Finite Abelian Categories and Eilenberg–Watts — Examples

By [L1] the quotient $S=A\mathrm{/}(\epsilon )$ is one-dimensional over $k$ with $\epsilon S=0$ and $S\cong k$, and by [L3] the $A$-submodules of $S$ are its $k$-subspaces, so $S\mathrm{\ne }0$ is a simple $A$-module.

The space $k^{\mathbb{N}}$ is infinite-dimensional: its vectors $e_n$ of finite support are linearly independent, so if it had a finite basis with $N$ elements then the $N+1$ vectors $e_0,\dots ,e_N$ would contradict the bound of [L6].

(ii) By [L7] the functor $T_S$ is $k$-linear and right exact, has the right adjoint ${\mathrm{Hom}}_A(S,-)$ which preserves finite-dimensional modules, and has kernel $T_S(A)\cong S$; the classification of [L7] uses finite free presentations of finite-dimensional modules and only the finite biproducts of $A\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$, so no infinite coproduct is formed.

Suppose a coproduct $X$ of the countably many copies of $A$ existed in $A\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$. Taking $Y=S$ in the universal property of [L5] gives a bijection ${\mathrm{Hom}}_A(X,S)\cong {\prod }_{n\in \mathbb{N}}{\mathrm{Hom}}_A(A,S)$, and ${\mathrm{Hom}}_A(A,S)\cong S$ as $k$-vector spaces by evaluation at $1$ from [L4], The comparison is $k$-linear, since it sends $f$ to $(f{\mathrm{ȷ}}_n{)}_n$ and therefore preserves pointwise addition and scalar multiplication. Hence ${\mathrm{Hom}}_A(X,S)\cong S^{\mathbb{N}}\cong k^{\mathbb{N}}$ as $k$-vector spaces by step 1.1.

But ${\mathrm{Hom}}_A(X,S)$ for objects $X,S$ of $A\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$ is finite-dimensional by [L2] and [L6]. This contradicts step 2.1 together with step 1.2, so no such coproduct exists; in particular $A\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$ is not cocomplete, so it does not satisfy the cocompleteness hypothesis of the arbitrary-ring setting.

The finite classification [L7] applies to $T_S$ without any additional hypothesis about infinite coproducts, despite the nonexistence of the countable coproduct in step 3.1. The arbitrary-ring theorem [L8] concerns the category of all modules, which has arbitrary coproducts; it is not applied to this finite category. Preservation of existing coproducts is still meaningful here (for example the countable coproduct of zero modules exists), so noncocompleteness alone does not make preservation vacuous. All presentations and biproducts used in the classification are finite, so no choice is used.

By [L1] the category $\mathcal{C}$ is a $k$-linear abelian category with finite-dimensional hom-spaces and finite-length objects; equivalently it is a locally finite $k$-linear abelian category in the sense of [L3].

Every simple object $S$ of $\mathcal{C}$ has a projective cover: $S$ is projective by [L1], so the identity $1_S:S\to S$ is an epimorphism with projective source whose kernel is the zero subobject, which is superfluous by [L2]; hence $1_S$ is an essential epimorphism and a projective cover of $S$.

Thus $\mathcal{C}$ satisfies all the hypotheses of the refuted statement: it is $k$-linear and abelian with finite-dimensional hom-spaces, all objects have finite length, and every simple object has a projective cover by step 2.1. But by [L1] its simple objects $S_m$ are pairwise non-isomorphic, one for each $m\in \mathbb{N}$, so $\mathcal{C}$ has infinitely many isomorphism classes of simple objects; by [L3] it is therefore not a finite $k$-linear abelian category. This refutes the statement.

The failure is exactly the failure of the finite-simple-classes clause of the intrinsic definition while all the other clauses hold, and [L1] additionally supplies that no object of $\mathcal{C}$ is a generator; the construction of [L1] uses no choice, so no choice is used here.

