Do endomorphisms of the adjoint representation of a Lie algebra commute? Given a field $k$ of characteristic $0$ and a finite dimensional simple Lie algebra $\mathfrak{g}$ over $k$. Consider the adjoint representation $(\mathfrak{g}, \mathrm{ad})$ of $\mathfrak{g}$ and let $\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ denote the ring of $\mathfrak{g}$-module endomorphisms with regard to this representation.
The claim is: $\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ is a field extension of $k$ and $\dim_k\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ is equal to the number of simple components of $g \otimes_k \overline{k}$ where $\overline{k}$ denotes the algebraic closure of $k$.
I've come this far: Since $\mathfrak{g}$ is simple, the adjoint representation must be irreducible. Thus, $\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ is a $k$ skew-field. But why do all elements from $\mathrm{End}(\mathfrak{g}, \mathrm{ad})$ commute?
If $A, B \in \mathrm{End}(\mathfrak{g}, \mathrm{ad})$, one can argue that $AB - BA$ is either $0$ or invertible. However, I wasn't able to rule out the latter case.
Also: How are $\mathfrak{g} \otimes_k \overline{k}$ and $\mathfrak{g}$ related with regard to (semi)-simplicity? I know that $\mathfrak{g}$ is semi-simple iff $\mathfrak{g} \otimes_k \overline{k}$ is semi-simple. Is there a corresponding result for the simple case?
 A: $\DeclareMathOperator{\g}{\mathfrak g}$
$\DeclareMathOperator{\ad}{\mathrm{ad}}$
$\DeclareMathOperator{\End}{\mathrm{End}}$
I tried to give a short intro to this theory in section 4.1 of my thesis which generally follows Jacobson, N.: A note on non-associative algebras. Duke Math. J. 3 (1937), no. 3, 544--548. doi:10.1215/S0012-7094-37-00343-0. Here is the part relevant to your question:
Re first question:
For a $k$-Lie algebra $\g$ define
$$K := K(\g) := \{ s \in \End_k(\g): s \circ \ad_{\g}(x) = \ad_{\g}(x) \circ s \text{ for all } x \in \g \}.$$
We view it as associative $k$-algebra and remark that as such it identifies with what you call $\End(\g, \ad)$.
If $\g$ is simple, then (as you remark) $K$ is a skew field by Schur's lemma.
In fact, it is a field; namely, since $\g = [\g, \g]$ it suffices to see that two elements $s, t \in K$ commute on a commutator $[x,y]$ for $x,y \in \g$. But
$$ s(t([x,y])) = s([x, ty]) = [sx, ty] = t([sx, y]) = t(s([x,y])) $$
where we have used, from left to right, that $t$ commutes with $\ad_{\g}(x)$, $s$ with $-\ad_{\g}(ty)$, $t$ with $\ad_{\g}(sx)$ and $s$ with $-\ad_{\g}(y)$.
One calls $K$ the centroid of $\g$ and remarks that $\g$ has a natural structure as Lie algebra over $K$. When viewed as such, write $^K \g$.
Re second question:
First, some notation. For a Lie algebra $\g$ over $k$, let $A(\g)$ be the (associative, unital) $k$-subalgebra of $\End_k(\g)$ generated by all $\ad_{\g}(x)$, $x \in \g$. Remark straightaway that for any field extension $L|k$, $a \otimes \ad_{\g}(x) \mapsto \ad_{\g_L} (a \otimes x)$ defines a natural isomorphism of associative $L$-algebras:
$$(*) \qquad L \otimes_k A(\g) \cong A(\g_L)$$
Also remark that $\g$ is a (left) $A(\g)$-module, and that an ideal of $\g$ is the same as an $A(\g)$-submodule.
Further, the inclusion $A(\g) \subseteq \End_k(\g)$ factors through natural maps $A(\g) \hookrightarrow \End_K(^K\g) \hookrightarrow \End_k(\g)$, and the first arrow is bijective by Jacobson's density theorem. (The theorem is absent from Jacobson's paper I quoted above, as he only proved it eight years later!) Consequently, the following are equivalent:

*

*$\g$ is simple and $K = k$

*$A(\g) = \End_k(\g)$.

In this case we call $\g$ central simple. So e.g. $^K\g$ is central simple if $\g$ is simple. It follows from $(*)$ that every scalar extension of a central simple Lie algebra is again central simple, a fortiori absolutely simple (A Lie algebra $\g$ over $k$ is called absolutely simple if $\g_{\bar k} := \g \otimes_k \bar k$ is simple over $\bar k$, or equivalently, $\g_K$ is simple over $K$ for every extension $K|k$.). But we have much more:
Proposition (4.1.2 in my thesis): Let $\g$ be a simple Lie algebra and $L|k$ a Galois extension containing the centroid $K$. Then $\g_L \simeq \g_1 \times ... \times \g_r$ where $r = [K:k]$ and the $\g_i$ are absolutely simple Lie algebras over $L$. In particular, $\g$ is central simple if and only if it is absolutely simple.
Proof:  Writing $K = k[X]/(f)$ where $f$ is a minimal polynomial of a primitive element of $K|k$, we have $L \otimes_k K \cong \prod_{i=1}^r L_i$ (as $L$-algebras) where the $L_i$ are all $L$ but with an $L$-action twisted via certain elements $\sigma_i : L \simeq L_i$ of the Galois group $Gal(L|k)$, permuting the zeros of $f \in L[X]$. In particular, $r = [K:k]$. Then with $(*)$,
\begin{align*}
A(\g_{L}) &\cong L \otimes_k \End_K(^K\g) \cong \End_{L\otimes_k K}((L \otimes_k K) \otimes_K (^K\g) ) \\
&\cong \End_{\prod_{i=1}^r L_i} (\bigoplus_{i=1}^r (^K\g)_{L_i}) \cong \prod_{i=1}^r \End_{L_i}((^K\g)_{L_i}).
\end{align*}
Calling $e_i$ the $i$-th idempotent in the last product, the $A(\g_L)$-module $e_i \cdot \g_L$ is a simple ideal $\g_i$ in $\g_L$, which is in fact the simple $L$-Lie algebra deduced from $(^K\g)_L$ by scalar extension (i.e. twisting the $L$-action) with $\sigma_i$.
