Is REQ a maximal left ideal with invertible matrix Q?

[xixi]

Let R = Mn(F) be the ring consists of all n*n matrices over a field F and
E = E11 + E22 + ... + En-1,n-1, where Eii is the elementary matrix (Eij is matrix whose ij th element is 1 and the others are 0) .
I know that RE is a maximal left ideal . Let Q be an invertible matrix . Can we say that REQ is a maximal left ideal ?

 

[Martin Rattigan]

E is the identity element so RE=R. MQ^{-1}Q=M, so REQ is also R, which wouldn't normally be called a maximal ideal because the definition of maximal ideal requires a proper ideal.

 

[xixi]

E is the identity element so RE=R. MQ^{-1}Q=M, so REQ is also R, which wouldn't normally be called a maximal ideal because the definition of maximal ideal requires a proper ideal.

Notice that E is not the identity element because its nth row and nth column are zero .

 

[Martin Rattigan]

Sorry - totally missed that. I though there was something awry.

 

[Martin Rattigan]

If R is any associative ring with 1 and q\in R has a multiplicative inverse, then \phi_q:R\rightarrow R defined by \phi_q:r\mapsto q^{-1}rq is a ring automorphism (with inverse \phi_{q^{-1}}).

If L\subset R is a maximal left ideal, then so is q^{-1}Lq. But then Lq=qq^{-1}Lq\subset q^{-1}Lq. Because L is a left ideal we have also q^{-1}Lq\subset Lq.

Hence Lq=q^{-1}Lq and therefore Lq is a maximal left ideal.

Using your assertion that RE in your question is a maximal left ideal and that Q is invertible, replacing L by RE and q by Q in the above gives the desired result.