Finitely generated ideal with special property

Question

Is there a ring with a finitely generated ideal $I$ which has an infinite subset $M\subseteq I$ such that $M$ generates $I$ but no finite subset of $M$ does it?

What I found out: If such a rings exists it can't be Noetherian because of the ascending chain condition.

Answer 1

Let $I=\langle a_1,\dotsc,a_n \rangle$. Every $a_i$ is a linear combination of elements in $M$. Only finitely many elements of $M$ occur. These generate $I$.

The same proof works for arbitrary algebraic structures (here we are in the case of modules). For example, every generating set of a finitely generated subring of a ring contains a finite generating set.

Answer 2

There is a finite subset of $M$ which generates $I$, write $I=(a_1,\ldots,a_m)$ and for each $i$, $a_i=\sum b_{i,k}$ (finite sum ) with $b_{i,k}\in M$, then $I$ is generates by $b_{i,k}\in M$ .