  Given a tournament $T$, a module of $T$ is a subset $X$ of $V(T)$ such that
for $x, y\in X$ and $v\in V(T)\setminus X$, $(x,v)\in A(T)$ if and only if
$(y,v)\in A(T)$. The trivial modules of $T$ are $\emptyset$, $\{u\}$ $(u\in
V(T))$ and $V(T)$. The tournament $T$ is indecomposable if all its modules are
trivial; otherwise it is decomposable. The decomposability index of $T$,
denoted by $\delta(T)$, is the smallest number of arcs of $T$ that must be
reversed to make $T$ indecomposable. The first author conjectured that for $n
\geq 5$, we have $\delta(n) = \left\lceil \frac{n+1}{4} \right\rceil$, where
$\delta(n)$ is the maximum of $\delta(T)$ over the tournaments $T$ with $n$
vertices. In this paper we prove this conjecture by introducing the co-modular
index of a tournament $T$, denoted by $\Delta(T)$, as the largest number of
disjoint co-modules of $T$, where a co-module of $T$ is a subset $M$ of $V(T)$
such that $M$ or $V(T) \setminus M$ is a nontrivial module of $T$. We prove
that for $n \geq 3$, we have $\Delta(n) = \left\lceil \frac{n+1}{2}
\right\rceil$, where $\Delta(n)$ is the maximum of $\Delta(T)$ over the
tournaments $T$ with $n$ vertices. Our main result is the following close
relationship between the above two indices: for every tournament $T$ with at
least $5$ vertices, we have $\delta(T) = \left\lceil \frac{\Delta(T)}{2}
\right\rceil$. As a consequence, we obtain $\delta(n) = \left\lceil
\frac{\Delta(n)}{2} \right\rceil = \left\lceil \frac{n+1}{4} \right\rceil$ for
$n \geq 5$, and we answer some further related questions.
