Answer to Question #17189 in Algebra for Hym@n B@ss

Question

Show that N ⊆ M ⇒ μ(N) ≤ μ(M).

Answer 1

We can use fact that μ(M) can be defined for anyfinitely generated left module M over a semisimple ring R. Thenwe can reduce to the case of modules over artinian simple rings. In this case,
the inequality μ(N)≤ μ(M)follows immediately from the formula μ(M) = [l(M)/n],since N ⊆ M implies l(N) ≤ l(M).Alternatively, since R is semisimple, N ⊆ M implies that there exists an R-epimorphismM → N, which gives μ(N) ≤ μ(M) right away.