Morita Bicategories and Projective Generators — Examples

(The four subspaces.) Since $b{E}_{11}$ is the matrix whose first column is the first column of $b$ and whose other columns vanish, $Be=\{{\sum }_i{a}_i{E}_{i1}:a_i\in k\}$; dually $eB=\{{\sum }_j{c}_j{E}_{1j}:c_j\in k\}$, and $eBe=k{E}_{11}$ with $e=E_{11}$ as identity, so $A=k{E}_{11}\cong k$ as a field, the isomorphism being $\lambda \mapsto \lambda e$. The products land where claimed because $E_{1j}{E}_{i1}={\delta }_{ji}{E}_{11}$ and $E_{i1}{E}_{1j}=E_{ij}$ by [F1].

(The bimodule structures.) Left multiplication by $B$ and right multiplication by $A\subseteq B$ make $M=Be$ a $(B,A)$-bimodule: both actions are $k$-linear, and associativity of matrix multiplication gives $b(ma)=(bm)a$ for $b\in B$, $m\in Be$, $a\in A$, with $A=ke$ acting by scalar multiplication by [F2]. Symmetrically $N=eB$ is an $(A,B)$-bimodule.

(${\mu }_N$.) The pairing $(x,y)\mapsto xy$ from $eB\times Be$ to $eBe$ is $k$-bilinear and $B$-balanced: $((xb)y)=x(by)$ by associativity for $b\in B$; by [F3] it descends to a homomorphism ${\mu }_N$ with ${\mu }_N(x\otimes y)=xy$, which is a map of $(A,A)$-bimodules because both the product and the tensor actions are induced from the two factors. The map $eBe\to eB{\otimes }_{B}Be$, $a\mapsto e\otimes a$, is well defined, and for $x\in eB$, $y\in Be$ one has $x=ex$ and hence $x\otimes y=(ex)\otimes y=e\otimes (xy)$ by $B$-balance, so the two composites are the identities: ${\mu }_N(e\otimes a)=a$ for $a\in eBe$ and $e\otimes xy=x\otimes y$. Hence ${\mu }_N$ is an isomorphism of $(A,A)$-bimodules.

(${\mu }_M$.) The pairing $(m,n)\mapsto mn$ from $Be\times eB$ to $B$ is $k$-bilinear and balanced over $A=k{E}_{11}$: $(m{E}_{11})n=m(E_{11}n)$ is a case of associativity, and scalar balancing holds because $A=k$ acts as scalars by [F2]. By [F3] it descends to a homomorphism ${\mu }_M$ with ${\mu }_M(m\otimes n)=mn$, a map of $(B,B)$-bimodules. Define the $k$-linear inverse on the basis $\{E_{ij}\}$ by $E_{ij}\mapsto E_{i1}\otimes E_{1j}$; this is well defined because the matrix units form a $k$-basis of $B$, and ${\mu }_M(E_{i1}\otimes E_{1j})=E_{i1}{E}_{1j}=E_{ij}$. Conversely, writing $m={\sum }_i{a}_i{E}_{i1}$ and $n={\sum }_j{c}_j{E}_{1j}$ one has $m\otimes n={\sum }_{i,j}{a}_i{c}_j{E}_{i1}\otimes E_{1j}$ and $mn={\sum }_{i,j}{a}_i{c}_j{E}_{ij}$, so the two composites are the identities on a spanning set and hence everywhere. Thus ${\mu }_M$ is an isomorphism of $(B,B)$-bimodules.

(Conclusion.) Step 3.1 gives the bimodule isomorphism $N{\otimes }_{B}M\cong eBe=A$ and step 3.2 gives $M{\otimes }_{A}N\cong B$, so by [F4] the rings $A\cong k$ and $B=M_n(k)$ are Morita equivalent with inverse pair of bimodules $(M,N)$; the explicit inverses are $a\mapsto e\otimes a$ and $E_{ij}\mapsto E_{i1}\otimes E_{1j}$ extended $k$-linearly, and the only elements used are the fixed matrix units, so no choice is used.

($P$ is projective and a generator.) As the free $k$-module on $\mathbb{N}$, $P$ is projective by [F1] under the declared Axiom of Choice. For generation, let $u\mathrm{\ne }v:X\to Y$ be distinct $k$-linear maps and pick $x\in X$ with $(u-v)(x)\mathrm{\ne }0$; the family with $f_0:k\to X$, $1\mapsto x$, and $f_n=0$ for $n\ge 1$ extends by [F2] to a $k$-linear $h:P\to X$ with $h(e_0)=x$ and $h(e_n)=0$ for $n\ge 1$. Then $(u-v)h\mathrm{\ne }0$, so $uh\mathrm{\ne }vh$, and by [F5] the object $P$ is a generator.

(Every map in the comparison image has finite-dimensional range.) An element of the source ${⨁}_n{\mathrm{Hom}}_k(P,k)$ is a family $({\phi }_n)$ with finite support by [F2], and by the universal property of the direct sum its image under the canonical comparison is the map $c(\phi ):x\mapsto {\sum }_n{\phi }_n(x)e_n$ with ${\phi }_n=0$ off a finite set $F$. For $x\in P$ one has $c(\phi )(x)\in {⨁}_{n\in F}k{e}_n$, a finite-dimensional subspace of $P$; hence $\mathrm{im}c(\phi )\subseteq {⨁}_{n\in F}k{e}_n$.

(${\mathrm{id}}_P$ is not in that image.) Suppose ${\mathrm{id}}_P=c(\phi )$ for some finite-support family $({\phi }_n)$ with support $F$. Then every basis vector $e_m$ with $m\mathrm{\notin }F$ would lie in $\mathrm{im}c(\phi )\subseteq {⨁}_{n\in F}k{e}_n$, so $e_m$ would be a finite $k$-linear combination of the finitely many vectors $e_n$, $n\in F$, contradicting the linear independence of the basis recorded in [F4]. Hence ${\mathrm{id}}_P$ is not in the image of the canonical comparison, and that comparison is not surjective.

(Conclusion.) By steps 1.1-2.1 the module $P$ is projective and a generator, but the canonical comparison ${⨁}_n{\mathrm{Hom}}_k(P,k)\to {\mathrm{Hom}}_k(P,P)$ fails to be surjective, so ${\mathrm{Hom}}_k(P,-)$ does not preserve the coproduct ${⨁}_{n}k$ and $P$ is not a small projective generator by [F6] and Small projective generators and progenerators. Equivalently $P$ is not finitely generated, in agreement with Small projective modules are exactly finitely generated projective modules; the progenerator identification. Thus "projective generator" cannot replace "small projective generator", and the only use of choice is the projectivity from [F1], so the failure is not choice-free.

(${\eta }^z$ 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.