First off, remember a root system is reducible if we can write it as the disjoint union of two collections of roots so that each root in is perpendicular to each one in . I assert that, for any base , is reducible if and only if can itself be broken into two collections in just the same way. One direction is easy: if we can decompose , then the roots in are either in or and we can define and .
On the other hand, if we write with each simple root in perpendicular to each one in , then we will find a similar decomposition of . But we know from our study of the Weyl group that every root in can be sent by the Weyl group to some simple root in . So we define to be the collection of roots whose orbit includes a point of , and to be the collection of roots whose orbit includes a point of . So a vector in is of the form for some Weyl group element and some simple root . The Weyl group element can be written as a sequence of simple reflections. A simple reflection corresponding to a root in adds some multiple of that root to the vector, and thus leaves the subspace spanned by invariant; while a simple reflection corresponding to a root in leaves the subspace spanned by fixed point-by-point. Thus any root in must lie in the subspace spanned by , and similarly any root in must lie in the subspace spanned by . This shows that is exactly the sort of decomposition we’re looking for.
If is irreducible, with base , then there is a unique maximal root relative to the partial ordering on roots. In particular, the height of is greater than the height of any other root in , for all simple roots , and in the unique expression
all of the coefficients are strictly positive.
First of all, if is maximal then it’s clearly positive, and so each is either positive or zero. Let be the collection of simple roots so that , and be the collection of simple roots so that . Then is a partition of the base. From here we’ll assume that is nonempty and derive a contradiction.
If , then for any other simple root . From this, we can calculate
Since is irreducible, there must be at least one and so that , and so we must have for this . This proves that must also be a root, which contradicts the maximality of . And so we conclude that is empty, and all the coefficients . In passing, we can use the same fact to show that for all , or else wouldn’t be maximal.
This same argument applies to any other maximal root , giving for all with the inequality strict for at least one . We calculate
which tells us that is a root unless . But if is a root, then either or , contradicting the assumption that both are maximal. Thus must be unique.