Reducible and irreducible root systems

Let $\mathrm{\Phi }\subseteq E$ be a reduced crystallographic root system in the finite-dimensional real inner product space $E$ (Reduced crystallographic Euclidean root system).

Then $\mathrm{\Phi }$ is reducible if there are linear subspaces $E_1,E_2\subseteq E$ with $E=E_1\oplus E_2$, $(E_1,E_2)=0$, both $E_i$ nonzero, and $\mathrm{\Phi }=(\mathrm{\Phi }\cap E_1)\bigsqcup (\mathrm{\Phi }\cap E_2),$ the union being disjoint. Otherwise $\mathrm{\Phi }$ is irreducible.

Equivalently, $\mathrm{\Phi }$ is reducible if it is the disjoint union of two nonempty subsets ${\mathrm{\Phi }}_1,{\mathrm{\Phi }}_2$ with $({\mathrm{\Phi }}_1,{\mathrm{\Phi }}_2)=0$, in which case one may take $E_i=\mathrm{span}{\mathrm{\Phi }}_i$: indeed if $\mathrm{\Phi }=(\mathrm{\Phi }\cap E_1)\bigsqcup (\mathrm{\Phi }\cap E_2)$ then $\mathrm{\Phi }$ spans $E_1$ and $E_2$ separately, because otherwise a nonzero vector of $E_i$ orthogonal to $\mathrm{\Phi }\cap E_i$ and to $E_{3-i}$ would be orthogonal to all of $\mathrm{\Phi }$ and hence zero. Consequently both $\mathrm{\Phi }\cap E_i$ are nonempty, and each of them is itself a reduced crystallographic root system in $E_i$ whose roots are those of $\mathrm{\Phi }$ lying in $E_i$.

A one-element root system is impossible: if $\alpha \in \mathrm{\Phi }$, reflection in $\alpha$ sends $\alpha$ to the distinct root $-\alpha$, because $\alpha \mathrm{\ne }0$. The zero vector space carries the empty root system under the stated root-system axioms; it is irreducible by the definition above, since the zero space has no orthogonal direct-sum decomposition into two nonzero subspaces. Every rank-one root system $\{\pm \alpha \}$ is likewise irreducible.