let $I$ be a finitely generated ideal of ring $R$.
Suppose $I/I^2$ as an $R/I$ module is generated by $r$ elements then question is to prove that $I$ is generated by $r+1$ elements..
I have tried induction on number of generators of $I$ but could not succeed..
I have then tried to prove for $r=0$..
Suppose $r=0$ i.e., $I/I^2=0$ i.e, $I=I^2$..
Now, $I$ is finitely generated ideal and we have $IM=M$ for $M=I$..
Then by nakayama lemma, we see that there exists $a\in I$ such that $(1+a)M=0$
Let $m\in I$ then $(1+a)m=0$ i.e., $m=-am$..
Thus, $I\subseteq (-a)$.. Other inclusion is trivial. so, we have $I=(-a)$ generated by $1$ element...
I am thinking of using same idea for $r\neq 0$..
Suppose $r=1$ then we have $I/I^2=\langle a+I^2\rangle$ i.e., $I=\langle a \rangle +I^2$.. Now,I want to write $I=\langle a \rangle +I^2=JI$ for some ideal $J$ and repeat the same as in the case of $r=0$. But could not think of any such $J$.. Please help me to solve this...
Please do not write answers, give some hints..
If $I/I^2$ is generated by $r$ elements, you can pick $r$ elements in $I$ which generates $I/I^2$. Let $J$ be the ideal generated by these $r$ elements and then consider $S=R/J$ and $I'=I/J$ an ideal in $S$. Then $I'/I'^2=0$ and you have your argument for this case, showing $I'$ is generated by one element and thus $I$ is generated by $r+1$ elements. My statement, which is stronger can also proved fairly easily and I can write it if you wish.