# Problem on idempotent finitely generated ideal

I have a question.

Let $\mathfrak a$ be a finitely generated ideal of $A$, commutative ring with identity, such that $\mathfrak a^2 = \mathfrak a$. Show that $\mathfrak a \subseteq Ae$ with $e^2 = e$, and $e$ is not identity.

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You should add the hypotheses and details about what you want. As things stand, you can simply take $e=1$... –  Mariano Suárez-Alvarez Mar 4 '11 at 18:24
Sounds more like a command than a question. Do you want to know if the statement is true? Or do you know it's true but can't prove it? In any case, it's false. Consider $a=A=F_2$. –  George Lowther Mar 4 '11 at 18:38
See also this question. –  Math Gems Feb 8 '13 at 3:52

What follows is partially excerpted from my Historia Matematica post of 2005\01\04 on this topic, in reply to questions by Colin McLarty and Martin Davis.

Below is a proof by Gilmer from a Monthly Classroom Note[1]. Gilmer mentions it's also presented on p.58 of his textbook[2].$\:$ Remark that the use of Cramer's rule in Gilmer's proof is simply a special case of the deduction of an equation of integral dependence over an ideal (vs. ring), see Kaplansky, Commutative Rings, p.11 Exer.1. or see his later Theorem 75, viz.

THEOREM $\rm\ 75.\ \$ Let $\rm\:R\:$ be a ring, $\rm\:J\:$ an ideal in $\rm\:R,\ B\:$ an $\rm\:R$-module generated by $\rm\:n\:$ elements,$\ \$ and $\rm\:r\:$ an element of $\rm\:R\:$ satisfying $\rm\ r\:B \subset JB\:.\$ Then $\rm\ (r^n - j)\ B = 0\$ for some $\rm\ j \in J\:.$

The desired proof now follows immediately, namely:

Specializing $\rm\ r=1,\ B=J\$ yields $\rm\ (1-j)\ J = 0\ \Rightarrow\ J = (j),\ \ j^2 = j\:.\quad$ QED

Note that this may be viewed as a generalization of the simpler Dedekind domain case.

The above proof doesn't work in the noncommutative case because the determinant trick no longer applies. However one can prove Theorem 75 without using determinants by instead appealing to Nakayama's Lemma. Namely, see Exercises 3.1, 3.2, p.43 in Atiyah and Macdonald, Introduction to Commutative Algebra. Below is said direct proof of Gilmer - using Cramer's rule.

LEMMA $\$ If $\rm\:B\:$ is a finitely generated idempotent ideal of a commutative ring $\rm\:T\:$ then $\rm\:B\:$ is principal and is generated by an idempotent element.

Proof $\$ We first assume that $\rm\:T\:$ has an identity and we let $\rm\:\{b_{\:i}\:\}\:$ be a finite set of generators for $\rm\:B\:.\:$ Then $\rm\:B = B^2 = \sum\: B\ b_{\:i}\:$ so that we obtain a system of equations $\rm\ b_k = \sum s_{\:k\:i}\ b_{\:i}\:,\$ where $\rm\:s_{\:k\:i} \in B\:.\$ This gives rise to a system of equations $\rm\ \sum\ (\delta_{\:k\:i}-s_{\:k\:i})\ b_{\:i} = 0\:,\:$ where $\delta =$ Kronecker delta.

By Cramer's rule $\rm\:d\ b_i = 0\:$ for all $\rm\:i\:,\:$ where $\rm\:d\:$ is the determinant of the matrix $\rm\ [\delta_{ki}-s_{ki}]\:.\$ It is easy to see that $\rm\:d\:$ has the form $\rm\: 1-b\:$ for some $\rm\:b \in B\:.\$ Since $\rm\ 0 = d\ b_i = b_i - b\ b_i\$ for each $\rm\:i,\ B\$ is the principal ideal generated by $\rm\:b\:.\$ And since $\rm\:1-b\:$ kills $\rm\:B\:,\:$ we conclude $\rm\ (1-b)\ b = 0\:,\$ or $\rm\ b = b^2\:.$

If $\rm\:T\:$ contains no identity element, we consider a commutative ring $\rm\:T'\:$ obtained by adjoining an identity element to $\rm\:T\:.\$ Then $\rm\:B\:$ is a finitely generated idempotent ideal of $\rm\:T',\:$ and hence is principal as an ideal of $\rm\:T',\:$ generated by an idempotent element $\rm\:v\:.\$ Since $\rm\:T'$ is obtained by adjoining an identity element to $\rm\:T\:,\:$ it follows $\rm\:v\:$ also generates $\rm\:B\:$ as an ideal of $\rm\:T\:.\quad$ QED

[1] Robert Gilmer, An Existence Theorem for Non-Noetherian Rings (in Classroom Notes),
The American Mathematical Monthly, Vol. 77, No. 6, pp. 621-623.

[2] Robert Gilmer, Multiplicative Ideal Theory,
Queens University, Kingston, Ontario, 1968. See p.58

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Seealso this question. –  Bill Dubuque Sep 17 '11 at 18:34