Exercise 5, § 1

This solution was written by gpt-5-6-mini and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

For a left ideal $L$ of $\mathbf M_n(A)$, let $N_i$ be the set of row vectors occurring as the $i$-th row of matrices belonging to $L$. Since multiplication on the left by the matrix units $e_{ji}$ carries the $i$-th row onto the $j$-th row and annihilates all the other rows, we have $N_i=N_j$ for all $i,j$. Moreover, each $N_i$ is a left $A$-submodule of the left $A$-module $A^n$. Thus, writing $N=N_1$, we have

It follows from Definition 2 that $\mathbf M_n(A)$ is left Artinian (resp. left Noetherian) if and only if the left $A$-module $A^n$ is Artinian (resp. Noetherian). If $A$ is left Artinian (resp. left Noetherian), then $A^n$ has the same property by Example 5 of §1. Conversely, if $A^n$ is Artinian (resp. Noetherian), its direct summand $A$ is Artinian (resp. Noetherian), by Proposition 3 of §1 applied to

Finally suppose that $A$ is left Artinian. By Theorem 1 of §1, $A$ has finite length. The inclusion-preserving bijection above identifies the lattices of submodules of the left $A$-module $A^n$ and of left ideals of $\mathbf M_n(A)$; hence the two modules have the same length. By additivity of length for finite direct sums,