Affine or projective · Qual Corpus

Projective varieties and homogeneous ideals (Tag D-CP2MH)

A subset of $\PP^n$ is closed if it is $V(T)$ for a set $T$ of homogeneous elements of $S \da k[x_0,\ldots,x_n]$. A projective variety is an irreducible closed subset of $\PP^n$; a quasi-projective variety is an open subset of one.

Let $p\in\PP^n$ have representative $v\in k^{n+1}\sm\ts{0}$, and let $f\in S$ be homogeneous of degree $d$. Then $f(\lambda v)=\lambda^d f(v)$ for $\lambda\in k^\times$, so the condition $f(v)=0$ does not depend on the representative $v$, although the value $f(v)$ does when $d>0$. For $f=x_0+x_1^2$ in $k[x_0,x_1]$ and $v=(-1,1)$, $f(v)=0$ but $f(2v)=2\ne0$ when $\operatorname{char}k\ne2$. Homogenizing with respect to a new variable $Z$ gives projective closures: $V(x^2 + y^2 - 1)$ becomes $V(X^2 + Y^2 - Z^2)$, and $V(y - x^3)$ becomes $V(YZ^2 - X^3)$.

The irrelevant ideal $S_+ = (x_0,\ldots,x_n)$ is homogeneous and radical, and $V(S_+) = \emptyset=V(S)$. The projective Nullstellensatz: for a homogeneous ideal $J$, $V(J) = \emptyset$ exactly when $\sqrt{J} \supseteq S_+$, and otherwise $I(V(J)) = \sqrt{J}$. Hence $V$ and $I$ are inclusion-reversing bijections, inverse to each other, between closed subsets of $\PP^n$ and homogeneous radical ideals of $S$ other than $S_+$ [@Har10a, Exercise I.2.4].

What is a projective variety?

Why must the defining polynomials be homogeneous?

Which homogeneous ideals correspond to which closed subsets of $\PP^n$?

Over an infinite field, show that $f$ is homogeneous of degree $d$ if and only if $f(\lambda \mathbf{x}) = \lambda^d f(\mathbf{x})$ for all $\lambda$.

Find the projective closures of $V(x^2 + y^2 - 1)$ and $V(y - x^3)$ in $\PP^2$.

Describe the line at infinity in $\PP^2$, and find the intersection of the projective closures of $V(y)$ and $V(y - tx - 1)$ when $t = 0$.

Show that a real conic $V(f) \subseteq \RR^2$ with infinitely many real points is a circle if and only if the projective closure of $V(f)$ passes through the circular points $[1 : \pm i : 0]$.

Show that conics in $\PP^2$ are parametrized by $\PP^5$, and that the conics through a fixed point form a hyperplane $H_p$.

Show that over an infinite field the conics through $5$ fixed points form a single point or an infinite set, and that there is a unique conic through $5$ points in linear general position.

Show that there is a unique circle through any $3$ non-collinear points of $\RR^2$.

Describe $\PGL_{n+1}(k)$ as a quotient, and show that for two sets of $n+2$ points in linear general position in $\PP^n$ there is a unique element of $\PGL_{n+1}(k)$ carrying one to the other.

Show that the $(k+1)$-dimensional linear subspaces of $\PP^n$ containing a fixed $k$-dimensional subspace $W$ form a $\PP^{n-k-1}$.

For $\PP^n = \PP(V)$, show that $\PP(\dualof{V})$ parametrizes the hyperplanes of $\PP^n$.

Show that there is a unique rational normal curve through $n+3$ points in linear general position in $\PP^n$.

Show that $\dim(V_1 \cap V_2) \geq \dim V_1 + \dim V_2 - n$ for linear subspaces $V_1, V_2 \subseteq \PP^n$.

Is it projective?

Projectivity is established by exhibiting a closed embedding into some $\PP^N$.

The Segre embedding, and why $\PP^m \times \PP^n$ is projective (Tag PR-VA4S3)

The Segre embedding identifies $\PP^m\times\PP^n$ with the rank-one tensor locus, cut out by the $2\times2$ minors.Suggest a slogan

The map \[ \psi: \PP^m \times \PP^n \to \PP^{(m+1)(n+1)-1}, \qquad ([x_i],[y_j]) \mapsto [x_i y_j] \] is well defined and injective, and its image is the closed set cut out by the $2 \times 2$ minors \[ z_{ij} z_{kl} - z_{il} z_{kj} = 0 . \] A product of projective varieties is therefore projective.

For $m = n = 1$ there is one minor, and the image is the quadric surface \[ z_{00} z_{11} - z_{01} z_{10} = 0 \] in $\PP^3$. The two rulings of that quadric are the images of $\ts{p} \times \PP^1$ and $\PP^1 \times \ts{q}$, and they make it a surface whose Picard group is $\ZZ^2$ rather than $\ZZ$.

The minors say the matrix $(z_{ij})$ has rank one, which is the coordinate-free statement: the Segre image is the locus of rank-one tensors in $\PP(V \tensor W)$.

Is $\PP^1 \times \PP^1$ a projective variety?

