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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 $\bar{k}^*$?

oleout
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1 Answers1

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Here's a reference:

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

  1. $A=H^0(X,\mathcal{O}_X)$ is a finite-dimensional $k$-algebra.
  2. $A=\prod_{i=1,\cdots,n} A_i$ is a product of Artinian local $k$-algebras, one factor for each connected component of $X$.
  3. If $X$ is reduced, then $A=\prod_{i=1,\cdots,n} k_i$ is a product of fields, each a finite extension of $k$.

...

  1. 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.

KReiser
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  • Thank you, the reference even has the geometrically integral condition, which I never knew was important in this case. For $X = \mathbb{P}^n_k$, the function field is the fraction field of $k[x_1,...,x_n]$ right? So in the case where $X$ is proper, do we get any difference in $\bar{k}(X)^*$? – oleout Aug 03 '21 at 04:20
  • You're welcome. $\Bbb P^n_k$ is proper over $k$, so you should have your answer: $\overline{k}(X)^\neq \overline{k}^$. – KReiser Aug 03 '21 at 04:25