Let $k$ be a field, and let $G$ be an algebraic group over $k$ (by which I mean a group scheme of finite type over $k$). Let $Z(G) \subseteq G$ be the center (which is an algebraic subgroup).
What are some conditions for which $Z(G)(k) = Z(G(k))$ holds (i.e. when is the rational points of the center equal to the center of the rational points)? The inclusion $Z(G)(k) \subseteq Z(G(k))$ is clear. We also have $Z(G)(k) = Z(G(k))$ whenever $G$ is reduced and $k$ is algebraically closed.
I am particularly interested in the case where $G$ is a reductive group. We have $Z(G)(k) = Z(G(k))$ when $G = \operatorname{GL}_n$ for example, since $Z(G)$ just consists of scalar matrices in this case.
Just to get this off the unanswered list. What you want holds true essentially always.
Proof: Evidently $Z(G)(k)\subseteq Z(G(k))$. To prove the converse it suffices to show that if $z$ is in $Z(G(k))$ then the conjugation map $c_z\colon G\to G$ given by $c_z(g)=zgz^{-1}$ is trivial. But, if $K:=\ker(c_z)$ we have, by assumption, that $G(k)\subseteq K$. But, as $G$ is a unirational $k$-variety and $k$ is infinite, we have that $G(k)$ is dense in $G$. Thus, $K$ is equal to $G$ as a topological space. As $G$ is reduced, we deduce that $G=K$ and so $c_z$ is trivial as desired. $\blacksquare$
The assumption that $G$ is unirational is automatic if $G$ is reductive or $k$ is perfect. Moreover, the assumption that $G$ is reduced is automatic if $k$ is of characteristic $0$. Thus, we see that if $k$ is of characteristic $0$ then $Z(G(k))=Z(G)(k)$ for all $k$.