Central elements as natural endomorphisms of the identity

(${\eta }^z$ is natural.) Let $z\in Z(A)$ and let $f:X\to Y$ be $A$-linear. The map ${\eta }_X^z$ is additive and $A$-linear since ${\eta }_X^z(ax)=a(zx)=(az)x=z(ax)$ for $a\in A$, using centrality; and ${\eta }_Y^z(f(x))=zf(x)=f(zx)=f({\eta }_X^z(x))$, so the naturality squares commute. Hence ${\eta }^z$ is a natural endomorphism of the identity.

(Center of a matrix ring.) Let $B=M_n(k)$ and let $C={\sum }_{p,q}{c}_{pq}{E}_{pq}\in B$ commute with every $E_{ij}$. Then $C{E}_{ij}={\sum }_p{c}_{pi}{E}_{pj}$ and $E_{ij}C={\sum }_q{c}_{jq}{E}_{iq}$ by [F3]; comparing the $(p,q)$-entries gives ${\delta }_{qj}{c}_{pi}={\delta }_{pi}{c}_{jq}$ for all $i,j,p,q$. Taking $p\mathrm{\ne }i$, $q=j$ gives $c_{pi}=0$, and taking $p=i$, $q\mathrm{\ne }j$ gives $c_{jq}=0$, so $C$ is diagonal; taking $p=i$, $q=j$ gives $c_{ii}=c_{jj}$ for all $i,j$, so all diagonal entries are equal. Hence $C=\lambda I_n$ is scalar, and every scalar matrix is central; thus $Z(M_n(k))=\{\lambda I_n:\lambda \in k\}\cong k$.

(Converse, uniqueness, and composition.) By [F2] every natural endomorphism $\eta$ of the identity has $\eta ={\eta }^z$ for the unique central element $z={\eta }_A(1)$, and conversely every central element arises this way; explicitly, naturality at the left $A$-linear map ${\mathrm{\ell }}_x:A\to X$, $a\mapsto ax$, gives ${\eta }_X(x)={\mathrm{\ell }}_x({\eta }_A(1))=zx$. For central $z,z^{\mathrm{\prime }}$ one has ${\eta }_X^{z{z}^{\mathrm{\prime }}}(x)=(z{z}^{\mathrm{\prime }})x=z(z^{\mathrm{\prime }}x)=({\eta }_X^z\circ {\eta }_X^{z^{\mathrm{\prime }}})(x)$, so ${\eta }^{z{z}^{\mathrm{\prime }}}={\eta }^z\circ {\eta }^{z^{\mathrm{\prime }}}$ componentwise. Therefore the bijection $z\mapsto {\eta }^z$ is a ring isomorphism $Z(A)\cong \mathrm{Nat}(1_{A\text{-}\mathrm{M}\mathrm{o}\mathrm{d}},1_{A\text{-}\mathrm{M}\mathrm{o}\mathrm{d}})$.

(Commutative rings.) If $A$ is commutative, then $Z(A)=A$ by [F1], so by step 2.1 the natural endomorphisms of the identity of $A\text{-Mod}$ are exactly the maps $x\mapsto zx$ for elements $z\in A$.

(Matrix rings.) For $A=M_n(k)$ step 1.2 identifies $Z(A)$ with the scalar matrices, so by step 2.1 the natural endomorphisms of the identity of $M_n(k)\text{-Mod}$ are exactly the multiplications by scalar matrices, i.e. the scalars, and this is the special case of the Morita-invariance statement The center is Morita invariant, via natural endomorphisms of the identity.

Steps 1.1 and 2.1 verify naturality, uniqueness, composition and the ring identification, step 1.2 computes the center in the matrix case, and steps 3.1 and 3.2 record the two announced specializations; the bijection is the one of The center is Morita invariant, via natural endomorphisms of the identity, and no choice is used.