y^{2} - y + x^{2} + 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 havem^{2} x^{2} - m x + x^{2} + x + 2 x m x = 0
or(m^{2} + 2m + 1) x^{2} + (1-m) x = 0
There are, therefore, two solutions. Either x = y = 0 or x = (m-1)/(m+1)^{2} and y = (m^{2}-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.The description we have just given of projective space using coordinates is external; next, we want to investigate an internal description. Let U_{0} = {(x_{0}:…,x_{n}) ∈ P^{n}_{k} : x_{0} ≠0}. Because the lead coordinate is nonzero, we can divide by it to normalize the representative of an equivalence class of a point in U_{0}. In other words,
(x_{0}:…:x_{n}) ∼ (1: [x_{1} / x_{0}]: [x_{2} / x_{0}] : … : [x_{n} / x_{0}]).
The normalized form of the homogeneous coordinates of points on U_{0} is unique, and we get a natural bijection between U_{0} and A^{n}_{k} = k^{n}.Let
Z_{0} = P^{n}_{k}\U_{0} = {(x_{0}:…:x_{n}) : x_{0} = 0} ≈ P^{n-1}_{k}.
Thus, we can think about building P^{n}_{k} by starting (internally) with affine n-space A^{n}_{k} and adding something at infinity; namely, Z_{0}. In particular,It's time to start thinking about a topology on projective space. Start with a polynomial f ∈ k[x_{0}, …, x_{n}]. 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.
f(ax_{0}, …, ax_{n}) = a^{d} f(x_{0}, …, x_{n})
for all a∈k^{*} and for all x_{0}, …, x_{n}.V(S) = { x=(x_{0},…,x_{n})∈P^{n}_{k} : f(x) = 0 ∀ f ∈ S }.
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 A^{n+1}_{k}. There are three possibilities:
The result follows from this case-by-case analysis.
[Y / Z] = [X^{2} / Z^{2}].
Since that equation doesn't give us a homogeneous form, we have to clear denominators in a minimal way to get YZ - X^{2}.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) = Z^{d} 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 - X^{2}, 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 - X^{2}). 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) = (X^{2}: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