Under what condition can we deduce that $\mathcal O_X(X)\otimes_k\bar k = \mathcal O(X_{\bar k})$? I'm reading Qing Liu's book corollary 3.21.
The condition of this corollay is 'X is a reduced connected algebraic variety proper over a field $k$'. Then the author conclude that $K\otimes_k\bar k = \mathcal O(X_{\bar k})$ where $K = \mathcal O_X(X)$ from a proposoition 2.24.
But this book does not have proposition 2.24.... The only theorem might be relevent is propositon 2.14 which shows a similar result but just for geometrically integral variety.
I'm confused how to get this identity.
I'm also wondering, in general, under what condition can we deduce that $\mathcal O_X(X)\otimes_k\bar k = \mathcal O(X_{\bar k})$?
Thank you in advance.



$\newcommand{\Spec}{\mathrm{Spec}}$This is true with essentially no conditions on $X$. Let us, for simplicity though, assume that $X$ is separated.
Suppose that $U_i$ is an affine open cover $X$. Then, $(U_i)_{\overline{k}}$ is an affine open cover $X_{\overline{k}}$. By the sheaf property, we have that
$$0\to H^0(X,\mathcal{O}_{X_{\overline{k}}})\to \prod_i \mathcal{O}((U_i)_{\overline{k}})\to \prod_{i,j}\mathcal{O}((U_i)_{\overline{k}}\cap (U_j)_{\overline{k}})$$
But, note that $(U_i)_{\overline{k}}\cap (U_j)_{\overline{k}}=(U_i\cap U_j)_{\overline{k}}$ and thus it suffices to check the equality
$$\mathcal{O}_X(X_{\overline{k}})=\mathcal{O}_X(X)_{\overline{k}}$$
holds functorially in affine schemes. But, this is clear since
$$\Spec(A)\times_{\Spec(k)}\Spec(\overline{k})=\Spec(A\otimes_k \overline{k})$$
