# Question on proof from Eisenbud's Commutative Algebra.

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.

• By the isomorphisms theorems you always have canonical isomorphisms $$\frac{\frac{M}{P}}{I\frac{M}{P}}=\frac{\frac{M}{P}}{\frac{IM}{IM\cap P}}=\frac{\frac{M}{P}}{\frac{IM+P}{P}}=\frac{M}{IM+P}=\frac{P}{IM\cap P}\quad.$$ In your case, $$\frac{P}{IM\cap P}=0\quad.$$ +1 to your question and to Georges's answer! Dec 11, 2011 at 16:23
• @Pierre: Why does $I\frac{M}{P}=\frac{IM}{IM\cap P}$, and how does $\frac{P}{IM\cap P}=0$? I guess that would follow if $P\subseteq IM$, but I don't see that containment since $P$ is sums of elements of $M$ with coefficients in $R$, not $I$.
– nknx
Dec 11, 2011 at 16:47
• By (implicit) definition: $P:=\sum_i Rm_i$. The submodule $$I\ \frac{M}{P}$$ is the image of $IM$ under the canonical projection $M\to M/P$, image which is $$\frac{IM}{IM\cap P}\quad.$$ Similarly, $$\frac{P}{IM\cap P}$$ is the image of $P$ under the canonical projection $M\to M/IM$, image which is $0$ by assumption. Dec 11, 2011 at 17:46
• @Pierre-YvesGaillard Thanks for clarifying. I suppose I just don't see the assumption that the image of $P$ under the canonical projection into $M/IM$ is $0$. Is it stated implicitly somewhere?
– nknx
Dec 11, 2011 at 18:00
• You're right! Let me try again: By the isomorphisms theorems you always have canonical isomorphisms $$\frac{\frac{M}{P}}{I\frac{M}{P}}=\frac{\frac{M}{P}}{\frac{IM}{IM\cap P}}=\frac{\frac{M}{P}}{\frac{IM+P}{P}}=\frac{M}{IM+P}\quad.$$ In your case, $$\frac{M}{IM+P}=0$$ because $M=IM+P$ since "$m_1,\dots,m_n\in M$ have images in $M/IM$ that generate it as an $R$-module". Dec 11, 2011 at 18:34

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.

• I was getting lost jumping back and forth from $M$ to $IM$ and the like. A warm thanks for your help, dear Georges Elencwajg!
– nknx
Dec 11, 2011 at 15:49