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

Question #17189

Show that N ⊆ M ⇒ μ(N) ≤ μ(M).

Expert's answer

We can use fact that

the inequality

*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.Need a fast expert's response?

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