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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.

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2 Answers 2

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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?

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  • $\begingroup$ In fact, the result you mentioned is Nakayama Lemma. $\endgroup$
    – user26857
    Mar 21, 2016 at 21:26
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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 $I=R$.

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