Maximal left ideal or not?

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  • #1
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Main Question or Discussion Point

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 ?
 

Answers and Replies

  • #2
[itex]E[/itex] is the identity element so [itex]RE=R[/itex]. [itex]MQ^{-1}Q=M[/itex], so [itex]REQ[/itex] is also [itex]R[/itex], which wouldn't normally be called a maximal ideal because the definition of maximal ideal requires a proper ideal.
 
  • #3
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[itex]E[/itex] is the identity element so [itex]RE=R[/itex]. [itex]MQ^{-1}Q=M[/itex], so [itex]REQ[/itex] is also [itex]R[/itex], 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 .
 
  • #4
Sorry - totally missed that. I though there was something awry.
 
  • #5
If [itex]R[/itex] is any associative ring with [itex]1[/itex] and [itex]q\in R[/itex] has a multiplicative inverse, then [itex]\phi_q:R\rightarrow R[/itex] defined by [itex]\phi_q:r\mapsto q^{-1}rq[/itex] is a ring automorphism (with inverse [itex]\phi_{q^{-1}}[/itex]).

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

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

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

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