5
$\begingroup$

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 $B \cong A \oplus C$

$\endgroup$
7
  • $\begingroup$ It's a general fact that a short exact sequence splits if and only if the the middle term is isomorphic to the direct sum of the outer ones. This is called the splitting lemma. $\endgroup$ Jul 1, 2016 at 9:25
  • 3
    $\begingroup$ @Mathematician42 This is false. See the related thread I've just included. $\endgroup$
    – Yai0Phah
    Jul 1, 2016 at 9:28
  • $\begingroup$ It is true in an abelian category, see for example: ncatlab.org/nlab/show/split+exact+sequence $\endgroup$ Jul 1, 2016 at 9:44
  • 2
    $\begingroup$ @Mathematician42 No, it's wrong. We should note the difference between the isomorphism of short exact sequences, and isomorphisms among the middle term. See the related post I gave, which includes a counterexample in the Abelian category of Abelian groups. $\endgroup$
    – Yai0Phah
    Jul 1, 2016 at 10:37
  • 1
    $\begingroup$ Could you expand a bit on the dimension counting argument for the case of bounded complexes of free finite rank modules over a PID ? After all, $\times 2:\mathbb{Z}\rightarrow\mathbb{Z}$ is not onto despite both source and target being of rank 1. $\endgroup$
    – Roland
    Jul 1, 2016 at 11:49

1 Answer 1

2
$\begingroup$

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 $C\cong C'\oplus C''$.

$\endgroup$
2
  • $\begingroup$ Thanks for the comment. Thanks to math.stackexchange.com/a/891527/23875, it seems to me the argument (if it's right) works for my question too. What if we add another restriction that $C',C,C''$ are bounded? In this case (which is the case for Roth's theorem), assuming that the ring is commutative, it seems to me that WLOG, we can assume that the ring is Noetherian, since we can restrict to a finitely generated subring. $\endgroup$
    – Yai0Phah
    Jul 11, 2016 at 1:37
  • $\begingroup$ I posted a question on MO. $\endgroup$
    – Yai0Phah
    Jul 13, 2019 at 13:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .