Prime correspondence on a Proj chart

The set $S_{(f)}$ of the Statement is a subring of $S_f$: it contains $0=0\mathrm{/}{f}^0$ and $1=f^0\mathrm{/}{f}^0$, and if $a\mathrm{/}{f}^k$ and $b\mathrm{/}{f}^m$ have $\mathrm{deg}a=kd$ and $\mathrm{deg}b=md$, then $(a\mathrm{/}{f}^k)+(b\mathrm{/}{f}^m)=(a{f}^m+b{f}^k)\mathrm{/}{f}^{k+m}$ has homogeneous numerator $a{f}^m+b{f}^k$ of degree $(k+m)d$, while $(a\mathrm{/}{f}^k)(b\mathrm{/}{f}^m)=ab\mathrm{/}{f}^{k+m}$ has homogeneous numerator $ab$ of degree $(k+m)d$. For a homogeneous $a$ of degree $kd$ the element $a^d\mathrm{/}{f}^{\mathrm{deg}a}=a^d\mathrm{/}{f}^{kd}=(a\mathrm{/}{f}^k{)}^d$ lies in $S_{(f)}$, since $a^d$ is homogeneous of degree $k{d}^2=kd\cdot d$ with $k{d}^2=(kd)\cdot d$; this power test is the basis of both constructions.

For a prime $\mathfrak{q}\subseteq S_{(f)}$, the set $P(\mathfrak{q})$ is closed under addition inside each degree: if $a,b\in P(\mathfrak{q})\cap S_e$, then expanding the $2d$-th power and grouping the terms of $(a+b{)}^{2d}\mathrm{/}{f}^{2e}$ by the number $i$ of $a$-factors, every monomial $a^i{b}^{2d-i}\mathrm{/}{f}^{2e}$ has $i\ge d$ or $2d-i\ge d$, and correspondingly equals $(a^d\mathrm{/}{f}^e)\cdot (a^{i-d}{b}^{2d-i}\mathrm{/}{f}^e)$ or $(b^d\mathrm{/}{f}^e)\cdot (a^i{b}^{d-i}\mathrm{/}{f}^e)$, a product of an element of $\mathfrak{q}$ (namely $a^d\mathrm{/}{f}^e$ or $b^d\mathrm{/}{f}^e$, which lies in $\mathfrak{q}$ because $a,b\in P(\mathfrak{q})$ and $e=\mathrm{deg}a=\mathrm{deg}b$) with an element of $S_{(f)}$ as in step 1.1; hence $((a+b{)}^d\mathrm{/}{f}^e{)}^2=(a+b{)}^{2d}\mathrm{/}{f}^{2e}\in \mathfrak{q}$, and since $\mathfrak{q}$ is prime and radical, $(a+b{)}^d\mathrm{/}{f}^e\in \mathfrak{q}$, that is, $a+b\in P(\mathfrak{q})\cap S_e$.

For a homogeneous prime $\mathfrak{p}\subseteq S$ with $f\mathrm{\notin }\mathfrak{p}$, membership in $Q(\mathfrak{p})$ is independent of a fraction representation: if $a\mathrm{/}{f}^k=b\mathrm{/}{f}^m$ in $S_f$, [F5] gives $f^N(f^{m}a-f^{k}b)=0$ for some $N$, and primality with $f\mathrm{\notin }\mathfrak{p}$ gives $a\in \mathfrak{p}$ if and only if $b\in \mathfrak{p}$. The set $Q(\mathfrak{p})$ of the Statement is therefore an ideal of $S_{(f)}$: it contains $0=0\mathrm{/}{f}^0$; if $a\mathrm{/}{f}^k,b\mathrm{/}{f}^m\in Q(\mathfrak{p})$ then their sum has numerator $a{f}^m+b{f}^k\in \mathfrak{p}$, homogeneous of degree $(k+m)d$, and multiplication by any fraction $c\mathrm{/}{f}^{\mathrm{\ell }}\in S_{(f)}$ gives numerator $ac\in \mathfrak{p}$. It is proper because $1=f^k\mathrm{/}{f}^k\in Q(\mathfrak{p})$ would force $f\in \mathfrak{p}$. It is prime because if a product represented by $ab\mathrm{/}{f}^{k+m}$ lies in $Q(\mathfrak{p})$, then $ab\in \mathfrak{p}$, so the homogeneous numerator $a$ or $b$ lies in $\mathfrak{p}$ and the corresponding factor lies in $Q(\mathfrak{p})$.

For a prime $\mathfrak{q}\subseteq S_{(f)}$, the span $\mathfrak{p}(\mathfrak{q})={⨁}_e(P(\mathfrak{q})\cap S_e)$ is a homogeneous ideal of $S$: it contains $0$ and is closed under addition componentwise by step 2.1. For homogeneous $s\in S_m$ and $a\in P(\mathfrak{q})\cap S_e$ one has $(sa{)}^d\mathrm{/}{f}^{e+m}=(a^d\mathrm{/}{f}^e)\cdot (s^d\mathrm{/}{f}^m)\in \mathfrak{q}$ because $s^d\mathrm{/}{f}^m\in S_{(f)}$, so $sa\in P(\mathfrak{q})$. Distributing products of finite sums of homogeneous elements extends this closure to arbitrary $s\in S$ and arbitrary elements of the direct sum. Thus the direct sum is an ideal, and it is exactly the ideal generated by the homogeneous set $P(\mathfrak{q})$.

