I'm reading the Gortz's Algebraic Geometry, proof of Theorem 15.18 and stuck at understanding some statement.
My first main question is as follows :
Q. Let $C$ be a connected and non-empty noetherian space (possibly of pure dimension $1$ ; i.e., all irreducible component is of dimension $1$ ) with finitely many irreducible components $C_1, \dots, C_n$. Let $S \subseteq C$ be a finite subset whose points lie on more than one irreducible component of $C$. Then for each $1\le j \le n$, $S \cap C_j \neq \varnothing$ ?
This question originates from the following proof of the Theorem 15.18 of the Gortz's book :
Theorem 15.18. Let $k$ be a field, and let $C$ be a separtaed curve over $k$. Then $C$ is quasi-projective over $k$.
At this point, we will prove the theorem under the additional assumption that $C$ is generically reduced so that $\operatorname{Sing}C$ is finite ( c.f. his book Remark 15.15-(4) ).
Proof. We may assume that $C$ is connected ( Remark 8.21 ).
(i) We first prove the theorem under the following assumption:
The curve $C$ has an open affine covering $C= \cup_{i\in I} U_i$ such that every $U_i$ contains the finite set $\operatorname{Sing}C$ of non-normal points of $C$.
We may assume that the index set $I$ is finite and that each $U_i \neq \varnothing$. Then $U_i$ meets every irreducible component of $C$ because among the singular pints are in particular those point which lie on more than one irreducible component. Thus $U_i$ is open and dense in $C$ and .. (Omitted). QED.
Why the bold statement is true? There, 'curve $C$ over a field $k$' means a non-empty $k$-scheme of finite type satisfying following equivalent conditions :
(i) For every closed point $x\in C$, $\operatorname{dim}\mathcal{O}_{C,x} =1$.
(ii) The closed irreducible subsets of $C$ are the $C_i$ ( $C_1, \dots C_n$ are the irreducible components ) and the closed points of $C$. And none of the $C_i$ consists of only one point.
(iii) The scheme $C$ is of pure dimension $1$ ( i.e., $\operatorname{dim}C_i =1$ for all $i$ ).
( Q. Interlude question : as the statement followed by the underlined statement, why the singular points are in particular those points which lie on more than one irreducible component )
If our question is true, then since $\operatorname{Sing}C$ meets every irreducible component of $C$ and each $U_i$ contains $\operatorname{Sing}C$, each $U_i$ meets every irreducible component of $C$. And is our question true? Why?
Can anyone help ?
Best Answer
Let me say a few things / address different issues.
A. It is not true that singular points are points that lie on more than one irreducible components e.g. take the nodal cubic. It is true that if a point lies on two components then it is a singular point. Think of it this way, the suppose $x$ is on two irreducible components. Then the local ring $O_x$ at that point is going to have zero divisors. This means it cannot be an integral domain and therefore certainly not a regular local ring (see this post.)
A. It is not true $Sing C$ meets each irreducible component. What is true that each $U_i$ will contain $Sing C$. So each $U_i$ will meet every irreducible component in at least one point due to connectedness.