$Proj S$ integral scheme with S not a domain?

Question

My question arises from the comment of this post Projective subschemes and their coordinate rings:

"For your first question, it is not true that $\mathrm{Proj} S$ integral implies $S$ is a domain. For example, consider any domain $S$, and $T=S[x]/(x^2)$, where $x$ has degree $1$. Then, $\mathrm{Proj} S\cong \mathrm{Proj}T$ by Exercise II.5.13, but $T$ is not a domain"

Which leads me to some questions:

Is the graded structure of $S$ simply just $S_0=S$ and $0$ else?
Why is $\mathrm{Proj}T$ not empty? Any homogeneous ideal $I$ not containing $x$ will not be prime, because of $x\cdot x=0\in I$, but $x\notin I$. Thus every prime ideal contains $(x)=T_1=T_+$. Here I am assuming that $T$ has the graded structure $T_0=S$, $T_1=<x>$ and $0$ else.
Is there an other example of an integral scheme $\mathrm{Proj}T$ with $T$ not being a domain (or another way to create an example)?

EDIT:

I am getting a bit stuck, as I am trying to apply the exercise, which states that $\mathrm{Proj} S\cong \mathrm{Proj} S^{(d)} $, if $ S^{(d)}=\oplus_{n \ge 0}S^{(d)}_n$, where $S^{(d)}_n=S_{nd}$.

I tried to calculate it with the polynomial ring $S=k[y]$, but this lead me to $T=k[y][x]/(x^2)$ and I don't see how $T\cong S^{(d)}$ (if this doesn't hold, I don't see how I would apply the exercise).

You can take $S$ to be any graded algebra here. Why do you want another example? This one looks quite satisfactory. — Sasha, Mar 30 '18 at 19:26
There is no need to apply that exercise: describe the proj directly. Every homogeneous prime ideal of S[x]/(x^2) has to contain x so it contains all the positive part of the ring. Using this you see that its proj is very similar to the spec of S. Now look at regular functions. — Mariano Suárez-Álvarez, Mar 30 '18 at 20:44
But if a prime ideal contains all the positive part of the (graded) ring, then it can't be in Proj by definition? Or what am I missing out on here — Notone, Mar 30 '18 at 20:55
I think I got it now. First I notice that there is a 1:1 correspondence between $\mathrm{Proj} S$ and $\mathrm{Proj} T$ via mapping an ideal $I$ in $\mathrm{Proj} S$ to $(I,x)$. Then one checks that the local rings are isomorphic, which I believe is true, since in the localisation of any prime ideal in $\mathrm{Proj} T$ $x$ is equivalent to $0$. — Notone, Mar 31 '18 at 13:33
I believe the example I gave in the linked comment does not quite work. Instead, please look at the answer I posted under the original question. — Takumi Murayama, Apr 05 '24 at 00:22

$Proj S$ integral scheme with S not a domain?

0 Answers0