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 *= *M*1 *⊕**· · ·**⊕**Mr*

be the decomposition of*M *intoits isotypic components. Let

*R *= *R*1 *⊕**· · ·**⊕**Rr,*

where*Ri *is the simplecomponent of *R *corresponding to *Mi*. Since *RjMi *=0 for *j<>* *i*, we have *μ*_{R}(*Mi*) = *μ*_{Ri}* *(*Mi*). We now accomplish the desiredreduction by proving that

*μ*(*M*) = max *{**μ*_{R}_{1} (*M*1)*, . . . , **μ*_{Rr}* *(*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 *= *μ*_{R}(*M*1), *m *= *μ*_{R}(*M*2), with *n ≥ m*. Let *{x*1*,. . . , xn} *be generators for *M*1, and *{y*1*, . . . , yn} *begenerators for *M*2. We finish by showing that *x*1 + *y*1*, .. . , xn *+ *yn *generate *M *= *M*1 *⊕**M*2. Indeed, for (*x, y*) *∈** M*1 *⊕**M*2, write *x*= (sum) *α*_{i}*x*_{i}(*α**i **∈** R*1)*, y*= (sum) *β*_{i}*y*_{i} (*β**i **∈** R*2)*. *Then,in the module *M*:

(sum)(*α**i *+ *β**i*)(*xi *+ *yi*) =(sum) *α**ixi *+(sum) *β**iyi *= *x *+ *y, *so *μ*(*M*) *≤ n*, as desired.

be the decomposition of

where

The inequality “

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