Definition of split exact sequence and splitness of some exact complex

commutative-algebra

First recall the definition of splitness of the short exact sequence :

Def. (Split short exact sequence) A short exact sequence of abelian groups or of modules over a fixed ring, or more generally of objects in an abelian category,
${\displaystyle 0\to A\mathrel {\stackrel {a}{\to }} B\mathrel {\stackrel {b}{\to }} C\to 0}$
is called split exact if it is isomorphic to the sequence where the middle term is the direct sum of the outer ones:

${\displaystyle 0\to A\mathrel {\stackrel {i}{\to }} A\oplus C\mathrel {\stackrel {p}{\to }} C\to 0}$

Now, Let $\mathcal{F} : 0 \to F_n \xrightarrow{{\varphi_{n}}} F_{n-1} \to \cdots F_1 \xrightarrow{{\varphi_{1}}} F_0 \xrightarrow{{\varphi_{0}}} F_{-1} \to 0$ be a exact sequence of (possibly free) $R$-modules.

Then what is the definition of 'splitness' of the $\mathcal{F}$?

My question originates from following proof (Eisenbud, p. 499, proof of the Theorem 20.9) :

enter image description here

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I'm now trying to understand the underlined statement. For the underlined statement, what the 'splitness of $\mathcal{F}_p$' exactly means? And why the rank formula for $M_p$ holds?

Note : In the Rotman's An introduction to homological algebra book, p.648, he defined that "A complex $(C,d)$ in an abelian category $\mathcal{A}$ is split if all its fundamental sequences are split" ; here fundamental sequences means

$$ 0\to B_n \xrightarrow{i_n} Z_n \to H_n(C)\to 0$$
$$ 0 \to Z_n \xrightarrow{j_n} C_n \xrightarrow{d'_n} B_{n-1}$$

($Z_n(C) := \operatorname{ker}d_n$, $B_n(C) := \operatorname{im}d_{n+1}$)
where $i_n, j_n$ are inclusions and $d'_{n}$ is just $d_n$ with its target changed from $C_{n-1}$ to $\operatorname{im}d_n=B_{n-1}$.

If we adopt this defition, it seems that we may prove the above rank formula : $\operatorname{rank}M_p = \Sigma_{i \ge 0}(-1)^{i}\operatorname{rank}F_i$

So my question is, the $\mathcal{F}_p$ (in the above image) is split exact in the above Rotman's sense? For every exact complex consisting of finite free modules ; i.e., of the form $\mathcal{F} : 0 \to F_n \xrightarrow{{\varphi_{n}}} F_{n-1} \to \cdots F_1 \xrightarrow{{\varphi_{1}}} F_0 \xrightarrow{{\varphi_{0}}} F_{-1} \to 0$, $\mathcal{F}$ is split? Why?

Best Answer

I think that I solved the problem, just using

  1. Any free module is projective module.

  2. A left $R$-module $P$ is projective if and only if every short exact sequence $0\to A\to B\to P \to 0$ splits.

  3. A left $R$-module $P$ is projective if and only if $P$ is a direct summand of a free left $R$-module.

  4. Exactness of the $\mathcal{F}$.

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