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Let $M$ be a finitely generated R-module and $I$ be an ideal of $R$ such that $IM=M$. Then prove there exists $a\in I$ such that $(1-a)M=0$.

Thanks

maryam
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2 Answers2

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This is a special case of a generalization of the Cayley-Hamilton theorem. It can be proved as follows:

Let $(m_1,\ldots,m_n)$ be a generating system for $M$. Since $M = IM$, we can write $m_j = \sum_i a_{ij} m_i$ with $a_{ij} \in I$. Considering $M^n$ as an $\operatorname{Mat}_n(R)$-module in the obvious way, we have $$(1_n-A) \left(\begin{matrix}m_1\\ \vdots \\ m_n \end{matrix}\right) = 0$$ where $A = (a_{ij})$. Multiply this with the adjugate matrix of $(1_n-A)$ to get $$\det(1_n -A)M = 0.$$ Now expand the determinant using the Leibniz Rule. This gives an $a \in I$ with $(1-a)M = 0$.

marlu
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  • I was thinking of the more general version that says that an endomorphism $\phi$ of $M$ with $\phi(M) \subset IM$ satisfies a monic algebraic relation over $I$. The above is the special case $\phi = 1_M$. I edited my answer and called it "a special case of a generalization" (these do not cancel out). – marlu Jan 24 '13 at 18:37
  • That makes sense, then :) – rschwieb Jan 24 '13 at 18:50
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This is a version of Nakayama's lemma, as stated in statement 1 here, and which you can find a proof for in zillions of places if you search for "proof of Nakayama's lemma". (That includes this link itself.)

To be clear, that statement says "There exists an $r$, $1-r\in I$ and $fM=0$. If you take $a=1-r$, you have your $a$.

rschwieb
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