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I don't understand the second part of the proof of Corollary 4.8 (Nakayama's Lemma) in Eisenbud's Commutative Algebra.

Let $I$ be an ideal contained in the Jacobson radical of a ring $R$, and let $M$ be a finitely generated $R$-module.

(a) If $IM=M$, then $M=0$.

(b) If $m_1,\dots,m_n\in M$ have images in $M/IM$ that generate it as an $R$-module, the $m_1,\dots,m_n$ generate $M$ as an $R$-module.

Proof of (b): Let $N=M/(\sum_i Rm_i)$. We have $ N/IN=M/(IM+(\sum_i Rm_i))=M/M=0, $ so $IN=N$ and $N=0$ by part (a), and the conclusion follows.

I don't understand why the first two equalities of the displayed line above are immediate. Could someone please explain why they hold? Thank you.

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    You're welcome! Sorry about my confusion... Welcome to MSE!!!2011-12-11

1 Answers 1

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The key point is to understand the $R$-submodule $IN\subset N$.
(For an element $m\in M$, I'll denote by $\bar m$ its image in $N$ in what follows.)

i) How do you write a typical element of that submodule $IN$ ?
Answer : $\Sigma i_k\bar p_k $ with $i_k\in I$ and $p_k\in M$.

ii) And how do you write a typical element of $N$ ?
Answer : $\bar m $ with $m\in M$ .
But the hypothesis of (b) says that you can write $m=\Sigma r_l m_l+\Sigma j_tm_t'$ (with $r_l\in R, m'_t\in M , j_t\in I$), so that $\bar m =\Sigma j_t\overline {m_t'}$, an element in $IN$ according to i).

So indeed $N=IN $ , as claimed.

By the way, you can find a mnemonic for Nakayama's lemma here.

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    I was getting lost jumping back and forth from $M$ to $IM$ and the like. A warm thanks for your help, dear Georges Elencwajg!2011-12-11