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I am trying to understand the proof of the following statement:

Let $A$ be a commutative ring, let $M$ be a finitely generated $A$-module and $I$ an ideal of $A$ such that $IM=M$. Then there is an $a\in I$ such that $(1-a)M=0$.

Proof. If $M=0$, there is nothing to prove. Let $M\neq0$ be generated by $m_1,..., m_n$. Since $IM=M$, there exist $x_{ij}\in I$ for $1\leq i,j\leq n$ such that

$$ \begin{pmatrix} 1-x_{11} & -x_{12} & ... & -x_{1n} \\ -x_{21} & 1-x_{22} & ... & -x_{2n} \\ \vdots & \vdots & \ddots & \vdots\\ -x_{n1} & -x_{n2} &...& 1-x_{nn} \end{pmatrix} \begin{pmatrix} m_1 \\ m_2 \\ \vdots \\ m_n \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ \vdots\\ 0 \end{pmatrix} $$ Multiply on the left by the adjoint of the matrix $T$ to get $$ \det T\begin{pmatrix} m_1 \\ m_2 \\ \vdots \\ m_n \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ \vdots\\ 0 \end{pmatrix} $$ which implies that $\det (T)M=0 $. But $T\equiv I_{n\times n} \mod I$, so $\det T\equiv 1 \mod I$, so $\det T=1-a$ for some $a\in I$.

I don't understand why do the $x_{ij}\in I$ exist and where does this matrix come from.

Jimmy R
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2 Answers2

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The matrix is encoding the statement $IM=M$. In fact, given that $\{m_i\}$ generate $M$, then we have $$IM=\{ \sum_i x_im_i : x_i\in I\}.$$ But since $m_i\in M = IM$ we get $m_j =\sum_ix_{ij}m_i$, giving the matrix by subtracting $$ \sum_i (\delta_{ij}-x_{ij})m_i = 0.$$

user26857
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Spencer Leslie
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Let $m_1,\ldots, m_n$ be a set of generators for $M$, then for any $m_i $ since $m_i \in M = IM$ there are $x_{ij} \in I $ such that $m_i = \sum x_{ij} m_j$, which gives rise to the system of liner equations $\sum(\delta_{ij} -x_{ij})m_j =0$.

user26857
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baharampuri
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