# Finite # of points, open affines and Quotients

Here’s some random stuff.  For an (abstract) variety, the condition ‘any finite set of points in contained in an open affine’ is equivalent to the variety being quasi projective.  The details of the proof of this assertion are presented in pages 111 – 116 of JMiline’s algebraic geometry notes, here is a brief outline:

First prove the result when the variety is projective:  $S \subset X$ a finite set of a projective variety, i.e. $X \cong \mathbb{P}^n$.  Then there is a linear form $L$ which ‘misses’ all the points in $S$, i.e. $L(P) \ne 0$ for all $P \in S$.

Now define a bijection with $D(L) = X - Z(L)$ and the affine variety $Z(L - 1) \subset \mathbb{A}^{n+1}$.  For $P = [p_0:\dotsc : p_n] \in D(L)$ set $P' = \frac{1}{L(P)}(p_0, \dotsc, p_n) \subset \mathbb{A}^{n+1}$, which makes sense since I’m picking $P$ so $L(P) \ne 0$.  Then

$P \in D(L) \Rightarrow L(P') = \frac{1}{L(P)} L([p_o: \dotsc :p_n]) = 1 \Rightarrow P' \in Z(L - 1)$

The map $P \mapsto P'$ is a bijection with inverse

$a = (a_1, \dotsc, a_{n+1}) \mapsto [a_1 : \dotsc : a_{n+1}]$.

Next, prove an analogous result with the linear form replaced by a homogeneous polynomial $f$, say of degree d.  This uses the Veronese map or d-uple embedding (see Hartshorne ex. I.2.12).  Basically this gives a map $\mathbb{P}^n \to \mathbb{P}^N$ where $N = \binom {n+d}{d} - 1$ where the hypersuface defined by $f$ maps isomorphically to a hyperplane.

Finally, deduce the general result: if $X$ is quasi projective then embed it as an open subset of projective space $X \subset \mathbb{P}^n$.  Consider $Z = \overline{X} - X$ (Note $Z$ is closed; look at its complement).  Recall $S$ is a finite set which doesn’t meet $Z$.   As $Z$ is closed, $Z = V(I(Z))$, so $S \cap Z = \emptyset$ means for all $p \in S$ there is a homogeneous $f_p \in I(Z)$ with $f_p(p) \ne 0$.  Then make all the $f_p$ have the same degree and argue that some linear combination of them don’t vanish at all points of $S$.  Then by the previous paragraph, $D(F) \cap \overline{X}$ is affine, contains all of $S$ and $D(F) \subset X$, so its an affine of $X$.

### Quotienting out by a group

One particular place where the above comes into play is in talking about quotienting out a scheme by an action of the group scheme.  For an affine scheme $X = \mbox{Spec} A$ with an action by a finite group $G$, the quotient is given by the ring of invariants: $X/G = \mbox{Spec} A^G$.  For a general scheme $X$, $X/G$ could be constructed by covering $X$ with affines $X_i$ and gluing together $X_i/G$, this requires $X_i$ to be G-invariant.  This can be guaranteed if $X$ is quasi projective.  Indeed, given $x \in X$, the G-orbit $Gx$ is a finite set of points, so contained in an affine $U$.  Then

$\bigcap_{g \in G} \mbox{ } gU$

is $G$ invariant and affine, as $X$ is separated.

With a little more work, I can define $X/G$ where $G$ is a group scheme with an action of $G$, i.e. a morphism $a\colon G\times X \to X$ which fits into some nice commutative diagrams.  When $X$ is affine the task boils down to defining the ring of invariants.  In this case $f \in \mathcal{O}_X^G$ if $p_2^\#(f) = a^\#(f)$.  Then $X/G := \mbox{Spec} \mathcal{O}_X^G$.  As above, for the general case, cover $X$ with affines and glue together the quotients.