Prime and local-ring correspondence on standard projective charts

Let $z\in S_{x_i}$ be homogeneous of degree $m$ and write $z=s\mathrm{/}{x}_i^e$ with $s\in S$ and $e\ge 0$; writing $s={\sum }_d{s}_d$ for the homogeneous components of $s$, the elements $s_d\mathrm{/}{x}_i^e$ are homogeneous of degree $d-e$ in $S_{x_i}$ and sum to $z$. Since $z$ is homogeneous of degree $m$, only the summand with $d-e=m$ can be nonzero, so $z=s_{m+e}\mathrm{/}{x}_i^e=(s_{m+e}\mathrm{/}{x}_i^{m+e})x_i^m$ with $a:=s_{m+e}\mathrm{/}{x}_i^{m+e}\in (S_{x_i}{)}_0=A_i$. Because the direct-sum expression in a graded ring is unique [L2], it follows that $S_{x_i}={⨁}_{m\in \mathbb{Z}}{A}_i{x}_i^m$, that $a_m{x}_i^m$ is the degree-$m$ component of ${\sum }_m{a}_m{x}_i^m$, and that $x_i$ is a unit; in particular the degree-$m$ part of $S_{x_i}$ is exactly $A_i{x}_i^m$.

Let $k\subseteq K$ be a field extension and $S_K=S{\otimes }_{k}K$, graded by $\mathrm{deg}(s\otimes \mu )=\mathrm{deg}s$. The $k$-algebra maps $S_{x_i}\to (S_K{)}_{x_i}$, $s\mathrm{/}{x}_i^e\mapsto (s\otimes 1)(x_i\otimes 1{)}^{-e}$, and $K\to (S_K{)}_{x_i}$, $\mu \mapsto 1\otimes \mu$, induce by the coproduct property [L8] a $k$-algebra map $S_{x_i}{\otimes }_{k}K\to (S_K{)}_{x_i}$; conversely the map $S_K\to S_{x_i}{\otimes }_{k}K$ built from $S\to S_{x_i}{\otimes }_{k}K$ and $K\to S_{x_i}{\otimes }_{k}K$ inverts $x_i\otimes 1$, whose inverse is $x_i^{-1}\otimes 1$, so by [L7] it induces $(S_K{)}_{x_i}\to S_{x_i}{\otimes }_{k}K$. Each composite is a map fixing the base ring and the inverted element, hence is the identity by the uniqueness in [L7] and [L8]; thus $(S_K{)}_{x_i}\cong S_{x_i}{\otimes }_{k}K$, compatibly with the gradings, since the maps send homogeneous elements to homogeneous elements of the same degree. As $K$ lies in degree zero, the degree-$m$ part of $S_{x_i}{\otimes }_{k}K$ is $(S_{x_i}{)}_m{\otimes }_{k}K=A_i{x}_i^m{\otimes }_{k}K$ by 1.1, so $((S_K{)}_{x_i}{)}_0=A_i{\otimes }_{k}K$ and $S_K$ is standard graded over $K$ with the images of the $x_i$ as degree-one generators.

If $\mathfrak{a}\subseteq S$ is homogeneous, then $\mathfrak{a}{S}_{x_i}$ is a homogeneous ideal of $S_{x_i}$: writing an element as a finite sum ${\sum }_{\mathrm{\ell }}{a}^{(\mathrm{\ell })}\mathrm{/}{x}_i^{e_{\mathrm{\ell }}}$ with $a^{(\mathrm{\ell })}\in \mathfrak{a}$ and decomposing each $a^{(\mathrm{\ell })}={\sum }_d{a}_d^{(\mathrm{\ell })}$ into components, which lie in $\mathfrak{a}$ by [L2], exhibits the element as a sum of homogeneous elements of $S_{x_i}$, so each of its homogeneous components lies in $\mathfrak{a}{S}_{x_i}$. If $\mathfrak{b}\subseteq S_{x_i}$ is a homogeneous ideal and $s={\sum }_d{s}_d\in {\lambda }_i^{-1}(\mathfrak{b})$, then ${\lambda }_i(s)={\sum }_d{\lambda }_i(s_d)$ with ${\lambda }_i(s_d)$ homogeneous of degree $d$ by [L1], so each ${\lambda }_i(s_d)\in \mathfrak{b}$ and hence each $s_d\in {\lambda }_i^{-1}(\mathfrak{b})$. Thus extension and contraction along ${\lambda }_i$ preserve homogeneity of ideals.

