$G$ is a connected, reductive group over a field $k$, $S$ is a maximal $k$-split torus of $G$, and $_k \Phi = \Phi(S,G)$ is the set of roots of $S$ in $G$. Equivalently, $_k \Phi$ is the restriction to $S$ of the set $\Phi = \Phi(T,G)$ of roots of $T$ in $G$, removing any roots which are trivial on $S$ (e.g. see here mathoverflow.net/questions/255283/two-definitions-of-restricted-roots).

This part in Borel, *Linear Algebraic Groups*, explains the relationship between certain bases of $\Phi$ and $_k \Phi$.

I've been staring at (3) for awhile.

It is clear to me that $_k\Delta \subseteq j(\Delta)$: by construction, every element of $_k\Phi^+$ is an nonnegative integral combination of elements of $j(\Delta)$, so $j(\Delta)$ must contain the unique base of $_k \Phi^+$, namely $_k \Delta$.

But I don't understand why $j(\Delta)$ should be contained in $_k \Delta \cup \{0\}$.

1more comment