Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

marked as duplicate by rschwieb Feb 5 at 20:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

6  
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
1  
Since you are new to this site, please consider reading this: How to ask a homework question?. In particular, you should use homework tag if your question comes from a homework. I wrote this comment because the question sounds homework-like. –  Julian Kuelshammer Jan 22 '13 at 6:34
1  
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
1  
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

1 Answer 1

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 $a\in Z(R)$.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.