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Jul 10, 2012 at 17:38 comment added Emerton Dear Justin, I think the main point is that there is a difference between what you are calling $k\mathbb P^1$, which is the set of $k$-valued points of the variety $\mathbb P^1$ over $k$, and the actual variety $\mathbb P^1$. This latter variety has many more points than just its $k$-valued points. A non-constant rational function on $\mathbb P^1$, defined over $k$ or not, will necessarily have poles, but it may happen that none of these poles are defined over $k$ (even if the function is), so that it induces a regular function when restricted to $k\mathbb P^1$; there is no mystery here.
Jul 10, 2012 at 12:28 history answered Justin Smith CC BY-SA 3.0