Let $A= \bigoplus_{n \ge 0} A_n$ be a commutative Noetherian graded ring and $f \in A_d$ a nonzero homogeneous element of degree $d>0$. The natural ring map $q:A \to A/(f)$ induces a well defined map $q^*: \operatorname{Proj}(A/(f)) \to \operatorname{Proj}(A)$ of Proj's.
Leading Question: Assume that $q^*$ is a homeomorphism, what can we say about algebraic properties $f$?
Especially, "how close" is $f$ to be nilpotent? (Recall, in "affine situation", if $A \to A/(f)$ induces a homeomorphism $\operatorname{Spec}(A/(f)) \to \operatorname{Spec}(A)$, then $f$ is nilpotent.
My initial motivation was to vastly generalize Hartshorne's Ex II.5.14(only part a) I intended to discuss here in MSE. Mostly, I wasn't happy with the requirement there the base field $k$ to assume to be closed (presumably to use there Hilbert's Nullstellensatz), because I'm not sure if that's really neccessary there.
Let's try to generalize it to arbitrary Noetherian graded rings, because I'm pretty sure that part (a) holds in much broader setting and is closely connected with the "Leading Question" I posted above.
Let $A= \bigoplus_{n \ge 0} A_n$ as before a Noetherian graded ring with irrelevant ideal $A_{+}=\bigoplus_{n \ge 1} A_n$ generated by homogeneous elements $t_1,t_2,\ldots, t_d$ and assume $A$ is saturated in the sense that if $a \in A$ with $a =0$ in every localization $ A_{t_i}$ (equiv $a$ killed by approp powers of $t_i$), then already $a = 0$ (in analogy to saturated ideal inside polynomial ring).
Now assume that $P:= \operatorname{Proj}(A)$ is a domain; equivalently all degree-$0$ localizations $A_{(t_i)}$ are domains; note that this a priori not mean that the (usual) localizations $A_{t_i}$ are domains, only their degree-$0$ part, at least a this argumentation stage!
But the claim is that then $A$ must be already a domain as ring itself.
My approach: Assume $A$ is not a domain, then there exist $f,g \in A$ with $f \cdot g=0$. Wlog, we can assume these to be homogeneous, ortherwise pass to highest degree summands $f_d, g_e$ of both which would also neccessarily multiply to zero, and we obtain the decomposition $P=V_+(0)= V_+(f) \cup V_+(g)$ (Note this is done only on pure topological level) which implies that wlog $P=V_+(f) $ on topological level, so $\operatorname{Proj}(A/(f)) \to \operatorname{Proj}(A)$ is homeo and thus we are in the context of the "Leading Question" above.
Then für each $t_i$ on affine level inside $D_+(t_i)=\operatorname{Spec}(A_{(t_i)})$ the element $f^{n_i}/t_i^d$ ($n_i$ degree of homog $t_i \in A$) would be nilpotent and thus zero in $A_{(t_i)}$ as the latter was assumed to be a domain. Thus there exist a $n >0$ such that $f^n$ inside $A$ is killed by powers of $t_i$
(…in general if $A$ a ring $f,s \in A$ and $A_s$ localization at $s$, then $f =0$ inside $A_s$ iff $fs^n=0$ in $A$ for appropriate power $s^n$. Note, that in our case there is a canonical inclusion of localizations $A_{(t_i)} \subset A_{t_i}$,the former is the degree-$0$-part of the latter),
and thus as $A$ saturated, $f \in A$ is nilpotent.
Is my argumentation correct so far? In turn this would mean that the answer to the "leading question" is that $f$ is nilpotent in $A$ as one has in affine situation.
(if even all $t_i$ are living in degree $1$, then $f/t_i^d$ is as nilpotent element inside a domain $A_{(t_i)}$ even zero and this zero in $A$ dua to saturation assumption). Is this argument also correct so far? The reason why I have a rather bad feeling about my argumentation is that I not completely understand if Hartshorne's assumption on algebraic closedness of base field $k=A_0$ is really neccessary (and my approach flaws somewhere), or just a didactical simplification and allows a more general statement, which I tried to prove above?
Remark: This question generalizes this.
Best Answer
In a commutative graded ring, the minimal prime ideals are homogeneous. See for example Lemma 10.57.8 of the Stacks Project at https://stacks.math.columbia.edu/tag/00JM. In any commutative ring, an element is nilpotent if and only if it is in every prime ideal. But it's sufficient to check the minimal primes, so an element is nilpotent if and only if it is in every homogeneous prime ideal.
Now the Proj of a commutative graded ring is defined to be the set of homogeneous prime ideals that do not contain all elements of strictly positive degree. So if $f$ is homogeneous, is not nilpotent, and has strictly positive degree, then there is some element of Proj($A$) that does not contain $f$, and so the map from Proj($A/(f)$) to Proj($A$) is not an isomorphism. On the other hand, if $f$ is nilpotent, then this map is an isomorphism.
For a graded commutative ring with Koszul signs, if $x$ has odd degree then $x^2=-x^2$ and so $2x$ is nilpotent, and is in fact in the nil radical. So modulo the nil radical, $x$ is congruent to $-x$. It is then easy to see that modulo the nil radical the ring is a strictly commutative graded ring, so the statement also transfers to these rings.
I hope this answers your question.