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.