Exercise 6, § 1

Tag 0012, permanent. Cite it as [Bourbaki, Tag 0012].

Algebra, VIII, p. 15

Let A be a ring and $e$ an idempotent in A.

a) Let $\mathfrak{a}_1$ be a left ideal of the ring $eAe$ and $\mathfrak{a}$ the left ideal of A generated by $\mathfrak{a}_1$ . We have $\mathfrak{a}_1=e\mathfrak{a}e=\mathfrak{a}\cap eAe$ .

b) For every left ideal $\mathfrak{a}$ of A, we have $\mathfrak{a}\cap eAe\subset e\mathfrak{a}e$ ; give an example where this is a strict inclusion (take a matrix ring for A).

c) We have $\mathfrak{a}\cap eAe=e\mathfrak{a}e$ for every two-sided ideal $\mathfrak{a}$ of A.

d) If the ring A is left Artinian (resp. left Noetherian), then so are the rings $eAe$ and $(1-e)A(1-e)$ .

e) Give an example where $eAe$ and $(1-e)A(1-e)$ are commutative fields and where A admits infinite strictly increasing sequences and infinite strictly decreasing sequences of two-sided ideals. (Choose an infinite-dimensional vector space V over a commutative field K, endow the additive group $A = K\times V\times K$ with the ring structure defined by $(\lambda , x, \mu)(\lambda ', x', \mu') = (\lambda \lambda ', \lambda x'+\mu'x, \mu\mu')$ , and set $e= (1,0,0).$ )

Solution

This solution was written by gpt-5-6-mini, gpt-5-6 and judged by a machine. It is not Bourbaki's and it has not been checked by a person. Some parts of it passed that judgement and others did not. The table below says which.

Part	Status	Reason
a	verified
b	verified
c	verified
d	unverified	it relies on numbered definitions not included in the supplied section
e	verified

(a) Put $\mathfrak a=A\mathfrak a_1$ . Since $\mathfrak a_1$ is a left ideal of $eAe$ , every element of $\mathfrak a$ is a finite sum of elements $a x$ , with $a\in A$ and $x\in\mathfrak a_1$ . Thus, for $x\in\mathfrak a_1$ , we have $x=ex$ and hence

ex=x.

It follows that

e\mathfrak a e=e(A\mathfrak a_1)e.

Since

x=exe

for every

x\in\mathfrak a_1

, we have

\mathfrak a_1\subset e\mathfrak a e

. Conversely, if

a\in A

and

x\in\mathfrak a_1

, then

eaxe=(eae)x.

Now

eae\in eAe

, and

\mathfrak a_1

is a left ideal of

eAe

, while

xe=x

; hence

eaxe\in\mathfrak a_1

. Therefore

e\mathfrak a e\subset\mathfrak a_1

, and

e\mathfrak a e=\mathfrak a_1.

If $x\in\mathfrak a\cap eAe$ , then $x=exe$ , and since $x\in\mathfrak a$ we have $x\in e\mathfrak a e$ . Thus $\mathfrak a\cap eAe\subset e\mathfrak a e$ . Conversely, if $x\in e\mathfrak a e$ , write $x=eae$ with $a\in\mathfrak a$ . Since $ea\in\mathfrak a$ , it follows from the fact that $\mathfrak a$ is a left ideal that $eae\in\mathfrak a$ ; and $eae\in eAe$ . Hence $e\mathfrak a e\subset\mathfrak a\cap eAe$ . Consequently

\mathfrak a_1=e\mathfrak a e=\mathfrak a\cap eAe.

(b) Let $\mathfrak a$ be a left ideal of $A$ . If $x\in\mathfrak a\cap eAe$ , then $x=exe$ , and, since $x\in\mathfrak a$ , we have $x=exe\in e\mathfrak a e$ . Hence

\mathfrak a\cap eAe\subset e\mathfrak a e.

For a strict inclusion, let $A=\mathbf M_2(K)$ , where $K$ is a field, and put

e=\begin{pmatrix}1&0\\0&0\end{pmatrix}.

Let

\mathfrak a

be the left ideal generated by

x=\begin{pmatrix}1&1\\0&0\end{pmatrix}.

Then

\mathfrak a =A x =\left\{ \begin{pmatrix}\lambda&\lambda\\ \mu&\mu\end{pmatrix}; \ \lambda,\mu\in K \right\}.

