Invertible functions on a proper variety It seems well-known that for a smooth proper geometrically integral variety $X$ over a number field $k$, the invertible functions on $\bar{X} := X \times _k \bar{k}$ are the nonzero constants, i.e., $\bar{k}[X]^* = \bar{k}^*$.
Does anyone have a proof or reference of this result? I've tried using the valuative criterion of properness but I couldn't get anywhere.
Furthermore, what can we say about the invertible rational functions $\bar{k}(X)^*$ in this case? Would it also be $\bar{k}^*$?
 A: Here's a reference:

Lemma (Stacks 0BUG): Let $k$ be a field. Let $X$ be a proper scheme over $k$.

*

*$A=H^0(X,\mathcal{O}_X)$ is a finite-dimensional $k$-algebra.

*$A=\prod_{i=1,\cdots,n} A_i$ is a product of Artinian local $k$-algebras, one factor for each connected component of $X$.

*If $X$ is reduced, then $A=\prod_{i=1,\cdots,n} k_i$ is a product of fields, each a finite extension of $k$.

...


*If $X$ is geometrically integral, then $A=k$.


This gives that $k[X]=k$ and $\overline{k}[X]=\overline{k}$, and therefore $\overline{k}[X]^*=\overline{k}^*$.
The key facts for the proof are the fact that the pushforward of a coherent sheaf along a proper morphism is coherent (02I5) and the fact that cohomology commutes with flat base extension (02KE). With those in hand, we can play a few little games with $A$ to get the results.

As for your follow-up question regarding $k(X)^*$: no, that's not true. Try taking a look at $X=\Bbb P^n_k$, for instance - you shouldn't find it too hard to compute $k(X)^*$ in that case.
