[proofplan] We prove the equality by checking membership in both sets element by element. The kernel condition says exactly that $\operatorname{ad}_x$ is the zero [linear map](/page/Linear%20Map) on $\mathfrak g$. Unwinding the definition of $\operatorname{ad}_x$ turns this into the condition $[x,y]=0$ for every $y \in \mathfrak g$, which is precisely the definition of the centre. [/proofplan] [step:Unwind the kernel condition for an arbitrary element] Let $x \in \mathfrak g$. By definition of the kernel of the linear map $\operatorname{ad}: \mathfrak g \to \mathfrak{gl}(\mathfrak g)$, \begin{align*} x \in \ker(\operatorname{ad}) \end{align*} if and only if \begin{align*} \operatorname{ad}(x) = 0_{\mathfrak{gl}(\mathfrak g)}, \end{align*} where $0_{\mathfrak{gl}(\mathfrak g)}: \mathfrak g \to \mathfrak g$ denotes the zero $F$-linear map. Since $\operatorname{ad}(x)=\operatorname{ad}_x$, this is equivalent to \begin{align*} \operatorname{ad}_x(y)=0 \end{align*} for every $y \in \mathfrak g$. [/step] [step:Translate the zero adjoint map condition into centrality] By the definition of $\operatorname{ad}_x: \mathfrak g \to \mathfrak g$, for every $y \in \mathfrak g$, \begin{align*} \operatorname{ad}_x(y) = [x,y]. \end{align*} Therefore \begin{align*} \operatorname{ad}_x(y)=0 \text{ for every } y \in \mathfrak g \end{align*} if and only if \begin{align*} [x,y]=0 \text{ for every } y \in \mathfrak g. \end{align*} The latter condition is exactly the defining condition for $x \in Z(\mathfrak g)$. Hence, for every $x \in \mathfrak g$, \begin{align*} x \in \ker(\operatorname{ad}) \iff x \in Z(\mathfrak g). \end{align*} Thus \begin{align*} \ker(\operatorname{ad}) = Z(\mathfrak g). \end{align*} [/step]