For a prime $\mathfrak{q}\subseteq S_{(f)}$ the ideal $\mathfrak{p}(\mathfrak{q})$ is prime and $f\mathrm{\notin }\mathfrak{p}(\mathfrak{q})$: if $a,b\in S$ are homogeneous with $ab\in \mathfrak{p}(\mathfrak{q})$, then $(ab{)}^d\mathrm{/}{f}^{\mathrm{deg}a+\mathrm{deg}b}=(a^d\mathrm{/}{f}^{\mathrm{deg}a})(b^d\mathrm{/}{f}^{\mathrm{deg}b})\in \mathfrak{q}$, so one of the two factors lies in $\mathfrak{q}$, that is, $a\in P(\mathfrak{q})$ or $b\in P(\mathfrak{q})$. This homogeneous test makes the homogeneous ideal prime: if arbitrary $u,v$ have nonzero images in $S\mathrm{/}\mathfrak{p}(\mathfrak{q})$, take their highest-degree nonzero homogeneous components; their product is the unique component of the highest possible degree in $uv$ and is nonzero by the homogeneous test. Thus $uv\mathrm{\notin }\mathfrak{p}(\mathfrak{q})$. Moreover $f^d\mathrm{/}{f}^d=1\mathrm{\notin }\mathfrak{q}$, so $f\mathrm{\notin }P(\mathfrak{q})$ and, $f$ being homogeneous, $f\mathrm{\notin }\mathfrak{p}(\mathfrak{q})$.

The composite $Q\circ \mathfrak{p}(\cdot )$ is the identity: for a prime $\mathfrak{q}\subseteq S_{(f)}$ and an element $y=a\mathrm{/}{f}^k\in S_{(f)}$ with $\mathrm{deg}a=kd$, one has $y\in Q(\mathfrak{p}(\mathfrak{q}))$ if and only if $a\in \mathfrak{p}(\mathfrak{q})$, if and only if $y^d=a^d\mathrm{/}{f}^{kd}\in \mathfrak{q}$, if and only if $y\in \mathfrak{q}$, the last step because $\mathfrak{q}$ is prime and hence radical in both directions.

The composite $\mathfrak{p}\circ Q$ is the identity: for a homogeneous prime $\mathfrak{p}$ with $f\mathrm{\notin }\mathfrak{p}$ and homogeneous $a\in S$, one has $a\in P(Q(\mathfrak{p}))$ if and only if $a^d\mathrm{/}{f}^{\mathrm{deg}a}\in Q(\mathfrak{p})$, if and only if $a^d\in \mathfrak{p}$, if and only if $a\in \mathfrak{p}$, the last step because $\mathfrak{p}$ is prime and hence radical. Since $\mathfrak{p}(\mathfrak{q})$ and both prime sets consist of homogeneous data, steps 4.1, 2.2, 4.2 and 5.1 exhibit mutually inverse bijections between the primes of $S_{(f)}$ and the homogeneous primes of $S$ avoiding $f$, which are exactly the points of $D_{+}(f)$ by [F1] and [F2].

The nilpotent case is consistent with the bijection and needs no prime-existence input: if $f^m=0$ for some $m\ge 1$, then $1\mathrm{/}1=f^m\mathrm{/}{f}^m=0$ in $S_f$ by [F5], so $S_f=0$ and $S_{(f)}=0$ has no prime ideals, while $f$ lies in every prime ideal of $S$ and hence $D_{+}(f)=\mathrm{\varnothing }$; both sides of the bijection are empty. If $f$ is not nilpotent then $S_f$ has $1\mathrm{\ne }0$ by [F5]; the bijection is between two sets that are simultaneously empty or nonempty by steps 4.2 and 5.1, and no converse emptiness statement is claimed here.

Distinguished opens: let $g\in S_{+}$ be homogeneous of positive degree $e$ and put $\tau =g^d\mathrm{/}{f}^e\in S_{(f)}$, a well-defined element because $g^d$ is homogeneous of degree $de=ed$. For $\mathfrak{p}$ a homogeneous prime with $f\mathrm{\notin }\mathfrak{p}$ and $\mathfrak{q}=Q(\mathfrak{p})$, step 5.1 gives $\tau \in \mathfrak{q}=Q(\mathfrak{p})$ if and only if $g^d\in \mathfrak{p}$, if and only if $g\in \mathfrak{p}$, the last because $\mathfrak{p}$ is prime and radical. Hence $\mathfrak{p}\in D_{+}(fg)$ if and only if $fg\mathrm{\notin }\mathfrak{p}$, if and only if $g\mathrm{\notin }\mathfrak{p}$ (as $f\mathrm{\notin }\mathfrak{p}$ and $\mathfrak{p}$ is prime), if and only if $\tau \mathrm{\notin }Q(\mathfrak{p})$, that is, $Q(\mathfrak{p})\in D(\tau )$. By [F2] we have $D_{+}(fg)=D_{+}(f)\cap D_{+}(g)$, so the bijection of step 5.1 restricts to a bijection $D_{+}(fg)\to D(\tau )$ onto the distinguished open of $\tau$ in $\mathrm{Spec}{S}_{(f)}$.

Summary. Steps 1.1-5.1 prove clauses (1)-(3) of the Statement and step 6.2 proves clause (4); the maps are given by explicit formulas in terms of the power test $a\mapsto a^d\mathrm{/}{f}^{\mathrm{deg}a}$, no choice is made in either construction, and [A1] enters only as the inherited hypothesis on the ambient affine spectrum, so the bijection itself is choice-free.