regular function projective variety

For a classical projective variety $X$, the homogeneous coordinate ring $S(X)$ is a graded domain by the choice-free irreducibility clause of projective irreducibility homogeneous prime. Its fraction field exists by $\mathrm{Frac}(D)$ is a field and $d\mapsto d\mathrm{/}1$ embeds the integral domain $D$. Define its field of rational functions to be the degree-zero subfield $k(X)=\{\frac{G}{H}\in \mathrm{Frac}(S(X)):G,H\text{ are homogeneous of the same degree and }H\mathrm{\ne }0\}\mathrm{.}$ This is a subfield: multiplying fractions adds numerator and denominator degrees equally, adding them uses a common homogeneous denominator, and a nonzero fraction is inverted by interchanging its numerator and denominator. Zero is represented by $0\mathrm{/}1$.

A rational function $f\in k(X)$ is regular at $p\in X$ if $f=G\mathrm{/}H$ for such homogeneous $G,H$ with $H(p)\mathrm{\ne }0$. It is regular on $U\subseteq X$ if it is regular at every point of $U$. Evaluation is independent of a homogeneous representative because the numerator and denominator scale by the same power; two expressions for the same fraction have equal cross products and hence equal values wherever both denominators are nonzero.

For an open subset $U\subseteq X$, write ${\mathcal{O}}_X(U)$ for the ring of actual functions $s:U\to k$ that are locally evaluations of such equal-degree fractions with nonvanishing denominator. For $U=\mathrm{\varnothing }$ this is the one-element zero ring. For nonempty $U$, these functions identify with the rational functions regular everywhere on $U$: any two local representatives agree as fractions, since their cross difference is homogeneous and vanishes on a nonempty open of the irreducible space $X$, hence on $X$. Nonempty opens intersect, and a homogeneous zero locus is closed, which justifies this assertion. Thus any one local representative gives the common rational function. This construction uses no Axiom of Choice.