The adele quotient V_X/(K + A_X) computes H^1 of the structure sheaf

Set up the two ambient objects. By [F1] the generic point $\eta$ has function field $K=k(X)$, and by [F2] and [F4] each closed point $p$ carries the valuation $v_p={\mathrm{ord}}_p$ whose valuation ring ${\mathcal{O}}_{X,p}$ lies in $K$, with $K\subseteq K_p$. By [F3] every $f\in K$ lies in ${\mathcal{O}}_{X,p}\subseteq A_p$ for all but finitely many $p$, so the diagonal map $K\to {\prod }_p{K}_p$ lands in $V_X$; it is injective because each map $K\to K_p$ is an inclusion of fields by [F4]. If $f\in K$ and $(g_p)\in V_X$, then $g_p\in A_p$ for all but finitely many $p$ and $f\in A_p$ for all but finitely many $p$ by [F3], so $f{g}_p\in A_p$ for all but finitely many $p$; hence componentwise multiplication makes $V_X$ a $K$-module containing $K$ as a $k$-subspace. This is part 1; the Axiom of Choice [F9] enters only through the cited suppliers.

Compute the quotient attached to a single $f\in K$. If $f=0$, then $(f{A}_X+A_X)\mathrm{/}{A}_X=0$. Now suppose $f\mathrm{\ne }0$ and let $S=\{p:{\mathrm{ord}}_p(f)<0\}$, finite by [F3], and put $n_p=-{\mathrm{ord}}_p(f)\ge 0$ for $p\in S$. Since $f{A}_p=t_p^{-n_p}{A}_p$ by [F4], inside $K_p$ one has $f{A}_p+A_p=t_p^{-n_p}{A}_p$, and $(f{A}_p+A_p)\mathrm{/}{A}_p=t_p^{-n_p}{A}_p\mathrm{/}{A}_p$ is a $k$-vector space of dimension $n_p[\kappa (p):k]$ by [F5] for $p\in S$, while $f{A}_p\subseteq A_p$ by [F4] for $p\mathrm{\notin }S$. Hence $f{A}_X+A_X\subseteq \left({\prod }_{p\in S}{t}_p^{-n_p}{A}_p\right)\times {\prod }_{p\mathrm{\notin }S}{A}_p$, and the quotient $(f{A}_X+A_X)\mathrm{/}{A}_X$ embeds into ${⨁}_{p\in S}{t}_p^{-n_p}{A}_p\mathrm{/}{A}_p$, a finite-dimensional $k$-vector space. Therefore $f{A}_X<A_X$ for every $f\in K$, and for each such $f$ the multiplication-by-$f$ endomorphism ${\theta }_f$ of $V_X$ satisfies ${\theta }_f(A_X)=f{A}_X<A_X$. This is the first assertion of part 2.

Verify the tail condition. Let $f,g\in K$ and let $S$ be a finite set of closed points outside which $f$, $g$ and, when $g\mathrm{\ne }0$, $g^{-1}$ are all integral; put $T=X\setminus S$. If $f=0$, all three tail images are zero, interpreting the $g^{-1}$ term as zero when $g=0$. If $f\mathrm{\ne }0$ and $g=0$, only $f{A}_T$ remains, and it lies in $A_T$ because $f$ is integral at every $p\in T$. If $f,g\mathrm{\ne }0$, integrality gives $f,g,g^{-1}\in A_p$ at each $p\in T$, so their products give $f{A}_p+fg{A}_p+f{g}^{-1}{A}_p\subseteq A_p$. Taking products proves the tail inclusion. Such finite sets $S$ exist by [F3] applied only to the nonzero functions among $f,g,g^{-1}$. This proves the second assertion of part 2 without taking the order of zero.

Identify $K\cap A_X$ with the global sections. Let $f\in K$. By [F4], $f\in A_p$ if and only if $f\in {\mathcal{O}}_{X,p}$, so $f\in K\cap A_X$ exactly when $f$ is regular at every closed point of $X$. Choose a finite affine cover of $X$ by opens $U=\mathrm{Spec}A$ (possible since $X$ is quasi-compact and of finite type over $k$ by [F1]) and let $f\in K=\mathrm{Frac}(A)$. If $f\mathrm{\notin }A$, the ideal of denominators $I=\{a\in A:af\in A\}$ is a nonzero proper ideal of the Noetherian domain $A$, so it is contained in a maximal ideal $\mathfrak{m}$ of $A$ by [F3]; the closed points of $U$ are exactly the maximal ideals of $A$ [F3], with ${\mathcal{O}}_{X,p}=A_{\mathfrak{m}}$ for the corresponding closed point $p$ by [F1] and [F6], and $f\mathrm{\notin }{A}_{\mathfrak{m}}$, because an expression $f=b\mathrm{/}s$ with $b\in A$ and $s\in A\setminus \mathfrak{m}$ would give $sf=b\in A$, hence $s\in I\subseteq \mathfrak{m}$, a contradiction. Passing to the contrapositive, $f$ regular at every closed point of $U$ forces $f\in A$; applying this on the finitely many charts shows that $f\in K\cap A_X$ if and only if $f\in \mathrm{\Gamma }(X,{\mathcal{O}}_X)=H^0(X,{\mathcal{O}}_X)$. This is part 3.

