# Centre of a matrix ring are $\operatorname{diag}\{ a, a, …, a \}$ with $a\in Z(R)$ [duplicate]

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

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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 $a\in Z(R)$.
• This answer assumes that $R$ contains 1, which we are not given. – Ceph Nov 14 '17 at 20:23
• 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. – Ben West Nov 14 '17 at 20:53
• I see the argument with $E_{ii}$, but not yet the one with $E_{ij}$. – Jos van Nieuwman Feb 26 at 21:38
• @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 $a_{ii}=a_{jj}$. – Ben West Feb 26 at 21:57