Let $A$ be a finitely generated algebra over an algebraically closed field $K$, and let $X$ be the ringed space of maximal ideals of $A$. Let $k$ be a subfield of $K$> Here is how $X$ acquires the structure of an affine $k$-variety, as described in Borel's Linear Algebraic Groups.
A $k$-structure on $A$ is a $k$-subalgebra $A_k$ of $A$ such that the natural map $A_k \otimes_k K \rightarrow A$ is an isomorphism of $K$-algebras. An ideal $J$ of $A$ is defined over $k$ (with respect to this $k$-structure) if the span of $J \cap A_k$ over $K$ is equal to $J$, or equivalently $J$ is generated by elements in $A_k$. A $k$-closed set $Z \subseteq X$ is a set which is the zero set of an ideal which is defined over $k$.
Let $U$ be a $k$-open set. Then $U$ has a finite open cover by principal $k$-open sets $D(f_i)$, where $f_i \in A_k$. Here $D(f) = \{ \mathfrak m \in X : f \not\in \mathfrak m \}$.
Since $\mathcal O_X$ is a sheaf, there is an exact sequence of $K$-vector spaces $$0 \rightarrow \mathcal O_X(U) \xrightarrow{\phi} \prod\limits_i \mathcal O_X(D(f_i)) \xrightarrow{\psi} \prod\limits_{i,j} \mathcal O_X(D(f_i) \cap D(f_j))$$ where here $D(f_i) \cap D(f_j) = D(f_if_j)$, and $\mathcal O_X(D(f))$ is canonically isomorphic to $A_f$. So we can interpret this as an exact sequence $$ 0 \rightarrow \mathcal O_X(U) \rightarrow \prod\limits A_{f_i} \rightarrow \prod\limits_{i,j} A_{f_if_j} $$ Since $A_k$ is a $k$-structure for $A$, one can show that $(A_k)_{f}$ is a $k$-structure for $A_f$, hence one has $k$-structures for the domain and codomain of $\psi$, and $\psi$ is defined over $k$ (it maps one $k$-structure into the other). Borel claims that the kernel $C$ of $\psi$ is defined over $k$, but I don't see why this is. This is equivalent to saying that the span of $C \cap \prod\limits_i (A_k)_{f_i}$ over $K$ is $C$. Does anyone see at a glance why this should be true?
In any case, one pulls back the $k$-structure on $C$ to $\mathcal O_X(U)$ by $\phi$, and that is what he defines to be the $k$-structure on $\mathcal O_X(U)$. This will be independent of the choice of open covering.
Wait this is trivial. Let $\psi'$ be the restriction of $\psi$ to $\prod\limits_i (A_k)_{f_i}$. Then we have an exact sequence $$ 0 \rightarrow \textrm{Ker } \psi' \rightarrow \prod\limits_i (A_k)_{f_i} \rightarrow \prod\limits_{i,j} (A_k)_{f_if_j}$$ so tensoring with $K$ gives us the original exact sequence, because $K/k$ is flat. So of course $\textrm{Ker } \psi = \phi^{-1}(\textrm{Ker } \psi \cap \prod\limits_i (A_k)_{f_i})$ is a $k$-structure for $\mathcal O_X(U)$.