$.

Let $M$ be a finitely generated faithful $R$-module and let $I$ be an ideal in $R$ such that $IM=M.$ Prove $I=R.$

From $M$ being finitely generated, I thought $\displaystyle M\cong\mathbb{Z}_{p_{1}^{e_{1}}}\times\cdots\mathbb{Z}_{p_{n}^{e_{n}}}\times\mathbb{Z}^{k}$, with $k\neq 0$ because otherwise $p_{1}^{e_{1}}\cdots p_{n}^{e_{n}}M=0$ and then $M$ won't be faithful. I tried to see if this would be helpful but I couldn't figure out. Let me know how to solve this.

If you look up for the "determinant trick", you will see that a corollary of it is that for finitely generated $R$-module, M, and ideal $I$ such that $IM=M$, there exists an element $y \in I$ so that $(1+y)M=0$. Is this enough to deduce what you want to?
If $$\{g_{1},...,g_{n}\}$$ generates $$M$$, then from $$IM=M$$, $$g_{j}=\sum i_{j,k}g_{k}\Rightarrow \sum(i_{j,k}-\delta_{j,k})g_{k}=0$$ for each $$j$$, for some $$A=\{i_{j,k}\}_{1\leq j,k\leq n}$$. Write the system in matrix form $$Ag=0$$ and multiply with the adjoint of $$A$$ to get $$\det(A)Ig=0$$. Because $$\{g_{1},...,g_{n}\}$$ generates $$M$$, this means $$\det(A)m=0$$ for any $$m\in M$$ and hence $$\det(A)=0$$. But 1 should appear as a term after expanding $$\det(A)$$, so $$1\in I$$ and this implies $$I=R$$.