1

I am interested in the unbounded case and looking for sufficient conditions on the modules of the complex or the ring for this to hold. Or even restricting the functor to some unbounded subcategory.

We know that since $(-) \otimes B$ is a left adjoint functor it preserves colimits and hence is right exact. That is given an short exact sequence \begin{equation} 0 \to X \to Y \to Z\to 0, \end{equation} of chain complexes $X,Y,Z$ over some ring $R$ we get an exact sequence \begin{equation} X\otimes B \to Y \otimes B \to Z\otimes B \to 0. \end{equation}

The question then comes down to if the map $X\otimes B \to Y\otimes B$ is a degreewise injection. Or put differently, conditions for monomorphisms to be preserved under $(-)\otimes B$.

If we look at the full category of chain complexes a necessary conditions on $B$ would be flatness in each degree but as shown in the question linked it is not sufficient.

In particular I am wondering if it would hold:

  • Over a field $R=k$
  • In the full subcategory of chain complexes with free or projective modules.

Edit: My original post also contained the the following but as pointed out in the comments it is different to my question. "In the question "Flat chain complex"? sufficient conditions in term of bounded complexes are given."

Bjorn
  • 180
  • 3
    The properties in the linked question ($-\otimes B$ takes acyclic complexes to acyclic complexes) and in this question ($-\otimes B$ takes exact sequences of complexes to exact sequences of complexes) are different. For example, a contractible complex of nonflat modules has the first property but not the second, and an unbounded complex of flat modules has the second property but, as is shown in the answer to the other question, might not have the first property. – Jeremy Rickard Jul 30 '21 at 20:22
  • Good catch! I should have read that question more carefully. My question is the one I stated. I make an edit to my post to make this clearer. – Bjorn Jul 30 '21 at 21:52

1 Answers1

2

Being flat in each degree is a necessary and sufficient condition.

The point is that the differentials of $B$, $X$, $Y$ and $Z$, and their grading by degree, are irrelevant.

Given a chain complex of modules $C$ with components $C_i$ ($i\in\mathbb{Z}$), let $\widetilde{C}$ denote the single module $\bigoplus_{i\in\mathbb{Z}}C_i$. Then $C\mapsto\widetilde{C}$ is a functor from complexes to modules.

Also, $\widetilde{C\otimes D}=\widetilde{C}\otimes\widetilde{D}$, and a sequence $0\to C\to D\to E\to0$ of complexes is exact if and only if $0\to \widetilde{C}\to \widetilde{D}\to \widetilde{E}\to0$ is exact.

So $0\to X\otimes B\to Y\otimes B\to Z\otimes B\to0$ is exact if and only if $0\to \widetilde{X}\otimes \widetilde{B}\to \widetilde{Y}\otimes \widetilde{B}\to \widetilde{Z}\otimes \widetilde{B}\to0$ is exact, which is the case for all short exact sequences $0\to X\to Y\to Z\to0$ if and only if $\widetilde{B}$ is flat, which is the case if and only if $B$ is flat in each degree.

Jeremy Rickard
  • 28,899