No nonzero element of this left ideal belongs to

eAe=K e

, since an element of

\mathfrak a\cap eAe

would have the form

\begin{pmatrix}\lambda&\lambda\\0&0\end{pmatrix} = \begin{pmatrix}\alpha&0\\0&0\end{pmatrix},

which implies

\lambda=\alpha=0

. Thus

\mathfrak a\cap eAe=0

. On the other hand,

exe=e,

so that

e\in e\mathfrak a e

. Hence

0=\mathfrak a\cap eAe\subsetneqq e\mathfrak a e=Ke.

(c) This follows from (b) and the additional hypothesis that $\mathfrak a$ is a two-sided ideal. Indeed, (b) gives

\mathfrak a\cap eAe\subset e\mathfrak a e.

x=eae\in e\mathfrak a e

, with

a\in\mathfrak a

, then

ae\in\mathfrak a

, because

\mathfrak a

is a right ideal, and therefore

eae\in\mathfrak a

, because

\mathfrak a

is also a left ideal. Thus

e\mathfrak a e\subset\mathfrak a\cap eAe

, and

\mathfrak a\cap eAe=e\mathfrak a e.

(d) Suppose first that $A$ is left Artinian. Let $(\mathfrak a_i)_{i\geqslant0}$ be a decreasing sequence of left ideals of $eAe$ . By (a), for each $i$ there is a left ideal $\mathfrak b_i=A\mathfrak a_i$ of $A$ such that

\mathfrak a_i=e\mathfrak b_i e.

Since

\mathfrak a_{i+1}\subset\mathfrak a_i

, we have

\mathfrak b_{i+1}=A\mathfrak a_{i+1}\subset A\mathfrak a_i=\mathfrak b_i.

By Definition 2 of § 1, a ring is left Artinian precisely when its left regular module

A_s

is Artinian; by Definition 1(ii), this means that every decreasing sequence of left ideals of

A

is stationary. Thus

(\mathfrak b_i)

is stationary, and consequently

(\mathfrak a_i)=(e\mathfrak b_i e)

is stationary. Hence

eAe

is left Artinian.

If $A$ is left Noetherian, the same argument with increasing sequences shows that $eAe$ is left Noetherian: by Definition 2, the left regular module $A_s$ is Noetherian, and by Definition 1(ii), every increasing sequence of left ideals of $A$ is stationary.

Replacing $e$ by the idempotent $1-e$ gives the same conclusions for $(1-e)A(1-e)$ . Thus, if $A$ is left Artinian (resp. left Noetherian), then both $eAe$ and $(1-e)A(1-e)$ are left Artinian (resp. left Noetherian).

(e) Let $V$ be an infinite-dimensional vector space over a commutative field $K$ , and let

A=K\times V\times K

with multiplication

(\lambda,x,\mu)(\lambda',x',\mu') =(\lambda\lambda',\,\lambda x'+\mu'x,\,\mu\mu').

Put

e=(1,0,0)

. Then

1-e=(0,0,1)

, and

eAe=\{(\lambda,0,0):\lambda\in K\},

with its induced multiplication, is canonically isomorphic to

K

. Likewise

(1-e)A(1-e)=\{(0,0,\mu):\mu\in K\},

and is canonically isomorphic to

K

. Thus both are commutative fields.

For every $K$ -linear subspace $W$ of $V$ , put

\mathfrak a_W=\{(0,x,0):x\in W\}.

(\lambda,x,\mu)\in A

and

(0,w,0)\in\mathfrak a_W

, then

(\lambda,x,\mu)(0,w,0)=(0,\lambda w,0)\in\mathfrak a_W

and

(0,w,0)(\lambda,x,\mu)=(0,\mu w,0)\in\mathfrak a_W.

Consequently

\mathfrak a_W

is a two-sided ideal of

A

for every subspace

W

V

Choose a sequence $(v_i)_{i\geqslant1}$ of linearly independent elements of $V$ , which exists because $V$ is infinite-dimensional. If

W_n=\operatorname{span}_K(v_1,\ldots,v_n),

then

\mathfrak a_{W_1}\subsetneqq\mathfrak a_{W_2}\subsetneqq\cdots

is an infinite strictly increasing sequence of two-sided ideals. If

W'_n=\operatorname{span}_K(v_n,v_{n+1},\ldots),

then

\mathfrak a_{W'_1}\supsetneqq\mathfrak a_{W'_2}\supsetneqq\cdots

is an infinite strictly decreasing sequence of two-sided ideals.

Cited by

Exercise 5, § 7

Cites

Nothing.