# Noetherian Local Ring

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.

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Dear jennifer, you got three good answers. Please accept one of them. – user26857 Feb 19 '13 at 9:15

If you don't get how Zev's hint goes here are three more hints to apply Nakayama's Lemma:

1. Since $R$ is Noetherian it follows that $I$ is finitely generated as an $R$ - module.

2. Since $R$ is a local ring, $\operatorname{Jac}( R) = \mathfrak{m}$, the maximal ideal of $R$.

3. Every ideal is contained in some maximal ideal. The containment could be proper or we can have equality.

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Hint: Nakayama's lemma.

(Also, since you didn't require $I$ to be a proper ideal, technically $I=R$ is also possible.)

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Thanks I see how the Lemma helps – Digital Gal Feb 4 '13 at 2:41
@jenniferage20 More generally one can show that any idempotent ideal which is finitely generated is principally generated by an idempotent. See here for instance. – JSchlather Feb 4 '13 at 2:43

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.

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