Projective Space

Before trying to define projective space, we consider an example.

Example Consider the affine plane curve X defined by

y2 - y + x2 + x + 2 x y = 0.

The point (x,y) = (0,0) clearly satisfies this equation, so it lies on X. Look at the intersection between X and a line y=mx passing through the origin. We have

m2 x2 - m x + x2 + x + 2 x m x = 0


(m2 + 2m + 1) x2 + (1-m) x = 0

There are, therefore, two solutions. Either x = y = 0 or x = (m-1)/(m+1)2 and y = (m2-m)/(m+1)2. For most lines, we get exactly two intersection points. (When m=1, we get a single point counted with multiplicity two.) This result makes sense since X is a curve of degree 2, and lines have degree 1. the only problem arises when x = -1, when the second intersection point disappears because we are trying to divide by zero. Algebraic geometers are bothered when a discrete invariant (like the number of points of intersection) jumps in the middle of a continuous process. In this case, the extra intersection point has escaped to infinity. One of the points of introducing the notion of projective space is to bring it back.

Definition We define projective n-space over an algebraically closed field k, and denote by Pnk, the set of lines through the origin in An+1k.

Remark We can give an explicit description of Pnk in terms of coordinates. To give a line through the origin, you must pick a point x = (x0, , xn)An+1k with at least one nonzero coordinate. Any other point (except the origin) on the line is related to x by being a scalar multiple of it. In other words, any other point has the form ax = (ax0, , axn) for some nonzero element ak. Thus, the points of projective space are equivalence classes arising from the equivalence relation on An+1k\{ (0,,0)} given by identifying xax. We will write (x0::xn) for one of these equivalence classes, and will refer to each coordinate as a homogeneous coordinate.

The description we have just given of projective space using coordinates is external; next, we want to investigate an internal description. Let U0 = {(x0:,xn) Pnk : x0 0}. Because the lead coordinate is nonzero, we can divide by it to normalize the representative of an equivalence class of a point in U0. In other words,

(x0::xn) (1: [x1 / x0]: [x2 / x0] : : [xn / x0]).

The normalized form of the homogeneous coordinates of points on U0 is unique, and we get a natural bijection between U0 and Ank = kn.


Z0 = Pnk\U0 = {(x0::xn) : x0 = 0} Pn-1k.

Thus, we can think about building Pnk by starting (internally) with affine n-space Ank and adding something at infinity; namely, Z0. In particular,
  1. P0k is just a point.
  2. P1k is obtained by adding a single point to the affine line.
  3. P2k is obtained by adding a copy of P1k to the affine plane. In other words, we add one point for each direction of a line through the origin of the affine plane.

It's time to start thinking about a topology on projective space. Start with a polynomial f k[x0, , xn]. It makes perfect sense to talk about where f takes on the value zero on affine space, but it makes no sense to discuss this on projective space. The problem, of course, is that f might vanish on one representative of an equivalence class, but not on other representatives.

Definition We say that f k[x0, , xn] is a homogeneous form if there exists a non-negative integer d such that

f(ax0, , axn) = ad f(x0, , xn)

for all ak* and for all x0, , xn.

Remark A homogeneous form cannot be thought of as a function on the points of projective space. However, the property f(x0, , xn)=0 is well-defined independently of the choice of a representative of an equivalence class in Pnk.

Definition Let S k[x0, , xn] be any set of homogeneous polynomials. Define the projective zero set associated to S to be

V(S) = { x=(x0,,xn)Pnk : f(x) = 0 f S }.

Remark The ideal <S> generated by a set of homogeneous elements is called a homogeneous ideal. Not every element of a homogeneous ideal is homogeneous. For example, the homogeneous ideal <x, y2> k[x,y] contains the nonhomogeneous element x+y2.

Remark It is not possible for every homogeneous ideal to pick out a nonempty set of points in projective space. For example, consider the maximal ideal I=<x0, , xn>. It is trivial to check that V(I)=, since each point of projective space must have at least one nonzero coordinate. We call I the irrelevant ideal.

Definition Define a closed algebraic set in Pnk to be the projective zero set associated to some homogeneous ideal. With these as the closed sets, we get a topology on projective space called the Zariski topology.

