I'm trying to solve a problem without the assumption of Nakayama's lemma, and this statement clearly implies the Nakayama's lemma.
The statement goes as follow: let $R$ be a ring, and let $M$ be an $R$-module with a finite generating set $T$. Let $T'$ be a subset of T. If there is an ideal $I \subset R$ contained in the Jacobson radical of $R$ such that $T' \subseteq IM$, show that $T \backslash T'$ generates M.
Can someone help me solving this problem please?
Note:$\;$If $y\in I$ then $1-y$ is not in any maximal ideal of $R$, hence $1-y$ is a unit of $R$.
Let $T=\{t_1,...,t_k\}$ and suppose $t_k\in T'$.
Let $N$ be the $R$-submodule of $M$ generated by $\{t_1,...,t_{k-1}\}$.
Claim:$\;N=M$.
Proof:
It suffice to show $t_k\in N$.
If $t_k=0$, the claim is trivially true, so assume $t_k\ne 0$.
First suppose $k=1$ (i.e., $T=T'=\{t_1\}$).
Then from $t_1\in IM$, we get $t_1=xt_1$ for some $x\in I$, hence $(1-x)t_1=0$.
But then since $1-x$ is a unit of $R$, we get $t_1=0$, contradiction.
Thus the claim holds for the case $k=1$.
Next suppose $k > 1$.
By hypothesis we have $t_k=xm$ for some $x,m$ where $x\in I$ and $m\in M$.
Since $T$ generates $M$, we can write $$ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! m=r_1t_1+\cdots+r_kt_k $$ for some $r_1,...,r_k\in R$, hence \begin{align*} & t_k=xm\\[4pt] \implies\;& t_k=x(r_1t_1+\cdots+r_kt_k)\\[4pt] \implies\;& t_k\equiv (xr_k)t_k\;(\text{mod}\;N)\\[4pt] \implies\;& (1-xr_k)t_k\equiv 0\;(\text{mod}\;N)\\[4pt] \implies\;& t_k\equiv 0\;(\text{mod}\;N)\qquad\text{[since $1-xr_k$ is a unit of $R$]}\\[4pt] \implies\;& t_k\in N\\[4pt] \end{align*} which completes the proof of the claim.
Thus $\{t_1,...,t_{k-1}\}$ generates $M$.
Iterating the process, eliminating the elements of $T'$ one at a time, it follows that $T{\,\setminus\,}T'$ generates $M$, as was to be shown.