By [L1] every element of $A$ is $a+b\epsilon$ with ${\epsilon }^2=0$, so $(\epsilon )=k\epsilon$, the quotient $S=A\mathrm{/}(\epsilon )$ is one-dimensional over $k$ with $\epsilon S=0$, and $A$ acts on $S$ through $k$; the $A$-submodules of $S$ are therefore exactly its $k$-subspaces, so $S\mathrm{\ne }0$ is simple. The quotient map $\pi :A\to S$ has kernel $(\epsilon )$, and the map $\phi :A\to (\epsilon )$, $\phi (a)=a\epsilon$, has image $(\epsilon )$ and kernel $(\epsilon )$, since $(a+b\epsilon )\epsilon =a\epsilon$ vanishes only for $a=0$; hence $\phi$ induces an $A$-module isomorphism $S=A\mathrm{/}(\epsilon )\cong (\epsilon )$ and $0\to (\epsilon )\to A\stackrel{\pi }{\to }S\to 0$ is a short exact sequence.

The sequence $0\to (\epsilon )\to A\to S\to 0$ does not split. If it split, then by [L4] there would be an $A$-module isomorphism $A\cong (\epsilon )\oplus S$, and $(\epsilon )\cong S$ by step 1.1, so $A\cong S\oplus S$; by step 1.1 the element $\epsilon$ acts as zero on each copy of $S$, hence as zero on $S\oplus S$ and therefore, through the isomorphism, as zero on $A$. But $\epsilon \cdot 1_A=\epsilon \mathrm{\ne }0$ in $A$. Contradiction, so the sequence does not split.

The functor $T_S=S{\otimes }_A-$ is right exact by [L2], so applying it to $0\to (\epsilon )\to A\to S\to 0$ gives the exact sequence $S{\otimes }_A(\epsilon )\to S{\otimes }_{A}A\to S{\otimes }_{A}S\to 0$ by [L3].

The three outer identifications of the statement hold: $S{\otimes }_{A}A\cong S$ by the unit isomorphism of [L3]; $S{\otimes }_A(\epsilon )\cong S{\otimes }_{A}S$ because $(\epsilon )\cong S$ by step 1.1; and $S{\otimes }_{A}S\cong S$, because applying the right exact functor $-{\otimes }_{A}S$ to $A\to S\to 0$ identifies $S{\otimes }_{A}S$ with the cokernel of $(\epsilon ){\otimes }_{A}S\to A{\otimes }_{A}S\cong S$, whose image is $\epsilon S=0$.

Under these identifications the first map $S{\otimes }_A(\epsilon )\to S{\otimes }_{A}A$ is zero: it is induced by the inclusion $(\epsilon )\rightarrowtail A$, and the generator $1_S\otimes \epsilon$ maps to $1_S\otimes \epsilon$, which corresponds under $S{\otimes }_{A}A\cong S$ to $1_S\epsilon =0$ because $\epsilon$ annihilates $S$; since $S{\otimes }_A(\epsilon )\cong S$ is generated as an $A$-module by $1_S\otimes \epsilon$, the map is zero. The second map $S{\otimes }_{A}A\to S{\otimes }_{A}S$ is induced by the quotient $A\twoheadrightarrow S$ and corresponds under the identifications to the identity of $S$, hence is an isomorphism. So the image sequence is $S\stackrel{0}{\to }S\stackrel{\cong }{\to }S\to 0$.

The sequence $S\stackrel{0}{\to }S\stackrel{\cong }{\to }S\to 0$ is exact, as the second map is an isomorphism and its kernel is zero, so right exactness of $T_S$ is exhibited directly. But $T_S$ is not left exact: the extended sequence $0\to S{\otimes }_A(\epsilon )\to S{\otimes }_{A}A\to S{\otimes }_{A}S\to 0$ fails to be exact at $S{\otimes }_A(\epsilon )$, because the map $S{\otimes }_A(\epsilon )\to S{\otimes }_{A}A$ is zero while $S{\otimes }_A(\epsilon )\cong S\mathrm{\ne }0$ by steps 1.1 and 2.3.

Consistently, $T_S$ has the right adjoint ${\mathrm{Hom}}_A(S,-)$ by [L2] but no left adjoint, since a functor with a left adjoint is left exact by [L4] and $T_S$ is not left exact by step 4.1; and its kernel $T_S(A)=S{\otimes }_{A}A\cong S$ is not a projective right $A$-module, since by [L4] a projective kernel would make $T_S$ exact, while $T_S$ is not left exact by step 4.1.

All modules and sequences above are finite-dimensional and the computations use only the finitely many structure maps of $A$ and $S$, so no choice is used.