# Answer to Question #25256 in Abstract Algebra for Tsit Lam

Question #25256

Which of the following implications are true? R left primitive ⇐⇒ Mn(R) left primitive.

Expert's answer

*Both implications here are true*.

In fact, assume

*S*: = M

*n*(

*R*)is left primitive (for some

*n*). For the idempotent

*e*=

*E*11(matrix unit),

*eSe*is also left primitive. Since

*eSe*=

*RE*

_{11}

*∼*

*R*(as rings)

*,*we conclude that

*R*is left primitive.Conversely, assume

*R*is left primitive, and fix a faithful simple left

*R*-module

*V*. Let

*U*be the group of column

*n*-tuples (

*v*1

*, .. . , vn*)

*, where*

^{T}*v*

_{i}*∈*

*V*. We make

*U*into a left

*S*-moduleby using “matrix multiplication” as the action. We are done if we can show that

*U*is a faithful simple

*S*-module. If (

*rij*)

*U*=0, wehave

*rijV*= 0, so

*rij*= 0, for all

*i, j*. To show that

*U*is simple, it suffices to show that, given any

*u*= (

*v*1

*, .. . , vn*)

*0 and*

^{T}<>*u'*= (

*v_*1

*, . .. , v_n*)

*T ,*

we have

*u’*

*∈*

*Su*. Say

*vi <>*0. Then there exist

*r*1

*,. . . , rn*

*∈*

*R*such that

*v_j*=

*rjvi*forall

*j*. For the matrix

*σ*

*∈*

*S*with

*i*thcolumn (

*r*1

*, . . . , rn*)

*and other columnszero, we have clearly*

^{T }*σ*

*u*=

*u'*.

Alternatively, we can use thecharacterization of left primitive rings :“

*R*is left primitive iff thereexists a left ideal A

*<>*

*R*which is comaximal with anynonzero ideal B.” Fix such a left ideal A in

*R*. Any nonzero ideal in

*S*has the form M

*n*(B) for some ideal B in

*R*. Since A + B =

*R*, the left ideal M

*n*(A) is clearly comaximal with M

*n*(B).

Therefore,

*S*is leftprimitive.

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