I have recently come across the following statement:
"A zero dimensional scheme $X$ in $\mathbb{P}^{n}$ is always affine"
I have the following vague argument:Since $X$ is a zero dimensional scheme it will correspond to a saturated ideal $I(X) \subset K[x_0,...x_n]=R$ such that $dim(R/I(X))=1$, from where I want to conclude that there exist a homogeneous polynomial $f \in R-I(X)$ and thus we will have a hypersurface $H=Z(f)$ such that $X \cap H= \emptyset$.
Then we have $X=X-(X \cap H)=X - H=X \cap D(f)$, which is affine.
Now I can't make this argument precise because of the following reason:
$(1)$ From $dim(R/I(X))=1$, how can I say that there exist a homogeneous $f \in R -I(X)$?,as all the elements in $R-I(X)$ could be non homogeneous.
$(2)$Even if there exist a homogeneous $f \in R -I(X)$,to show $X \cap H= \emptyset$, we have to show $I(X) \cup (\sqrt f) =R$ and I don,t see why that should be the case?
Instead without using these arguments can we say that there exists a hypersurface $H$ such that $X \cap H= \emptyset$.
$(3)$ we know that $(Proj(R) \cap D(f) =Spec(R_{(f)})$ and hence affine .But why is the same statement true with $Proj(R)$ replaced by $X$?
Can anyone give any explicit example of the statement mentioned at the beginning (I mean or any references)?
Any help from anyone is welcome.
NOTE:Scheme here means closed subscheme.
If you have any locally noetherian scheme $X$ of dimension zero, then it has the discrete topology. In partcular, if $X$ is a non-empty $k$-scheme of finite type with $\mathrm{dim}(X)=0$, then $X$ is discrete and has only finitely points. So it is affine and is the spectrum of $\prod_{x \in X} \mathcal{O}_{X,x}$.
Now $\mathbb{P}^n_k$ over any field $k$ is of finite type. Therefore, any closed susbscheme of it is also of finite type.
Reference: page $125$, Chapter $5$, Algebraic Geometry 1 by Görtz, Wedhorn.