Let ${\mathfrak{p}}_0\subseteq A_i$ be a prime. By 1.1 the ideal it generates in $S_{x_i}$ is ${\mathfrak{p}}_0{S}_{x_i}={⨁}_{m\in \mathbb{Z}}{\mathfrak{p}}_0{x}_i^m$, which is homogeneous and satisfies ${\mathfrak{p}}_0{S}_{x_i}\cap A_i={\mathfrak{p}}_0$; its quotient is ${⨁}_m(A_i\mathrm{/}{\mathfrak{p}}_0)x_i^m$, a ring in which $x_i$ is a unit, and a product of two nonzero elements there has nonzero coefficient at the lowest occurring power of $x_i$ because $A_i\mathrm{/}{\mathfrak{p}}_0$ is a domain by [L6]. Hence ${\mathfrak{p}}_0{S}_{x_i}$ is a homogeneous prime. Conversely, if $\mathfrak{q}\subseteq S_{x_i}$ is a homogeneous prime with ${\mathfrak{q}}_0:=\mathfrak{q}\cap A_i$, then every homogeneous $z\in \mathfrak{q}$ of degree $m$ equals $a{x}_i^m$ with $a\in A_i$ by 1.1, and $a=x_i^{-m}z\in \mathfrak{q}$ because $x_i$ is a unit, so $z\in {\mathfrak{q}}_0{S}_{x_i}$; since homogeneous ideals are generated by their homogeneous elements [L2] we get $\mathfrak{q}={\mathfrak{q}}_0{S}_{x_i}$. Therefore ${\mathfrak{p}}_0\mapsto {\mathfrak{p}}_0{S}_{x_i}$ and $\mathfrak{q}\mapsto \mathfrak{q}\cap A_i$ are inverse bijections between $\mathrm{Spec}(A_i)$ and the homogeneous primes of $S_{x_i}$, both given by extension respectively contraction of ideals and therefore inclusion-preserving.

Applying [L3] to $S$ and the multiplicative set of powers of $x_i$: contraction along ${\lambda }_i$ is an inclusion-preserving bijection from $\mathrm{Spec}(S_{x_i})$ onto the primes of $S$ avoiding $x_i$, with inverse $\mathfrak{p}\mapsto \mathfrak{p}{S}_{x_i}$. By 2.2 this bijection and its inverse carry homogeneous primes to homogeneous primes, and since the two maps are given by contraction and extension of ideals, they restrict to inverse inclusion-preserving bijections between the homogeneous primes of $S$ avoiding $x_i$ and the homogeneous primes of $S_{x_i}$.

Composing the bijections of 3.1 and 2.3 gives inverse bijections between the homogeneous primes $\mathfrak{p}$ of $S$ with $x_i\mathrm{\notin }\mathfrak{p}$ and the points ${\mathfrak{p}}_0$ of $\mathrm{Spec}(A_i)$: the composite sends $\mathfrak{p}$ to $\mathfrak{p}{S}_{x_i}\cap A_i$ and ${\mathfrak{p}}_0$ to ${\lambda }_i^{-1}({\mathfrak{p}}_0{S}_{x_i})$, both maps are inclusion-preserving, and each composite is the identity because ${\mathfrak{p}}_0{S}_{x_i}\cap A_i={\mathfrak{p}}_0$ by 2.3 and because for a homogeneous prime $\mathfrak{p}$ of $S$ avoiding $x_i$, the ideal $\mathfrak{p}{S}_{x_i}$ is a homogeneous prime of $S_{x_i}$ by 3.1 to which $\mathfrak{q}=(\mathfrak{q}\cap A_i)S_{x_i}$ applies, while its contraction to $S$ is $\mathfrak{p}$ by the inverse property in [L3]. A homogeneous prime with $x_i\mathrm{\notin }\mathfrak{p}$ satisfies $S_{+}⊈\mathfrak{p}$ since $x_i\in S_{+}$, and it lies in $D_{+}(x_i)$ by [L1]; these are exactly the points of the chart. This is claim 1.