Build the two auxiliary sheaves. Let $K^0$ be the constant sheaf with value $K$ on $X$, and let $K^1$ be the sheaf $U\mapsto {⨁}_{p\in U}{K}_p\mathrm{/}{A}_p$, the direct sum over the closed points of the skyscraper sheaves with values $K_p\mathrm{/}{A}_p$; this is a sheaf because every open subset of $X$ is quasi-compact by [F1], so a compatible family over a cover is already determined by its members over a finite subcover, and it is flasque because a section over an open $U$ extends by zero to $X$ by [F7]. Define the map of sheaves $\delta :K^0\to K^1$ on sections over a nonempty open $U$ by sending $f\in K=K^0(U)$ to the class $(f+A_p{)}_{p\in U}$; this is well defined because $f\in A_p$ for all but finitely many $p$ by [F3], so the family has finite support. Then $\ker \delta$ has sections over a nonempty open $U$ equal to $\{f\in K:f\in A_p$ for every closed point $p\in U\}$, and the cokernel of $\delta$ on global sections is $V_X\mathrm{/}(K+A_X)$.

Prove that $0\to {\mathcal{O}}_X\to K^0\stackrel{\delta }{\to }{K}^1\to 0$ is exact. The map ${\mathcal{O}}_X\to K^0$ sends a regular function to itself viewed in $K$; it is injective on every nonempty open because $\mathrm{\Gamma }(U,{\mathcal{O}}_X)\subseteq K$, and its image is killed by $\delta$. By construction $\ker \delta$ has sections over a nonempty open $U$ equal to $\{f\in K:f\in A_p$ for all closed points $p\in U\}$, which by the chart computation of step 1.4 (applied to the members of a finite affine cover of the quasi-compact open $U$) is exactly $\mathrm{\Gamma }(U,{\mathcal{O}}_X)$; hence $\ker \delta ={\mathcal{O}}_X$. For the surjectivity of $\delta$ it suffices to check stalks: at a closed point $p$ the stalk of $K^0$ is $K$ and the stalk of $K^1$ is $K_p\mathrm{/}{A}_p$, and the map $K\to K_p\mathrm{/}{A}_p$ is surjective because $K+A_p=K_p$: since $K_p=A_p[t_p^{-1}]$ by [F4], every $x\in K_p$ equals $c{t}_p^{-m}$ with $c\in A_p$ and $m\ge 0$, and the image of the completion map ${\mathcal{O}}_{X,p}\to A_p$ being dense supplies $b\in {\mathcal{O}}_{X,p}$ with $b\equiv c<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>t_p^m{A}_p$, so that $b{t}_p^{-m}\in K$ and $x-b{t}_p^{-m}\in A_p$. At the generic point $\eta$ the stalk of $K^1$ is $0$ because the class of a local section of the direct sum of skyscrapers is killed after removing its finitely many support points, so $\delta$ is surjective there as well. Thus the sequence of sheaves is exact.

Compute the cohomology of $K^0$ and $K^1$. Since $K^0$ is the constant sheaf with value $K$ on the irreducible space $X$ [F1], [F7] gives $H^0(X,K^0)=K$ and $H^q(X,K^0)=0$ for every $q>0$. Since $K^1$ is flasque, [F7] gives $H^q(X,K^1)=0$ for every $q>0$, and its global sections are $H^0(X,K^1)={⨁}_p{K}_p\mathrm{/}{A}_p=V_X\mathrm{/}{A}_X$, the direct sum over all closed points, again by [F7].

Apply the long exact sequence. The short exact sequence of step 2.1 yields, by [F6], the exact sequence $0\to H^0(X,{\mathcal{O}}_X)\to K\to V_X\mathrm{/}{A}_X\to H^1(X,{\mathcal{O}}_X)\to 0$, where the middle map $K\to V_X\mathrm{/}{A}_X$ sends $f$ to $(f+A_p{)}_p$ and its cokernel is $V_X\mathrm{/}(K+A_X)$; here the cohomology of $K^0$ and $K^1$ is that computed in step 2.2. Hence $H^1(X,{\mathcal{O}}_X)\cong V_X\mathrm{/}(K+A_X)$ canonically as $k$-vector spaces, and the injective map $H^0(X,{\mathcal{O}}_X)\to K$ identifies $H^0(X,{\mathcal{O}}_X)$ with the kernel of $K\to V_X\mathrm{/}{A}_X$, which is $K\cap A_X$ in agreement with step 1.4. This is part 4, except for the finiteness of $V_X\mathrm{/}(K+A_X)$.

Finiteness and the constant-field case. Since $X$ is proper over $k$ and ${\mathcal{O}}_X$ is coherent, [F8] gives ${\dim }_k{H}^1(X,{\mathcal{O}}_X)<\mathrm{\infty }$, so step 3.1 makes $V_X\mathrm{/}(K+A_X)$ finite-dimensional, which completes part 4. For part 5 let $L:=K\cap A_X=H^0(X,{\mathcal{O}}_X)$ by step 1.4; by [F8] the ring $L$ is a finite-dimensional $k$-algebra and a domain, and $L\mathrm{\ne }0$ because it contains the unit of $K$. If $X$ is geometrically integral and $\bar{k}\mathrm{/}k$ is an algebraic closure, then $K{\otimes }_k\bar{k}$ is a domain by [F8], and the inclusion $L\hookrightarrow K$ remains injective after tensoring with the flat $k$-module $\bar{k}$, so $L{\otimes }_k\bar{k}$ is a domain of dimension $[L:k]$ over the algebraically closed field $\bar{k}$. A finite-dimensional commutative domain over a field is a field, since a nonzero element has injective and hence bijective multiplication; thus $L{\otimes }_k\bar{k}$ is a field of degree $[L:k]$ over $\bar{k}$, forcing $[L:k]=1$ and $L=k$, which is part 5.