Nakayama's Lemma

Question

When we prove Nakayama's Lemma, which states that if $M$ is a finitely generated $R$-module, where $R$ is a commutative ring and if $I$ is an ideal of $R$ contained in the Jacobson radical of $R$, if $IM=M$ then $M=0$.

We take the contradiction that $M$ is not equal to zero and the set $\{g_1,\dots,g_n\}$ {$n>1$) is a minimal set of generators of $M$. Since $IM=M$ and we write $g_n=a_1g_1+\dots+a_ng_n$ where $a_1,a_2,\dots,a_n$ are elements of $I$, we rearrange it $(1-a_n)g_n=a_1g_1+\dots+a_{n-1}g_{n-1}$ and since $1-a_n$ is a unit because $a_n$ is in $I\subseteq J(R)$ and, for any element $a$ in $J(R)$, $1-a$ is a unit by the theorem. So how can we say here if $1-a_n$ is unit so $g_n$ belongs to the generating set $\{g_1,\dots,g_{n-1}\}$ which obviously contradicts the minimality of $\{g_1,\dots,g_n\}$.

Can any one assist me to give this concept of $g_n$ belongs to generating set of $\{g_1,\dots,g_{n-1}\}$?

Answer 1

Since $1-a_n$ is a unit, you can multiply by the inverse and get $$ g_n=b_1g_1+\dots+b_{n-1}g_{n-1} $$ If $x\in M$, you know that \begin{align} x&=c_1g_1+\dots+c_{n-1}g_{n-1}+c_ng_n\\ &=c_1g_1+\dots+c_{n-1}g_{n-1}+c_n(g_n=b_1g_1+\dots+b_{n-1}g_{n-1})\\ &=(c_1+c_nb_1)g_1+(c_2+c_nb_2)g_2+\dots+(c_{n-1}+c_nb_{n-1})g_{n-1} \end{align} Thus every element of $M$ is a linear combination of $\{g_1,\dots,g_{n-1}\}$, which is a contradiction.

Answer 2

Simply multiply both sides of your equations by $(1-a_n)^{-1}$ to see that $g_n$ is in the submodule generated by $g_1...g_{n-1}$.