$0\to C'\to C\to C''\to0$ splits if $C\cong C'\oplus C''$ as a chain complex?

Question

Question

Given a unitary ring $A$ and an exact sequence $$0\to C'\xrightarrow iC\xrightarrow pC''\to0$$ in the Abelian category of chain complexes over $A$, where $C,C',C''$ are chain complexes of finitely-generated free modules (I don't know whether this could be replaced by projective modules). If $C\cong C'\oplus C''$ as chain complexes, is it true that the original exact sequence splits in the Abelian category of chain complexes?

Results

If $A$ is a field or a PID, and that the complexes (the total complexes seen as $A$-modules) are of finite rank, then the statement could be proved as follows:

Let $\mathcal A$ be the Abelian category of chain complexes over $A$. Take $\operatorname{Hom}_{\mathcal A}(C'',-)$, we have an exact sequence:

$$0\to\operatorname{Hom}(C'',C')\xrightarrow{i_*}\operatorname{Hom}(C'',C)\xrightarrow{p_*}\operatorname{Hom}(C'',C'')$$

Since $C\cong C'\oplus C''$, we have $\operatorname{Hom}(C'',C)\cong\operatorname{Hom}(C'',C')\oplus\operatorname{Hom}(C'',C'')$, and we should note that all these $\operatorname{Hom}$'s are submodules of free modules, hence free ($A$ is a PID). It follows from dimension counting that $p_*$ is surjective, hence the original exact sequence splits.

Backgrounds

It's a generalization of Roth's theorem. Given matrices $A,B,C$ over a commutative ring $R$ and let $$P=\begin{bmatrix}B&0\\&C\end{bmatrix}$$ and $$P_A=\begin{bmatrix}B&A\\&C\end{bmatrix}$$ Then

• If $P,P_A$ are equivalent, then there exists $X,Y$ such that $A=BX-YC$.
• If $B,C$ are square matrices and $P,P_A$ are similar, then there exists $X$ such that $A=BX-XC$.

The first statement follows from the question (if it's true): we consider two complexes $$K\colon\to0\to K_0=R^\bullet\xrightarrow CR^\bullet\to0\to$$ and $$L\colon\to R^\bullet\xrightarrow BL_0=R^\bullet\to0\to0\to$$ The chain homomorphism $f$ is given by the matrix $A\colon K_0\to L_0$ (and zero on any other degree). Consider the canonical exact sequence involving a mapping cone: $$0\to L\to\operatorname{cone}(f)\to K[-1]\to0$$ Note that the matrix associated to the boundary operator of $\operatorname{cone}(f)$ is $P_A$ (up to some signs), which means that $\operatorname{cone}(f)\cong L\oplus K[-1]$. We apply the result of the question, and it follows directly that $f$ is null homotopic, hence we can solve the matrix equation.

The second statement follows from the first statement. If $P,P_A$ are similar, then $T-P,T-P_A$ are equivalent over the ring $R[T]$, hence there exists $P(T)\in\operatorname{Mat}(R[T])$ and $Q(T)\in\operatorname{Mat}(R[T])$ (we omit the computation of the magnitude of matrices) such that $(T-B)P(T)+Q(T)(T-C)=A$. If we factor $Q(T)=(T-B)Q_1(T)+Q_0$, we have $$(T-B)(P(T)+Q_1(T)(T-C)+Q_0)+BQ_0-Q_0C=A.$$ Compare the remainder term, we obtain $BQ_0-Q_0C=A$.

Maybe related

I just found this post: A nonsplit short exact sequence of abelian groups with $B \cong A \oplus C$

Answer 1

I think it may be true if the ring is noetherian, but not for general rings.

I'll build a counterexample in a few steps.

First, there are easy examples of non-split short exact sequences $0\to A'\to A\to A''\to0$ of bounded complexes of finitely generated vector spaces over a field $k$. For example, there's an obvious such sequence with $$A'=\dots\to0\to0\to k\to0\to\dots,$$ $$A=\dots\to0\to k\stackrel{\sim}{\to} k\to0\to\dots,$$ $$A''\dots\to 0\to k\to 0\to0\to\dots.$$

Next, by taking the direct sum with countably many copies of the split short exact sequence $$0\to A'\oplus A\oplus A''\to(A'\oplus A\oplus A'')^2\to A'\oplus A\oplus A''\to0,$$ we can construct a non-split short exact sequence $0\to B'\to B\to B''\to0$ of complexes of vector spaces with all the non-zero terms isomorphic to a countably infinite dimensional vector space $V$, and with $B\cong B'\oplus B''$.

For any object $V$ of an additive category, with $E=\operatorname{End}(V)$, the functor $\operatorname{Hom}(V,-)$ is an equivalence of categories from the category of finite direct sums of copies of $V$ to the category of finitely generated free right $E$-modules.

So finally, applying the functor $\operatorname{Hom}_k(V,-)$ to $0\to B'\to B\to B''\to0$, we get a non-split short exact sequence $0\to C'\to C\to C''\to0$ of complexes of finitely generated free $E$-modules with $C\cong C'\oplus C''$.