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I came across this old exam problem. If $R$ is a local Noetherian ring and $I$ is an ideal in $R$ such that $I^2=I$ then $I =0$.
Any hint would be appreciated. I'm only familiar with what the definition of local and Noetherian mean. I'm not sure why these properties are useful.
If you don't get how Zev's hint goes here are three more hints to apply Nakayama's Lemma:
Since $R$ is Noetherian it follows that $I$ is finitely generated as an $R$ - module.
Since $R$ is a local ring, $\operatorname{Jac}( R) = \mathfrak{m}$, the maximal ideal of $R$.
Every ideal is contained in some maximal ideal. The containment could be proper or we can have equality.
Hint: Nakayama's lemma.
(Also, since you didn't require $I$ to be a proper ideal, technically $I=R$ is also possible.)
How familiar are you with the Jacobson radical $J(R)$? It is the intersection of all the maximal ideals of a ring $R$ with 1. If $r\in J(R)$ we have $1-r$ cannot lie in any maximal ideal and so must be invertible.
To answer your question, you must show what was actually, for $M$ an ideal of $R$, in Jacobson's thesis years before it became known as Nakayama'a Lemma (with Nakayama denying he (Nakayama) was the originator).
Nakayama's Lemma Let $M_R$ be a finitely generated module over any ring $R$ with identity. Then $MJ(R)\ne M$.
To show this, start with $\{a_1,\ a_2,\ \cdots\, \ a_n\}$ a set of generators for $M$ of smallest cardinality $n$. Express $a_1$ as a sum of the form $a_1=\sum_{i=1}^n a_ij_i$ where $\{j_i\}\subseteq J(R)$. Now hopefully you can complete the proof of the theorem in your posting and edit the posting to make sure the theorem is mathematically correct.
