Does a long exact sequence of flat modules remain exact after tensoring with an arbitrary module?

Question

In Liu's Algebraic Geometry and Arithmetic Curves, Proposition 1.2.6 states that given any short exact sequence $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ with $M''$ flat, taking the tensor product of this with any module $N$ gives an exact sequence $0 \rightarrow M' \otimes N \rightarrow M \otimes N \rightarrow M'' \otimes N \rightarrow 0$.

Later in the book, in Exercise 5.2.8, page 193, he suggests using this Proposition to show that given a long exact sequence which is eventually $0$:

$$ 0 \rightarrow M_0 \rightarrow M_1 \rightarrow \cdots $$

of flat modules, taking the tensor product with an arbitrary module still gives an exact sequence

$$ 0 \rightarrow M_0 \otimes N \rightarrow M_1 \otimes N \rightarrow \cdots$$

How does this generalization follow? It appears that in the proof of the Proposition, he doesn't really use the surjectivity of $M \rightarrow M''$, but then again, all he proves is that $M' \otimes N \rightarrow M'' \otimes N$ is injective; the full exactness being something that always happens for tensor products of short exact sequences.

Answer 1

In fact, Liu says (on page 193, Exercise 2.8(b)) that the sequence is "zero from a finite rank on". Then you start the breaking from the rightmost side of the sequence: $0\to X_{n-1}\to M_{n-1}\to M_n\to 0$, and so on. You know that $M_{n-1}$ and $M_n$ are flat, so $X_{n-1}$ is also flat, and so on.

Answer 2

let $$0 \rightarrow M_0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow \cdots$$ be exact. then there is exact sequences (break):
$$0 \rightarrow M_0 \rightarrow M_1 \rightarrow C_1 \to 0$$ and $$0\to C_1 \rightarrow M_2 \rightarrow C_2 \to 0$$ and so on...

now tensor with $N$ and join sequences again.

Answer 3

If the base ring is Noetherian, then the answer is yes:

1. It is sufficient to show that all syzygies $Z^k := \text{ker}(M^k\to M^{k+1})$ of the given long exact sequence $$(\ddagger)\quad 0\to M^0\to M^1\to\ldots$$ are again flat, since then, as user1 pointed out, one can break the sequence into short exact sequences of flat modules and apply the exercise you referred to.

2. It suffices to treat the case of a Noetherian local ring $R$, since flatness is a local property.

3. Over a Noetherian local ring, any module of finite projective dimension has projective dimension at most $\text{depth}_RR$ by the Auslander-Bridger formula. In particular, the big finitistic projective dimension of $R$ is finite, and by a result of Jensen, any module of finite flat dimension has finite projective dimension. Hence $\text{fl.dim}_RM<\infty$ implies $\text{fl.dim}_RM\leq\text{depth}_RR$ for any $R$-module $M$.

4. Viewing $0\to M^0\to M^1\to\ldots\to M^{k-1}\to Z^k\to 0$ as a flat resolution of $Z^k$ shows that $Z^k$ has finite flat dimension $\text{fl.dim}_R Z^k\leq k$, and 3. shows that even $\text{fl.dim}_R Z^k\leq\text{depth}_RR$.

5. Any interval $0\to Z^k\to M^k\to\ldots\to M^{k+l-1}\to Z^{k+l}\to 0$ can be viewed as a partial flat resolution of $Z^{k+l}$ of length $l$. Combining this with 4. allows to conclude that $Z^k$ is flat by choosing $l>\text{depth}_RR$.

Of course, if we know that $(\ddagger)$ is bounded, user26857's argument is the one to apply and does not need any Noetherianness hypothesis.

Note also the result is wrong if $M^{\ast}$ is unbounded to both sides: Take $R = k[x]/(x^2)$ and consider $$\ldots\to R\xrightarrow{\cdot x} R\xrightarrow{\cdot x} R\to \ldots$$ It is exact, but tensoring with $k := R/(x)$ yields the non-exact sequence $$\ldots\to k\xrightarrow{0} k\xrightarrow{0}\ldots$$