Empty Proj and irrelevant torsion

Empty Proj from all-nilpotent. Suppose every homogeneous element of $S_{+}$ is nilpotent and let $\mathfrak{p}\subseteq S$ be a homogeneous prime. All nilpotents lie in $\mathfrak{p}$ by [F1], so $S_{+}\subseteq \mathfrak{p}$ and $\mathfrak{p}$ is not a point of $\mathrm{Proj}S$; hence $\mathrm{Proj}S=\mathrm{\varnothing }$, which is the easy half of (1).

Finite generation forces a nilpotency exponent. Assume $S_{+}=(f_1,\dots ,f_m)$ is finitely generated as an ideal with each $f_i$ homogeneous of positive degree, and assume every homogeneous element of $S_{+}$ is nilpotent. If $m=0$, then $S_{+}=0$ and $S_{+}^1=0$, so take $N=1$. Otherwise choose exponents $e_i\ge 1$ with $f_i^{e_i}=0$ and put $N=e_1+\cdots +e_m$. The ideal $S_{+}^N$ is spanned by the monomials $f_{i_1}\cdots f_{i_N}$: expanding each of $N$ factors of a product of elements of $S_{+}$ as an $S$-combination of the generators exhibits every element of $S_{+}^N$ as an $S$-combination of such monomials. For a monomial let $c_i$ be the number of occurrences of $f_i$; then $c_1+\cdots +c_m=N$, so if $c_i\le e_i-1$ for all $i$ we would get $N\le N-m<N$, a contradiction; hence some $c_i\ge e_i$, the monomial is divisible by $f_i^{e_i}=0$, and the monomial vanishes. Thus $S_{+}^N=0$.

Irrelevant torsion is invisible on every chart. Let $m\in M$ with $S_{+}^{r}m=0$ and let $f\in S_{+}$ be homogeneous of positive degree. Then $f^r\in S_{+}^r$, so $f^{r}m=0$ in $M$, hence the class of $m$ in the full localisation $M_f$ is zero. If a degree-zero fraction $m\mathrm{/}{f}^k$ is formed from such a homogeneous $m$, it too is zero in the degree-zero component $M_{(f)}=(M[f^{-1}]{)}_0$, which by [F5] is $\mathrm{\Gamma }(D_{+}(f),\tilde{M})$. As $f$ was arbitrary, claim (3) follows on every standard chart.

Nonempty chart from a nonnilpotent element. Suppose $f\in S_{+}$ is homogeneous of positive degree and not nilpotent. Then $1=f^{\mathrm{deg}f}\mathrm{/}{f}^{\mathrm{deg}f}\mathrm{\ne }0$ in the degree-zero localisation $S_{(f)}$, because $S_{(f)}=0$ would force $f$ to be nilpotent; hence $S_{(f)}$ is a nonzero commutative ring, its proper ideals form a nonempty poset in which every chain has an upper bound, and Zorn's lemma [F2] provides a maximal ideal $\mathfrak{q}\subseteq S_{(f)}$, which is prime by [F3]. By [F4] there is a homogeneous prime $\mathfrak{p}(\mathfrak{q})\subseteq S$ with $f\mathrm{\notin }\mathfrak{p}(\mathfrak{q})$; since $f\in S_{+}$ this gives $S_{+}\subseteq ̸\mathfrak{p}(\mathfrak{q})$, so $\mathfrak{p}(\mathfrak{q})\in \mathrm{Proj}S$ and $\mathrm{Proj}S\mathrm{\ne }\mathrm{\varnothing }$. Together with step 1.1 this proves (1).

Converse and zero cases. If $S_{+}^N=0$ then every element of $S_{+}$ is nilpotent, so the two conditions of (2) are equivalent; this argument also covers $m=0$, where $S_{+}=0=S_{+}^1$, and the zero ring $S=0$, where $\mathrm{Proj}S=\mathrm{\varnothing }$ by [F1] and $S_{+}$ is nilpotent.

Conclusion. Steps 1.1 and 2.1 prove the emptiness criterion (1), steps 1.2 and 2.2 the finite-generation form (2), and step 1.3 the local vanishing of irrelevant torsion (3). The Axiom of Choice [A1] is used exactly once, through Zorn's lemma [F2], to produce a prime ideal in the nonzero ring $S_{(f)}$ in step 2.1; steps 1.2 and 1.3 are choice-free. No converse of (3) is claimed: without degree-one generation an element can vanish in every chart localisation without being annihilated by a power of $S_{+}$, and this boundary is not decided here. [A1, F2, step 1.2, step 2.1, step 1.3] \qed