Question #17189

Show that N ⊆ M ⇒ μ(N) ≤ μ(M).

Expert's answer

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*-epimorphism*M → N*, which gives *μ*(*N*) *≤ **μ*(*M*) right away.

the inequality

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