Let $RVar$ be the category which has complex algebraic verieties as objects and rational maps as morphisms [Edit: for $RVar$ to be a category, the rational maps have to be dominant]. Let's consider a group object $G$ in this category, i.e. a tuple $(G,\mu,e,\iota)$ where $\mu:G\times G \rightarrow G$ , $e:* \rightarrow G$, and $\iota: G \rightarrow G$ are rational maps satisfying the usual commutative diagrams defining a group structure. [Edit: in the light of the comments, e.g. the observation that rational maps have better be dominant, the identity $e:*\rightarrow G$ doesn't seem to make much sense; same for the diagram for the inverse, then; what people, among whom Weil, actually considered were "rational group chunks" in which there's only an associative rational $\mu$, and you can ask the same question(s)].

Just out of curiosity, two natural questions:

Is such a $G$ necessarily an algebraic group? That is: is it the case that for any $(G,\mu,e,\iota) \in Grp(RVar)$, there exists an algebraic group $(G',\mu',e',\iota')$ and a birational map $\varphi: G\rightarrow G'$ such that "$\varphi$ intertwines the operations of $G$ and $G'$"?

Analogous question in the holomorphic/meromorphic setting.

noidentity axiom and so no inverses! (The empty scheme is a birational group law.) It's all about translations and assoc. There is a finer notion of "strict birational group law", and thm is that (i) any non-empty bir. gp law on a smooth septd scheme of finite type contains dense open that is strict (for "induced" bir. gp law structure), and (ii) any non-empty strict bir. gp. law is dense open in unique actual smooth group. No control on affineness in the result. – BCnrd Apr 22 '10 at 4:02strictbirational group law involve the "inverse" in some way? The problem in defining the identity instead, as far as I've understood from the comments above, is that $*\rightarrow G$ is almost never dominant. Right? – Qfwfq Apr 22 '10 at 4:17