Manish Kumar Singh

ohh i see thanks for helping me out, I have to rewire my brain :-(

so the error from my part was it might not exist but if it did then unique

*might not

ohh does it mean if it exist, it might exist?

arent we able to decompose it in terms of ideal?

@arctictern I understand your point, even I had my doubts about having same structure (and hence the question) but then I couldn't make sense of lemma 3.8. I wish you could read that (I think you should recheck the link ; pg 33 and 34 are not in preview but pg 35 is atleat in my country)

I read the pg 35 lemma 3.8 of the link and understood the following, would you re read and tell me what it means from your point of view

books.google.co.in/…

you could have a look at this link

@arctictern because left semisimple iff right semisimple,

because left iff right

but should they be same

No I am not, to arrive at a contradiction to your claim (of having multiple component)

and hence becomes ideal

because left semisimple implies right semisimple

both

yes

contradiciton because Mn(D) is simple because D is division ring

@arctictern Suppose Mn(D) has two simple components U1 and U2 ie Mn(D)= U1 direct sum U2. Now Mn(D) is both left and right semisimple, this implies U1 and U2 are ideals which should be simple. This Forces Mn(D) not to be simple which is a contradiction. hence it should have only 1 component. (Now could you find a flaw in this)

@arctictern : I will rephrase it again just give me 2 minutes

@arctictern If that's true then would u find a flaw in my argument mentioned; (which i repeared previously) Since you agree that Mn(D) is left semisimple, by wedderburn it also implies that its right semisimple. So r_R = U1 circle+ U2 = R_r, (where R_r and r_R is right and left semisimple ring) then it implies U1 and U2 are simple ideal which implies R is not simple-a contradiction, hence r_R has only one component. (Can you find a flaw in my argument if you believe its wrong)

@EricStucky I think it does if you see corollary 3.7 on next page

@EricStucky Why I thought that's the whole point of Weddernburn theorem (left iff right)

@EricStucky Since you agree that Mn(D) is left semisimple, by wedderburn it also implies that its right semisimple. So r_R = U1 circle+ U2 = R_r, (where R_r and r_R is right and left semisimple ring) then it implies U1 and U2 are simple ideal which implies R is not simple-a contradiction, hence r_R has only one component. (Can you find a flaw in my argument if you believe its wrong)

@EricStucky: Okay think I can counter that, just give me 2 minutes

@EricStucky Ok I think I am little confused, so are you suggesting that Mn(D) when seen as left semisimple ring has more than one simple component.

(just look down the page)

because its semisimple

@EricStucky I mean to say left ideal is Mn(D)

@ EricStucky Also If you scroll down a bit on that page you would discover that Mn(D) is also left semisimple (with one component) and hence should not have any left submodule. The example you suggested is not submodule because of matrix multiplication (am i correct on this)

@EricStucky Where is Mn(D)*r*Mn(D) comming from, am I missing something?

@EricStucky books.google.co.in/… (SEE THIS LINK FROM THE BOOK I AM READING)

@EricStucky Are you saying Mn(D) not simple? If yes then the prove is Ideal of Mn(D) are in 1-1 correspondance with ideal of D.

@EricStucky:(in reply to your 1st comment) In TY LAM its stated that Mn(D) is simple, left semisimple etc. This is used to prove Wedderburn theorem. I included Weddernburn theorem to give you a context about the question, I didn't mean to say it comes from that.

@EricStucky : I tried but couldn't understand where I am going wrong.

let me know about some clarification

R = Mn(D) where D is a division ring then we know that R is left semi simple with one component (see weddernburn structure theorem). Now if I choose a random element 'r' in R, then the left ideal generated by it is whole of R. so I can say 'r' has a left inverse. But I know for example E_(1,1) ie 1 at (1,1) and 0 elsewhere is non invertible. So there is contradiction and hence some mistake in my reasoning, so can anyone point out it to me. (sorry couldn't find a way to use latex in chat)

ok guys give me 5 minutes to post it properly and clearly

ok so if Mn(D) is left semisimple with one component (basically left simple), the my assertion Mn(D) should be division ring because and hence Ra = R and hence I can find left inverse. But If you consider E11 (ie 1 at (1,1) and 0 everywhere matrix) it has got no inverse. So I know I am making a mistake but dont know where exactly

few typos

ok so if Mn(D) is left semisimple with one component (basically left simple), the my assertion Mn(D) should be division ring because and Ra = R and hence I can find left inverse. But If you consider E11 (ie 1 at (1,1) and 0 everywhere matrix) it has got no inverse. So I now I am making a mistake but dont know where exactly

could some1 ask it from my behalf

yes every1 of them is because i am currently studying, so naturally i will always have doubts (sometimes silly)

which one

possibly beacuse they dont find my question of high quality

problem is my account wont accept new questions

yes are you intreseted

do you know where should I ask such questions?

yes

Matrix with elements from D