PropositionThere is a bijection between closed algebraic sets in Pnk and radical homogeneous ideals in k[x0, , xn], except for the irrelevant ideal.

Proof:For each subset YPnk, we let J(Y) be the ideal generated by all homogeneous polynomials that vanish at all points of Y. It is easy to see that J(Y) is always a radical ideal. It is also easy to check that V(J(Y))=Y- is the closure in the Zariski topology.

The main point is now to check that I = J(V(I)) whenever I is a radical homogeneous ideal. To proceed, we can forget the grading and homogeneity for a moment, and look instead at the affine variety Z(I) that it defines in An+1k. There are three possibilities:

  1. Z(I) is empty. By the weak Nullstellensatz, this can only happen if I=k[x0, , xn], in which case we also have that V(I) is empty.
  2. Z(I) only contains the origin. Again using the affine results, this can only happen if I is the irrelevant maximal ideal, and so V(I) is again empty.
  3. Z(I) is a union of lines through the origin. In this case, we have J(V(I)) = I(Z(I)). By Hilbert's Nullstellensatz, we have I(Z(I)) = rad(I) = I.

The result follows from this case-by-case analysis.

Definition The affine cone over a projective variety Y Pnk is the closed subset of An+1k defined by Z(J(Y)), where we forget that J(Y) is a homogeneous ideal.

Remark A nonempty projective algebraic set Y is irreducible if and only if its affine cone is irreducible.

Corollary Under the bijection of the proposition, nonempty irreducible closed algebraic sets correspond to nonirrelevant proper homogeneous prime ideals.

Corollary Every projective algebraic set can be writtten uniquely as a finite irredundant union of irreducible projective algebraic sets.

Example Let's return to the example with which we opened this discussion. Recognizing that we are looking at a parabola, let's straighten it out, and look at the affine plane curve X = Z(y-x2). When you look at the intersections of this parabola with the lines y = mx, one of the two intersection points escapes to infinity as the line becomes vertical. But we can put the affine plane A2k (with coordinates (x,y)) into the projective plane P2k (with coordinates (X:Y:Z)) as the open subset where Z0. Normalizing projective coordinates, this means that x = X/Z and y = Y/Z, and the affine equation of the curve transforms into

[Y / Z] = [X2 / Z2].

Since that equation doesn't give us a homogeneous form, we have to clear denominators in a minimal way to get YZ - X2.

More generally, given any affine ideal I k[x,y], you can associate to each polynomial f(x,y) I, of degree d, a homogeneous form f*(X,Y,Z) = Zd f(X/Z, Y/Z). We let I* k[X,Y,Z] denote the homogeneous ideal generated by all the homogeneous forms associated to polynomials in I. We then refer to V(I*) as the projective closure of the affine algebraic set Z(I).

In this example, the projective closure of the affine parabola is the curve defined by YZ - X2, and the projective closure of a line y = mx is the projective line defined by Y = mX. Letting m go to infinity, this line becomes the "vertical" line X=0, which intersects the projective closure of the parabola in two points:

The next natural question to ask is: what are the maps between projective algebraic sets? In the affine case this was easy: we simply took lists of polynomials (that clearly gave maps from one affine space to another), restricted them to an algebraic subset of the domain, and looked at conditions that ensured that the image landed in another algebraic set. In the projective case, things are not so simple. For example, let's continue looking at the projective conic C = V(YZ - X2). We can try to map this to the projective line by the formula (X:Y:Z) |→ (X:Z). (This is an attempt to extend to the projective setting the map that projects the affine parabola straight down onto the x-axis.) The problem with this formula, however, is that it takes the point (0:1:0) (which we conveniently added to solve the intersection-counting problem) to the nonexistent point (0:0)! Perservering, we find for points of C that

(X:Z) = (X2:XZ) = (YZ:XZ) = (Y:X).

Using this formula, the point (0:1:0) simply maps to (1:0). However, this formula takes the perfectly good point (0:0:1) to the nonexistent point (0:0)!

This example illustrates a typical phenomenon with maps between projective algebriac sets. It is almost always the case that there is no single formula (defining a map on all of projective space or even on all of the projective set) that works everywhere to define morphisms. Instead, you have to define everything locally and verify that the formulas patch together in a reasonable way. Solving this problem is the subject of the next chapter.

Comments on this web site should be addressed to the author:

Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210