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Show that $Z(M_n(R))$ consist of $ \operatorname{diag}\{ a, a, ..., a \} $ with $ a\in Z(R) $

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marked as duplicate by rschwieb abstract-algebra Feb 5 '15 at 20:32

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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 $a\in Z(R)$.

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  • $\begingroup$ This answer assumes that $R$ contains 1, which we are not given. $\endgroup$ – Ceph Nov 14 '17 at 20:23
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    $\begingroup$ Fair point. I guess if it's just a rng the claim isn't always true. One could take $R$ the rng of square zero, then $M_n(R)$ also has zero multiplication, so is commutative. But then the center is not just the diagonal matrices. $\endgroup$ – Ben West Nov 14 '17 at 20:53
  • $\begingroup$ Interesting. I thought the claim was true but merely a bit more tedious to prove, but I see now you're right, we must have 1. $\endgroup$ – Ceph Nov 14 '17 at 20:56
  • $\begingroup$ I see the argument with $E_{ii}$, but not yet the one with $E_{ij}$. $\endgroup$ – Jos van Nieuwman Feb 26 at 21:38
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    $\begingroup$ @JosvanNieuwman You can use the formula for finding the entries of a product of matrices. Let $B=AE_{ij}$. Then the $(k,l)$ entry of $B$ is $b_{kl}=\sum_{s=1}^na_{ks}e_{sl}=a_{kk}e_{kl}$ since $A$ is diagonal. In particular, when $k=i$ and $l=j$, one concludes $b_{ij}=a_{ii}e_{ij}=a_{ii}$. Similarly, since $B=E_{ij}A$, the same argument shows that $b_{ij}=a_{jj}$, so $a_{ii}=a_{jj}$. $\endgroup$ – Ben West Feb 26 at 21:57

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