# Centre of a matrix ring are diagonal matrices [duplicate]

Show that $Z(M_n(R))$ consist of diagonal matrices with entries $Z(R)$.

-

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 5 '15 at 20:32

We do not like taking orders. Instead, please indicate how you came across this problem; why it interests you; what you already know about it, and about the concepts mentioned in it; how far you got in your efforts to solve it; where you got stuck; and so on, and so on. Engage with us, so we can engage with you. – Gerry Myerson Jan 22 '13 at 6:32
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Jan 22 '13 at 6:35
For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Jan 22 '13 at 6:37

Here are some things to think about which should put you in the right direction.

Suppose $A=(a_{ij})\in Z(M_n(R))$. Let $E_{ij}$ be the matrix whose $i,j$ entry is $1$, and all other entries are $0$. Then the equations $$E_{ii}A=AE_{ii}$$ for $1\leq i\leq n$ implies that $A$ is necessarily diagonal. (Why?) Furthermore, $$AE_{ij}=E_{ij}A$$ for $1\leq i,j\leq n$ implies that $a_{ii}=a_{jj}$ for all $i$ and $j$. (Why?) Hence $A=aI_n$ for some $a\in R$. But notice that $$aI_n(bI_n)=bI_n(aI_n),\quad \forall b\in R$$ implies that $a\in Z(R)$.

-