Let $A= \oplus_{n \ge 0} A_n$ a 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 $A= \oplus_{n \ge 0} A_n$ as before a Noetherian graded ring with irrelevant ideal $A_{+}=\oplus_{n \ge 1} A_n$ generated by homogeneous elements $t_1,t_2,..., t_d$ 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 not says a priori that the (usual) localizations $A_{t_i}$ are domains, only about their degree-$0$ part! But the claim is that then $A$ must be already a domain as ring itself.

My approach: Assume $A$ not 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 $P=V_+(0)= V_+(f) \cup V_+(g)$ (on 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 $D_+(t_i)$ the element $f^{n_i}/t_i^d$ ($n_i$ degree of homog $t_i \in A$) is nilpotent and thus zero as $A_{(t_i)}$ is domain, this there exist a $n >0$ such that $f^n$ inside $A$ is killed by powers of $t_i$, and thus as $A$ saturated, $f \in A$ is nilpotent.
Is my argumentation correct? 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.