(when) is a map of schemes determines by values on Zariski dense subset? The motivation for this question is to justify/understand a particular abuse of notation in the world of group schemes, namely people write things like $m:(g_1,g_2) \mapsto g_1g_2$ when discussing the multiplication or action maps $G \times_k G \to G$ when $G$ is a group scheme over $k$. This bothers me because over a general base field $k$, identifying points in the fibered product $G \times_k G$ as pairs $(g_1,g_2)$ only makes sense on $k$-rational points.
However, suppose we are in the situation where the $G(k) \hookrightarrow G$ is Zariski dense, and $G$ is a variety, so in particular (geometrically) reduced, separated, etc. Then I would think that a map on $k$-rational points, given by polynomials, determines uniquely a map of schemes. 
So the question is, suppose $X$ is a variety of a general base field $k$ such that $X(k)$ is Zariski dense in $X$. Then does a regular map $X(k) \to Y$, $Y$ separated, determine uniquely a morphism $X \to Y$.
 A: Assuming $X$ and $Y$ are geometrically reduced, separated, finite type $k$-schemes with $X(k)$ dense in $X$, then yes, all of these things are the same:


*

*"Regular maps" $X(k) \rightarrow Y(k)$ (nonstandard terminology to define such things on $k$-rational points when $k$ is not algebraically closed, but w/e)

*Regular maps of classical varieties $X(\overline{k}) \rightarrow Y(\overline{k})$ which are "defined over $k$" in the sense that affine locally, they are polynomial functions defined via polynomials with coefficients in $k$

*Morphisms of $k$-schemes $X \rightarrow Y$
When your schemes are not geometrically reduced, you have to be careful because a regular map $X(\overline{k}) \rightarrow Y(\overline{k})$ might not give you a morphism of schemes.  This is because $X(\overline{k})$ is the classical variety associated to the scheme $X_{\textrm{red}}$.
When I'm dealing with group schemes $G, H$ of finite type over a field $k$, with $k$ is perfect, and I want to define a morphism of group schemes $f: G \rightarrow H$, I usually go the "classical" route of considering the map on $\overline{k}$ points $G(\overline{k}) \rightarrow H(\overline{k})$.  I check that this is a regular map and a group homomorphism.  Then I check whether it is "defined over $k$" by checking that it commutes with the action of $\operatorname{Gal}(\overline{k}/k)$ on $G(\overline{k})$ and $H(\overline{k})$.  That ensures everything coming from a unique morphism of group schemes $G \rightarrow H$.
