# Do we get this lucky property in Noetherian rings?

One definition of a Noetherian ring is that every ideal is finitely generated. If $R$ is a Noetherian ring and $I$ and idempotent ideal, that is, $I^2=I$, would it be too far-fetched to guess that $I$ is generated by an idempotent element?

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I am pretty sure this question has been asked here at least a couple of times before... –  Mariano Suárez-Alvarez Oct 22 '11 at 20:36
Yes, Mariano, as Google just told me, you answered it yourself ! math.stackexchange.com/questions/29454/… [I selfishly wish you had written this comment before I wrote my answer :)] –  Georges Elencwajg Oct 22 '11 at 21:09

It is true, and a consequence of Nakayama's lemma: since $I$ is a finitely generated $R$-module such that $I\cdot I = I$, there is an $x \in I$ such that for all $y \in I$, $(1-x)y=0$, that is $y=xy$. Hence, $x$ is the required idempotent.

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What if $R$ is not commutative? –  Arturo Magidin Oct 22 '11 at 20:29
I added a "Hence" in the final sentence. It almost looked like you had the equation "$y = xy.x$ instead of a sentence that ended in $y=xy$, and then one that began with $x$. (Note: In general, one should avoid starting sentences with mathematical symbols) –  Arturo Magidin Oct 22 '11 at 20:32
Thanks for this. –  Nicole Oct 22 '11 at 21:00

Yes, if an ideal $I\subset R$ satisfies $I=I^2$ and if $I$ is finitely generated, then there is an idempotent $i\in I$ which generates $I$ ( i.e. $I=iR$).

Proof
Recall Nakayama's lemma: if a finitely generated $R$-module $M$ satisfies $M=IM$ for some ideal $I\subset R$ (not assumed finitely generated!), then there is an $i\in I$ such that for all $m\in M$ we have $m=im$.
In your situation take $M=I$ and conclude first that $i=i^2$ and second that for all for all $m\in I$ you have $m=im\in I$, so that $I\subset iR$. If you remember that $i\in I$ you see that conversely $iR\subset I$, so that finally $I=iR$ for an idempotent $i\in I$, as requested.

Remarks
1) The ring $R$ doesn't have to be noetherian, i.e. you needn't assume that ideals of $R$ different from $I$ are finitely generated.
2) In the ring $R=\mathbb F_2^{\mathbb N}$ of sequences from the field with two elements, the ideal $I=\mathbb F_2^{(\mathbb N)}$ of sequences ultimately zero satisfies $I=I^2$ (since we even have $i=i^2$ for every $i\in R$ !) but is not finitely generated.
The reason is that $I$ contains sequences with arbitrary many non-zero entries, whereas finitely many sequences $s_1,s_2,...,s_r\in I$ can only generate an ideal all of whose elements $(a_0,a_1,\ldots, a_n, \ldots)$ satisfy $a_n=0$ for $n\geq$ some fixed $N\in \mathbb N$

A geometric interpretation
The condition $I=I^2$ may look artificial. Let me show that it comes up in real life, that is in scheme theory.
Suppose $Y\hookrightarrow X$ is a closed immersion of schemes; is it flat?
The problem is local, so assume $X=Spec(R), Y=Spec (R/I)$. The question reduces to: Is $R/I$ an $R$-flat module?
Consider the exact sequence $0\to I\to R$. If $R/I$ is flat, the tensored sequence $0\to I\otimes _R R/I \to R \otimes _R R/I$ must remain exact . But it is, up to canonical isomorphisms,
$0\to I/I^2 \stackrel {f}{\to} R/I$ where $f(\bar x)=\tilde x=0$ for $x\in I$. The only way for $f$ to be both zero and injective is to have $I=I^2$!
Conversely, if $I$ is finitely generated and $I=I^2$, we have seen that $I=(i)$ for some idempotent $i$.
Then $Spec (R/I)\to Spec(R)$ is flat because it is an open immersion since $Spec (R/I)=V(I)=V(i)=Spec(R)\setminus V(1-i)=$ open subset of $Spec(R)$ .
So in particular for a noetherian ring $R$, the closed immersion $Spec(R/I) \hookrightarrow Spec(R)$ is flat iff $I=(i)$ for some idempotent $i\in R$. In other words closed immersions of schemes are essentially never flat!

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This makes sense. Thanks. –  Nicole Oct 22 '11 at 20:59