# Answer to Question #17187 in Algebra 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

be the decomposition of

where

The inequality “

(sum)(

*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.Need a fast expert's response?

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