Exercise 2, § 1

Tag 000Y, permanent. Cite it as [Bourbaki, Tag 000Y].

Algebra, VIII, p. 14

Let A be a ring, M an A-module of finite length $n$ , and R a set of nilpotent endomorphisms of M that is stable under composition.

a) If $n= 1$ , then every element of R is zero. If $n >1$ , prove that there exists a submodule N of M, distinct from 0 and M, that is stable under all elements of R. (We may assume that R has an element $s\not= 0$ ; choose one such that $s(M)$ has mimimum length, and prove that we have $srs= 0$ for every $r\in R$ . Deduce that we can take $N =s(M)$ if $Rs=\{0\}$ and $N =\sum_{r\in R}r(s(M))$ if $Rs\not=\{0\}.$ )[^1]

b) Deduce from a) that there exists a Jordan–Hölder series $(M_i)_{0\leqslant i\leqslant n}$ of M such that we have $r(M_i)\subset M_{i+1}$ for all $r\in R$ and $0\leqslant i\leqslant n-1$ . Deduce that we have $r_1\cdots r_n= 0$ for every sequence $(r_i)_{1\leqslant i\leqslant n}$ of $n$ elements of R.

[^1]: This proof was communicated to us by A. Rosenberg.

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.

Part	Status	Reason
a	verified
b	verified

(a) Suppose first that $n=1$ . If $r\in R$ is non-zero, then $r(M)$ is a non-zero submodule of $M$ and is therefore equal to $M$ . Thus $r$ is surjective. Since $r$ is nilpotent, this is impossible. Hence every element of $R$ is zero.

Suppose now that $n>1$ . If every element of $R$ is zero, any submodule of $M$ distinct from $0$ and $M$ will do. We may therefore suppose that there exists an element $s\in R$ with $s\neq0$ . Choose such an element $s$ for which the length of $s(M)$ is minimum.

For every $r\in R$ , the endomorphism $srs$ belongs to $R$ . We shall prove that $srs=0$ . The image of $srs$ is contained in $s(M)$ . If $srs\neq0$ , then the image of $srs$ is a non-zero submodule of $s(M)$ . If it were equal to $s(M)$ , then

srs(M)=s(M).

Put

S=s(M)

. The equality above gives

S=s(r(S)).

The module

r(S)

is an image of the finite length module

S

, and hence

\operatorname{length}(r(S))\leqslant \operatorname{length}(S).

On the other hand, the restriction of

s

r(S)

is a surjective map onto

S

, so

\operatorname{length}(S)\leqslant \operatorname{length}(r(S)).

Consequently these lengths are equal, and the restriction of

s

r(S)

is an isomorphism. In particular,

(sr)(S)=s(r(S))=S.

But

sr\in R

, and

sr

is nilpotent. Its restriction to the finite length module

S

is therefore a surjective nilpotent endomorphism, which is impossible. Hence

srs(M)\neq s(M)

It follows that, if $srs\neq0$ , then $srs(M)$ is a non-zero proper submodule of $s(M)$ . Therefore

\operatorname{length}((srs)(M))<\operatorname{length}(s(M)).

This contradicts the choice of

s

. Therefore

srs=0

for every

r\in R

If $Rs=\{0\}$ , we take $N=s(M)$ . This submodule is non-zero, since $s\neq0$ . It is distinct from $M$ , since otherwise $s$ would be surjective, which is impossible for a non-zero nilpotent endomorphism. Moreover, for every $r\in R$ we have

r(N)=r(s(M))=0,

so that

N

is stable under all elements of

R

Assume now that $Rs\neq\{0\}$ and put

N=\sum_{r\in R}r(s(M)).

This submodule is stable under all elements of

R

, because for

r,r'\in R

one has

r'(r(s(M)))=(r'r)(s(M))\subset N.

It is non-zero because

Rs\neq\{0\}

. Finally, for every

r\in R

we have

s(r(s(M)))=srs(M)=0,

and therefore

N\subset\ker s.

Since

s\neq0

, we have

\ker s\neq M

, and hence

N\neq M

. Thus in all cases there exists a submodule

N

distinct from

0

and

M

which is stable under all elements of

R

(b) We prove the assertion by induction on the length $n$ of $M$ . If $n=1$ , the assertion follows from (a), since every element of $R$ is zero.

Suppose that $n>1$ . By (a), there exists a proper non-zero submodule $N$ of $M$ stable under all elements of $R$ . Let $m$ be the length of $N$ . The endomorphisms induced by the elements of $R$ on $M/N$ are again nilpotent endomorphisms, and the set of these induced endomorphisms is stable under composition. By the induction hypothesis applied to $M/N$ , there exists a Jordan--Hölder series

M/N=\overline{M}_0\supset \overline{M}_1\supset\cdots\supset \overline{M}_{n-m}=0

such that the induced endomorphism defined by every

r\in R

sends

\overline{M}_i

into

\overline{M}_{i+1}

Taking inverse images in $M$ , we obtain a series

M\supset M_1\supset\cdots\supset M_{n-m}=N

with the same property. Applying the induction hypothesis to

N

, we obtain a Jordan--Hölder series

N=M_{n-m}\supset M_{n-m+1}\supset\cdots\supset M_n=0

such that

r(M_i)\subset M_{i+1}

for every

r\in R

. Joining the two series gives a Jordan--Hölder series

M=M_0\supset M_1\supset\cdots\supset M_n=0

satisfying

r(M_i)\subset M_{i+1}

for every

r\in R

and

0\leqslant i\leqslant n-1

Let now $(r_i)_{1\leqslant i\leqslant n}$ be a sequence of elements of $R$ . Since

r_n(M_0)\subset M_1,\quad r_{n-1}r_n(M_0)\subset M_2,\quad\ldots,\quad r_1\cdots r_n(M_0)\subset M_n,

we obtain

r_1\cdots r_n(M)=0.

Hence

r_1\cdots r_n=0.

Cited by

Exercise 10, § 10

Cites

Nothing.