I have several questions around a part of Exercise 2.5.17(a) from Hartshorne's Algebraic Geometry (p 126). The setting:

Let $k$ be a field, and let $X$ be a closed subscheme of of $\mathbb{P}^r_k$. We define the homogeneous coordinate ring $S(X)$ of $X$ for the given embedding to be $k[x_0, ..., x_r]/I$, where $I=I(X)$ is the unique saturated ideal constructed in the proof of (5.16) corresponding to $X$. (note that there are many homog ideals $I(X)$ defining same $X \subset \mathbb{P}^r_k$, but one one s saturated; the "biggest" one).

Now assume for this exercise that $k$ is an algebraically closed field, and that $X$ is a connected, normal closed subscheme of $\mathbb{P}^r_k$.
(a) Show that $S(X)$ is a domain (ie contains no zero divisors) [...]

Two questions:
(1) Can the condition that $k$ is algebraically closed be dropped for showing that $S(X)$ is domain?
#EDIT: Here(see part 3.) I tried to elaborate an argument which seemingly not depends on algebraic closedness of $k$; not sure if it's gap free

(2) What about more general setting, namely say $S$ is a graded ring, such that $\text{Proj}S=:Y$ is integral (=reduced & irreducible). How far away could $S$ be from beeing a domain?

On (1): Since $X$ is normal, it is reduced and irreducible (because for normal schemes connected= irreducible), so integral. Assume $S(X)$ is not a domain, therefore there exist nonzero $f,g \in S(X)$ with $f \cdot g=0$.
Considering the degree expansion of $f$ and $g$ we can wlog assume that $f$ and $g$ are homogeneous: just take the summands of highest degrees. Then we have equalities

$$ X=V_+(0)=V_+(f) \cup V_+(g) $$

Wlog $X=V_+(g)$, so by (projective) Nullstellensatz $ f$ is nilpotent in $S(X)$. But that's absurd, because if $f \neq 0$ there exist a $x_i$ such that $f/x_i^d \in S_{(x_i)}$ is nonzero, but still nilpotent, that's a contradiction as $X$ integral.
(Note, that to deduce that $f$ is nilpotent we used Nullstellensatz, and so exploited that $k$ is algebraically closed.)

But the question of (1) is if that's possible to still deduce using probably different techniques in this setting that $S(X)$ is domain without assumption on algebraic closedness of $k$.

On (2): In a comment here Takumi Murayama gave seemingly a counterexample on point (2):
Take $S$ be any domain and $T:=S[x]/(x^2)$, where $x$ has degree $1$. Then, $\text{Proj} S \cong \text{Proj} T$ but $T$ is not a domain.

If I understand the example correctly, then it is rather peculiar: In which sense is $S$ considered to be a graded ring (...in order to form the Proj)? Do I understand it correctly that $S$ is considered to be endowed with "trivial" grading, namely $S= \oplus_k S_k$ with $S_0=S$ and $S_k=0$ for $k >0$? And then, $\text{Proj}(S)= \text{Spec}(S)$?
(Otherwise, if $S_1 \neq 0$, then $T$ localized at any nonzero $s \in S_1$ would be still not a domain, right, as I not see any reason why $\overline{x}$ should be zero in $T_{(s)}$? Or do I misunderstand the example?)

In turn, is this kind of pathology the only one which could happen preventing (2) or are reasons for failure of (2) much more "natural"? Ie, say if $S$ a graded ring, and $\text{Proj}(S)$ is integral and not affine (ie there exist no ring with $\text{Proj}(S) \cong \text{Spec}(R)$, is then $S$ is a domain?

Formulated more philosophically, I'm wondering if it's possible to track down which obstructions prevent the graded ring $S$ with integral Proj from beeing a domain. If I understood the counterexample quoted before correctly, then seemingly it worked only with Proj which turned out to be affine, which is a rather "special" situation.



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