By [L1] the chart $D_{+}(x_i)$ is the affine scheme $\mathrm{Spec}(A_i)$ and, by 4.1, the point of $D_{+}(x_i)$ corresponding to $\mathfrak{p}$ is the prime ${\mathfrak{p}}_0=\mathfrak{p}{S}_{x_i}\cap A_i$ of $A_i$. Since the stalk of a scheme at a point may be computed in any open chart containing it and the stalk of an affine scheme at a prime is the localization at that prime [L4], we get ${\mathcal{O}}_{X,\mathfrak{p}}\cong (A_i{)}_{{\mathfrak{p}}_0}$. This is claim 2.

Assume now $x_i,x_j\mathrm{\notin }\mathfrak{p}$ and set ${\mathfrak{p}}_{0,i}=\mathfrak{p}{S}_{x_i}\cap A_i$, ${\mathfrak{p}}_{0,j}=\mathfrak{p}{S}_{x_j}\cap A_j$. By [L3] applied to $S$ and the multiplicative set generated by $x_i$ and $x_j$, which $\mathfrak{p}$ avoids, the ideal $\mathfrak{p}{S}_{x_i{x}_j}$ is a prime of $S_{x_i{x}_j}$, so ${\mathfrak{p}}_{0,ij}:=\mathfrak{p}{S}_{x_i{x}_j}\cap A_{ij}$ is a prime of $A_{ij}$ by [L8]. Its contraction $Q:=\mathfrak{p}{S}_{x_i{x}_j}\cap S_{x_i}$ is a prime of $S_{x_i}$ containing $\mathfrak{p}{S}_{x_i}$, and $Q\cap S=\mathfrak{p}$: an element $z\in Q$ satisfies $x_j^{N}z\in \mathfrak{p}{S}_{x_i}$ for some $N\ge 0$, because $\mathfrak{p}{S}_{x_i{x}_j}$ is the localization of $\mathfrak{p}{S}_{x_i}$ at $x_j$ and $\frac{z}{1}$ is a fraction with numerator in $\mathfrak{p}{S}_{x_i}$; so for $s\in S\cap Q$ we get $x_j^{N}s\in \mathfrak{p}{S}_{x_i}\cap S=\mathfrak{p}$ by the inverse property in [L3], whence $s\in \mathfrak{p}$ as $x_j\mathrm{\notin }\mathfrak{p}$ and $\mathfrak{p}$ is prime. Contraction is injective on primes of $S_{x_i}$ by [L3], so $Q=\mathfrak{p}{S}_{x_i}$, and intersecting with $A_i\subseteq S_{x_i}$ gives ${\mathfrak{p}}_{0,ij}\cap A_i={\mathfrak{p}}_{0,i}$. By [L1] the ring $A_{ij}$ is the localization of $A_i$ at the element $x_j\mathrm{/}{x}_i$, which does not lie in ${\mathfrak{p}}_{0,i}$: otherwise $x_j=x_i\cdot (x_j\mathrm{/}{x}_i)\in \mathfrak{p}{S}_{x_i}\cap S=\mathfrak{p}$, since the contraction of $\mathfrak{p}{S}_{x_i}$ to $S$ is $\mathfrak{p}$. So [L3] applied to $A_i$ and the powers of $x_j\mathrm{/}{x}_i$ shows that ${\mathfrak{p}}_{0,i}{A}_{ij}$ is the unique prime of $A_{ij}$ contracting to ${\mathfrak{p}}_{0,i}$; since ${\mathfrak{p}}_{0,ij}$ is such a prime, ${\mathfrak{p}}_{0,ij}={\mathfrak{p}}_{0,i}{A}_{ij}$, and by the same argument with $i,j$ exchanged, ${\mathfrak{p}}_{0,ij}={\mathfrak{p}}_{0,j}{A}_{ij}$. Hence ${\mathfrak{p}}_{0,i}{A}_{ij}={\mathfrak{p}}_{0,j}{A}_{ij}$.

