Existence of toral subalgebra

Question

If $L$ is a semi-simple Lie algebra every element $x \in L$ has a Jordan decomposition into $x = x_s + x_n$ where $[x_s,x_n] = 0$ and $ad(x_s)$ is semi-simple while $ad(x_n)$ is nilpotent.

Thus if $L$ isn't nilpotent there is an $x \in L$ such that $ad(x_s) \neq 0$ and so the sub-algebra span$\{x_s\}$ is a toral sub-algebra.

In Humphrey's book he says that span$\{x_s : x \in L, x_s \neq 0\}$ is also a toral sub-algebra.

However, if $x, y \in L$, does it necessarily follow that $x_s + y_s$ is an ad-semi-simple element? I can't find a reason, for instance, that $x_s + y_s = (x+y)_s$.

Why is span$\{x_s : x \in L, x_s \neq 0\}$ a toral sub-algebra?

Screen shot from Humphrey's book:

Answer 1

As I read it, the text in the book just claims that the 1-dimensional subalgebra spanned by one nonzero $x_s$ is a non-zero subalgebra consisting of semisimple elements. (I agree the formulation is not completely clear.)

Answer 2

I believe that is a small mistake. As Andreas Cap said, that observation only implies that there is a 1-dimensional subalgebra consisting of semisimple elements.

The mistake is not even in the last errata of the book. Too bad Humphreys died from COVID 19 and that errata won't be updated anymore :(

However, this explained at some lenght and correctly in Erdmann, Karin; Wildon, Mark J. Introduction to Lie algebras, section 10.2 (note that the book calls maximal toral subalgebras Cartan subalgebras, but in the semisimple finite dimensional case these are the same):

Let $L$ be a complex semisimple Lie algebra. We shall show that $L$ has a non-zero Cartan Subalgebra. We first note that $L$ must contain semisimple elements. If $x\in L$ has Jordan decomposition $x=s+n$, then by Theorem 9.15 both $s$ and $n$ belong to $L$. If the semisimple part $s$ were always zero, then by Engel's Theorem (in its second version), $L$ would be nilpotent and therefore solvable. Hence we can find a non-zero semisimple element $s\in L$. We can now obtain a non-zero Cartan Subalgebra $L$ by taking any subalgebra which contains $s$ and which is maximal subject to being abelian and consisting of semisimple elements. (Such a subalgebra must exists because $L$ is finite-dimensional.)