Relation between number of generators of $I$ and $I/I^2$ for a finitely generated ideal $I$ in commutative ring

Question

Let $R$ be a commutative ring with unity and let $I$ be a finitely generated ideal of $R$ such that the ideal $I/I^2$ can be generated by $r$ elements . Then how to show that $I$ can be generated as an ideal by $r+1$ elements ? .

I tried to proceed like this : Let $a_1,...,a_r \in I$ such that $\bar a_1 ,..., \bar a_r$ generates the ideal $I/I^2$ . Let $J=(a_1,...,a_r)$ . Then $I=J+I^2$ ; then I think we need to use Nakayma lemma somehow . But I am unable to see how .

Please help . Thanks in advance

Answer 1

Consider $K=(J:I)$, the set of elements which take $I$ into $J$. Then, I claim that $I+K=R$. If not, it is contained in a maximal ideal $M$.But, when you localize at $M$, we have $I_M=J_M$, since $I=J+I^2$ and Nakayama. So, there exists an $s\not\in M$ with $s\in K$, a contradiction.

So, we can find $h\in I$ with $1-h\in K$. I claim that $I=J+hR=I'$, which will prove what you need. So, let $a\in I$. So, we an write, $a=ah+a(1-h)$. Then, $ah\in I'$ and $a(1-h)\in I\subset I'$, since $1-h\in K$. Thus we are done.

Answer 2

From $I=J+I^2$ , we have $(I/J)^2=I/J$ ; now $I$ is a finitely generated ideal in $R$ , so $I/J$ is a finitely generated ideal in $R/I$ ; so by Nakayama lemma , we have $I/J$ is a principal ideal in $R/J$ ( generated by an idempotent) , say $(e+J)R/J = I/J$ , so then $I=J+Re$ , hence the claim .