What equations cut out the Segre image?

Show that the Segre variety is the categorical product of projective varieties $X \subseteq \PP^n$ and $Y \subseteq \PP^m$.

What is the graph $\Gamma_f$ of a morphism $f \colon X \to B$ of quasi-projective varieties?

Give a precise definition of a morphism of projective varieties, in terms of homogeneous coordinates locally.

Describe the image of $[x:y] \mapsto [x^2 : xy : y^2]$, and give equations for the Veronese surface and its ambient $\PP^N$.

What is a determinantal variety?

The $d$-uple Veronese embedding $\nu_d\colon\PP^n\to\PP^{\binom{n+d}{d}-1}$ sends each degree-$d$ hypersurface $V(f)$ to a hyperplane section of $\nu_d(\PP^n)$. So $\PP^n\setminus V(f)$ is isomorphic to $\nu_d(\PP^n)$ minus a hyperplane section, a closed subvariety of $\AA^{\binom{n+d}{d}-1}$, and is affine.

Is it affine?

Regular functions on a projective variety are constant (Tag PR-EFW6B)

A connected projective variety over an algebraically closed field has no nonconstant global regular functions.Suggest a slogan

Let $X$ be a connected projective variety over $k = \bar{k}$. Then $\OO_X(X) = k$.

This is the algebraic shadow of the maximum principle, and it is the fastest separator of the two categories. An affine variety is determined by its ring of global functions and has as many of them as it has coordinates; a projective one has only the constants, so it can be affine only if it is a point. In particular $\PP^n$ is not affine for $n \geq 1$.

The same statement is the reason a morphism from a projective variety to an affine one is constant on each connected component, which is how one usually shows some proposed map cannot exist.

What are the global regular functions on a projective variety?

Why is $\PP^n$ not affine?

A closed subvariety of an affine variety is affine, and a complete connected affine variety is a point, so a variety containing a complete curve is not affine. In particular $\PP^n$, $\PP^m \times \PP^n$, and every projective variety of positive dimension are not affine.

Hypersurface complements are affine (Tag PR-WZGOQ)

A hypersurface complement is affine: localize by its equation in affine space, or use Veronese to make it a hyperplane complement projectively.Suggest a slogan

Let $f$ be nonconstant. Then $\AA^n \sm V(f)$ is affine, with coordinate ring $k[x_1,\ldots,x_n]_f$. So is $\PP^n \sm V(f)$ for $f$ homogeneous of degree $d$.

Affine case: $D_f$ is isomorphic to the closed subvariety $V(1 - y f) \subseteq \AA^{n+1}$, the graph of $1/f$. The extra coordinate $y$ is the inverse that was missing, which is the same trick that reduces the strong Nullstellensatz to the weak one.

Projective case, in two steps. For a hyperplane, $\PP^n \sm V(x_i)$ is the standard chart $\AA^n$. For $\deg f = d$, the $d$-uple Veronese $\PP^n \injects \PP^N$ carries $V(f)$ into a hyperplane section, so the complement becomes a closed subvariety of $\PP^N$ minus a hyperplane, hence a closed subvariety of $\AA^N$.

The projective case shows that affineness of a complement is intrinsic rather than an artifact of an affine presentation.

Is the complement of a hypersurface in $\AA^n$ affine?

Is the complement of a hypersurface in $\PP^n$ affine?

Is it complete?

Complete varieties and projectivity (Tag D-VARCOMP)

A variety $X$ over $k$ is complete, equivalently proper, if the structure morphism $X \to \Spec k$ is proper: separated, of finite type, and universally closed.

(Consequences of completeness)

Let $k$ be algebraically closed and $X$ a complete variety over $k$. Then $\OO_X(X) = k$, the image of $X$ under any morphism of varieties is closed, and any morphism from $X$ to an affine variety is constant. Every projective variety is complete; over $\CC$, a variety is complete if and only if it is compact in the Euclidean topology.

$\AA^1$ is not complete, although $\AA^1\to\Spec k$ is a closed map: after base change, the projection $\AA^1 \times \AA^1 \to \AA^1$ sends the closed hyperbola $V(xy-1)$ to $\AA^1 \sm \ts{0}$, which is not closed.

Complete varieties need not be projective: Hironaka constructed a nonsingular complete threefold that is not projective. Every complete curve and every nonsingular complete surface is projective, while there are complete normal surfaces that are not projective.

What is a complete variety?

What is a proper variety?

Give an example of a non-proper variety.

Projective implies complete and not conversely, so the three classes nest: projective inside complete inside separated of finite type.

Examples

$\PP^1 \times \PP^1$ is projective by the Segre embedding and not affine because it contains the complete curve $\PP^1\times\{p\}$. The complement of a curve in $\PP^2$ is affine by the Veronese argument, and it is not projective, because it is a proper open subset of an irreducible projective surface.

Serre's criterion characterizes affineness cohomologically: a noetherian scheme $X$ is affine exactly when $H^1(X,\mcf)=0$ for every quasicoherent $\mcf$.

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