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Answer on Algebra Question for Hym@n B@ss

Question #17187
Generalize the computation of μ(M) to the case of a finitely generated left module M over a semisimple ring R.
Expert's answer
We shall solve by reducing thecalculation of μ(M)to the case of modules over artinian simple rings. Let
M = M1 ⊕· · ·⊕Mr
be the decomposition of M intoits isotypic components. Let
R = R1 ⊕· · ·⊕Rr,
where Ri is the simplecomponent of R corresponding to Mi. Since RjMi =0 for j<> i, we have μR(Mi) = μRi (Mi). We now accomplish the desiredreduction by proving that
μ(M) = max {μR1 (M1), . . . , μRr (Mr)} .
The inequality “≥” is easy,since each Mi may be viewed as an epimorphic image of M. To provethe inequality “≤”, let us assume (in order to simplify the notation)that r = 2 .Say n = μR(M1), m = μR(M2), with n ≥ m. Let {x1,. . . , xn} be generators for M1, and {y1, . . . , yn} begenerators for M2. We finish by showing that x1 + y1, .. . , xn + yn generate M = M1 ⊕M2. Indeed, for (x, y) ∈ M1 ⊕M2, write x= (sum) αixi(αi ∈ R1), y= (sum) βiyi (βi ∈ R2). Then,in the module M:
(sum)(αi + βi)(xi + yi) =(sum) αixi +(sum) βiyi = x + y, so μ(M) ≤ n, as desired.

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