Let $\mathfrak{q}$ be a homogeneous prime of $S_K$ with $x_i\mathrm{\notin }\mathfrak{q}$ and $\mathfrak{p}=\mathfrak{q}\cap S$. Since $S\to S_K$, $s\mapsto s\otimes 1$, is degree-preserving, the degree-$d$ component of $\mathfrak{p}$ is the contraction of the degree-$d$ component of $\mathfrak{q}$; hence $\mathfrak{p}$ is homogeneous [L2], and $x_i\mathrm{\notin }\mathfrak{p}$ is clear. Let ${\mathfrak{q}}_0=\mathfrak{q}(S_K{)}_{x_i}\cap (A_i{\otimes }_{k}K)$ and ${\mathfrak{p}}_0=\mathfrak{p}{S}_{x_i}\cap A_i$ be the chart primes of $\mathfrak{q}$ for $X_K=\mathrm{Proj}{S}_K$ and of $\mathfrak{p}$, both given by claim 1 (4.1) applied to the standard graded $K$-algebra $S_K$ of 2.1 and to $S$. Under the identifications of 2.1 we have $A_i\subseteq A_i{\otimes }_{k}K\subseteq (S_K{)}_{x_i}$ and $S_{x_i}\subseteq (S_K{)}_{x_i}$. The contraction $Q:=\mathfrak{q}(S_K{)}_{x_i}\cap S_{x_i}$ is a prime of $S_{x_i}$ containing $\mathfrak{p}{S}_{x_i}$ whose contraction to $S$ is $\mathfrak{p}$: if $s\in S\cap Q$ then $x_i^{N}s\in \mathfrak{q}$ for some $N\ge 0$, so $x_i^{N}s\in \mathfrak{q}\cap S=\mathfrak{p}$ and hence $s\in \mathfrak{p}$. Contraction is injective on primes of $S_{x_i}$ by [L3], so $Q=\mathfrak{p}{S}_{x_i}$ and therefore ${\mathfrak{q}}_0\cap A_i=Q\cap A_i=\mathfrak{p}{S}_{x_i}\cap A_i={\mathfrak{p}}_0$.

With the notation of 5.2, the point $\mathfrak{p}$ lies in both charts, so by 5.1 its stalk is $(A_i{)}_{{\mathfrak{p}}_{0,i}}$ when computed in chart $i$ and $(A_j{)}_{{\mathfrak{p}}_{0,j}}$ when computed in chart $j$. Since $A_{ij}$ is the localization of $A_i$ at $x_j\mathrm{/}{x}_i$, the general form of [L5] applied to the multiplicative set $A_i\setminus {\mathfrak{p}}_{0,i}$ gives $(A_i{)}_{{\mathfrak{p}}_{0,i}}\cong (A_{ij}{)}_{{\mathfrak{p}}_{0,i}{A}_{ij}}$; indeed, put $f=x_j\mathrm{/}{x}_i\mathrm{\notin }{\mathfrak{p}}_{0,i}$. Every element outside ${\mathfrak{p}}_{0,i}{A}_{ij}$ is a fraction $a\mathrm{/}{f}^n$ with $a\mathrm{\notin }{\mathfrak{p}}_{0,i}$. After inverting the images of $A_i\setminus {\mathfrak{p}}_{0,i}$, such a fraction is a unit, with inverse $f^n\mathrm{/}a$. Conversely every image of an element of $A_i\setminus {\mathfrak{p}}_{0,i}$ is outside the extended prime. Thus inverting these images and inverting the entire prime complement have the same universal property [L7]. By 5.2, ${\mathfrak{p}}_{0,i}{A}_{ij}={\mathfrak{p}}_{0,ij}={\mathfrak{p}}_{0,j}{A}_{ij}$, so both charts give the localization of $A_{ij}$ at ${\mathfrak{p}}_{0,ij}$; in particular $(A_i{)}_{{\mathfrak{p}}_{0,i}}\cong (A_j{)}_{{\mathfrak{p}}_{0,j}}$, which completes claim 3.

