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Conversely, suppose J(R) = {0}. Let {Mi | i ∈ I} be a set of twoby-two non-isomorphic simple left R-modules so that this set contains a representative of every isomorphism class of the simple left R-modules. Then, M = i∈I Mi is a semisimple left R-module and annR (M ) = ∩i∈I annR (Mi ) = J(R) = {0}. So, M is a faithful simple left R-module. 2 A ring R is left (respectively right) primitive if R has a faithful simple left (respectively right) module. There exists examples (due to Bergman and Jategaonkar) of left primitive rings that are not right primitive.

Further, because of Hamilton’s theorem we know that g is a root of its characteristic polynomial P (recall that we consider g as an element in Mrn (k0 )). So mg (X)|P (X) and hence deg(mg (X)) ≤ deg(P (X)) = rn. Now, one can consider the coefficients of mg (X) as elementary symmetric functions in the roots of mg (X). 6. LINEAR GROUPS AND THE BURNSIDE PROBLEM 43 roots of unity in C, the norm of these elementary functions (in the roots) is bounded by 2rn . Hence the coefficients of mg (X) ∈ Z[X] are bounded by the number 2rn .

To prove part two, assume R is left primitive. 1, J(R) = {0}. Let M be a faithful simple left R-module. Then, for every non-zero ideal I of R, the set IM is a non-zero submodule of M . As M is simple we get IM = M . In particular, for any two non-zero ideals J and I of M , IJM = IM = M. Hence IJ = {0}. So we get that R is prime. ✷ So we have introduced several classes of rings that are linked as follows: semisimple ⇒ semiprimitive ⇒ semiprime ⇑ ⇑ simple ⇒ left primitive ⇒ prime In general, none of the converse implications is valid.

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Associative algebra by Eric Jespers

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