# Answer on Algebra Question for Mohammad

Question #17363

If R is a commutative principal ideal domain, show that the von Neumann regular elements of Mn(R) are precisely matrices of the form UAV , where U, V ∈ GLn(R), and A = diag(1, . . . , 1, 0, . . . , 0).

Expert's answer

First observe that, if

diag(

and so

*u*,*v*are units in any ring*S*, then*an element a**∈**S is von Neumann regular iff uav is*. (To see this, it suffices to provethe “only if” part. Suppose*a*=*asa*for some*s**∈**S*; we call any such*s*a “*pseudo-inverse*”of*a*. To find a pseudo-inverse*t*for*uav*, we need to solvethe equation:*uav*= (*uav*)*t*(*uav*). This amounts to*a*=*a*(*vtu*)*a*, so it suffices to choose*t*=*v−*1*su−*1,and*all pseudo-inverses for uav arise in this way*.) Applying this remarkto M*(*_{n}*R*), it follows that, for*U*,*V*,*A*,*UAV*is von Neumann regular (since*A*=*A*^{2}=*A*^{3}is). Conversely, let*B*be any von Neumann regular element in M*(*_{n}*R*).By the Smith Normal Form Theorem, there exist*P*,*Q**∈**GLn(**R*) such that*PBQ*= diag(*b*1*, . . . , bn*)*,*where*b*_{i}_{+1}*∈**b*for each_{i}R*i < n*. By the remark above,diag(*b*1*, . . . , bn*) is von Neumann regular. Thus, upon interpretingM*(*_{n}*R*) as End*R*(*R*), im(diag(^{n}*b*1*,. . . , bn*)) =*b*1*R**⊕**· · ·**⊕**bnR**⊆**Rn*must be a direct summand of*R*Thisis possible only if each^{n}.*b*is either 0 or_{i}R*R*. Sincethe “elementary divisors”*bi*are determined up to units anyway, we mayassume thatdiag(

*b*1*, . . . , bn*) =diag(1*, . . . ,*1*,*0*, . . . ,*0) :=*A,*and so

*B*=*UAV*for*U*=*P*^{−}^{1}and*V*=*Q*^{−}^{1}in GL_{n}(*R*).Need a fast expert's response?

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