With the notation of 5.3 and 2.1, 5.1 applied to the chart $(A_i{\otimes }_{k}K{)}_{{\mathfrak{q}}_0}$ of $X_K$ gives ${\mathcal{O}}_{X_K,\mathfrak{q}}\cong (A_i{\otimes }_{k}K{)}_{{\mathfrak{q}}_0}$, while ${\mathcal{O}}_{X,\mathfrak{p}}{\otimes }_{k}K\cong (A_i{)}_{{\mathfrak{p}}_0}{\otimes }_{k}K$. Put $B=A_i{\otimes }_{k}K$ and $S_0=A_i\setminus {\mathfrak{p}}_0\subseteq B$. The $k$-algebras $S_0^{-1}B$ and $(A_i{)}_{{\mathfrak{p}}_0}{\otimes }_{k}K$ are canonically isomorphic: by [L7] and [L8], homomorphisms from either of them into a commutative $k$-algebra $C$ correspond naturally to a pair consisting of a $k$-algebra map $A_i\to C$ inverting $S_0$ and a $k$-algebra map $K\to C$, so both are localizations of $B$ at $S_0$ and [L7] identifies them. By 5.3 we have ${\mathfrak{q}}_0\cap A_i={\mathfrak{p}}_0$, so $S_0\subseteq B\setminus {\mathfrak{q}}_0$; the general form of [L5] applied to the multiplicative sets $S_0$ and $B\setminus {\mathfrak{q}}_0$ in $B$ therefore gives ${\mathcal{O}}_{X_K,\mathfrak{q}}\cong B_{{\mathfrak{q}}_0}\cong {\overline{(B\setminus {\mathfrak{q}}_0)}}^{-1}\left(S_0^{-1}B\right),$ the localization of ${\mathcal{O}}_{X,\mathfrak{p}}{\otimes }_{k}K\cong S_0^{-1}B$ at the multiplicative set generated by the image of $B\setminus {\mathfrak{q}}_0$. Every element outside the prime ${\mathfrak{q}}_0{S}_0^{-1}B$ is a fraction $b\mathrm{/}s$ with $b\mathrm{\notin }{\mathfrak{q}}_0$ and $s\in S_0$. Inverting the image of $B\setminus {\mathfrak{q}}_0$ makes $b\mathrm{/}s$ a unit, with inverse $s\mathrm{/}b$. Conversely those images lie outside the extended prime, so the two localizations have the same universal property [L7]. Hence ${\mathcal{O}}_{X_K,\mathfrak{q}}$ is the localization of ${\mathcal{O}}_{X,\mathfrak{p}}{\otimes }_{k}K$ at the prime induced by ${\mathfrak{q}}_0$, which completes claim 4.

Claim 1 is 4.1 and claim 2 is 5.1; claim 3 is 5.2 together with 6.1; claim 4 is 2.1, 5.3 and 6.2. The degenerate case is covered: if $x_i$ is nilpotent in $S$, then $1=0$ in $A_i$, so $\mathrm{Spec}(A_i)$ is empty, and no prime of $S$ avoids $x_i$, so both sides of the bijection in claim 1 are empty, consistently with $D_{+}(x_i)=\mathrm{Spec}0$. AC is inherited through the affine chart sheaves in [L1]; the algebraic correspondences use no additional choice, and no hypothesis on $I$ beyond homogeneity is needed. The arguments are valid over an arbitrary field $k$.