1 Blowup

1.1 Example

The curve y^{2} = x^{3} + x^{2} has a singularity (double point) at the origin, but it feels as if nothing is wrong there: if one traces the curve then the curve appears nice and smooth - it is just that we visit the same point twice.

So, with an additional coordinate t (time) the singularity is gone. If we study curves up to birational equivalence, then the new coordinate should be a rational function of the old coordinates, and t = y/x works: the two passes through the origin have t = 1 and t = −1.

Now the pair of equations y^{2} = x^{3} + x^{2}, y = tx in (x, y, t)-space defines the union of a lifted version of the planar y^{2} = x^{3} + x^{2}, and the vertical line x = y = 0. Substitute y = tx in the first equation and divide by x^{2} to get the pair t^{2} = x + 1, y = tx defining a smooth space curve, birationally equivalent to the original curve. The maps are (x, y, t) ↦ (x, y) and (x, y) ↦ (x, y, y/x), well-defined and inverses of each other outside the origin.

1.2 Blowing up a point in affine space

This can be done more generally. Given affine space with coordinates (x_{1}, ..., x_{n}), introduce new projective coordinates (t_{1}, ..., t_{n}) restricted by x_{i}t_{j} = x_{j}t_{i}. This defines a subvariety of A^{n} × P^{n−1}. The map ϕ given by (x_{1}, ..., x_{n}, t_{1}, ..., t_{n}) ↦ (x_{1}, ..., x_{n}) is 1-1 outside the inverse image of the origin, with inverse (x_{1}, ..., x_{n}) ↦ (x_{1}, ..., x_{n}, x_{1}, ..., x_{n}). (At the origin this inverse is not defined since a projective point cannot have all coordinates zero.) The inverse image of the origin is the entire projective space {0} × P^{n−1}.

This variety is irreducible, since the inverse image of (the irreducible variety) A^{n} \ {0} is dense.

1.3 Blowing up a point in a subvariety of affine space

Now if Y is a subvariety of A^{n}, then the blow-up of Y at the origin is by definition the closure in A^{n} × P^{n−1} of ϕ^{−1}(Y \ {0}).

(The example we started with is the special case where n = 2 and instead of projective coordinates (s, t) with sy = xt we used the affine part with s = 1.)

1.4 Blowing up a subvariety of a variety

If X is a variety with coordinates x_{1}, ..., x_{m} and Y is a subvariety with ideal I(Y) generated by functions f_{1}, ..., f_{n}, then we have a map

(x_{1}, ..., x_{m}) ↦ (x_{1}, ..., x_{m}, f_{1}(x_{1}, ..., x_{m}), ..., f_{n}(x_{1}, ..., x_{m}))

from X \ Y into A^{m} × P^{n−1}. Now the blowup of X along Y is defined as the closure in A^{m} × P^{n−1} of the image of this map.

1.5 Resolution of singularities

It turns out to be possible to show that an arbitrary algebraic variety is birationally equivalent to a smooth one (i.e., one without singularities), and that nonsingular model can be reached by a series of blowups.

Theorem 1.1 (Hironaka) Let X be any variety over a field of characteristic zero. Then there exists a variety Y and a regular map ϕ : Y → X that is a birational equivalence.

The analog of this theorem in char p is open, but has been proved for curves and surfaces.

2 Quadratic transformations

Blowup has the disadvantage that the dimension goes up. For the case of curves in the plane one can find a reasonably good model while staying in the plane.

The transformation used is the map ϕ given (in projective coordinates) by (X, Y, Z) ↦ (Y Z, XZ, XY). It is defined on P^{2}\{P, P′, P′′} where P = (0, 0, 1), P′ = (0, 1, 0) and P′′ = (1, 0, 0). Let U be the open set P^{2} \ V (XY Z). Then ϕ is 1-1 on U, and its own inverse. Hence ϕ is a birational isomorphism. It maps the line Z = 0 to the point P, so ϕ^{−1} looks like a blowup: it blows up the point P into a line.

Theorem 2.1 Any curve in the plane can be brought into a form where the only singularities are ordinary multiple points.

(An ordinary multiple point is a multiple point without multiple tangent.)

We give a full proof.

Let C be an irreducible curve, defined by the equation F (X, Y, Z) = 0. Let the points P, P′, P′′ have multiplicities m, m′, m′′ on C, respectively. Then the image C^{ϕ} under ϕ is given by the equation F^{ϕ}(X, Y, Z) = 0, where

F^{ϕ}(X, Y, Z) = F (Y Z, XZ, XY)/(X^{m′′}Y^{m′}Z^{m}).

Indeed, by definition of multiplicity, if P has multiplicity m on C, and F has degree n, then F (X, Y, Z) = F_{m}(X, Y)Z^{n−m} + ... + F_{n}(X, Y), with F_{j}(X, Y) homogeneous of degree j. Now

F (Y Z, XZ, XY) = F_{m}(Y Z, XZ)(XY)^{n−m} + ... + F_{n}(Y Z, XZ)

has precisely m factors Z. This shows that F^{ϕ} is a polynomial, homogeneous of degree 2n − m − m′ − m′′, and this polynomial is irreducible (since F is).

The point P has multiplicity n − m′ − m′′ on C^{ϕ}:

Indeed, expanding F^{ϕ} in powers of Z we find

F^{ϕ}(X, Y, Z) = F_{m}(Y, X)X^{n−m−m′′}Y^{n−m−m′} + ... + F_{n}(Y, X)X^{−m′′}Y^{−m′}Z^{n−m}

with highest Z-exponent in the last term.

We have (F^{ϕ})^{ϕ} = F.

Indeed, F^{ϕϕ}(X, Y, Z) = F^{ϕ}(Y Z, XZ, XY)/(X^{n−m−m′}Y^{n−m−m′′}Z^{n−m′−m′′}) = F (XY ZX, XY ZY, XY ZZ)/(X^{n−m−m′}Y^{n−m−m′′}Z^{n−m′−m′′}(XY)^{m}(XZ)^{m′}(Y Z)^{m′′}) = F (X, Y, Z).

That settles the relation between F and F^{ϕ}. The goal is to choose coordinates in such a way that things improve around P and do not get worse elsewhere.

The quadratic transformation is 1-1 outside the triangle V (XY Z), so nothing happens there. In order to control what happens near the corners of the triangle we must choose the triangle in good position. That is, by definition, choose them such that none of the three lines X = 0, Y = 0, Z = 0 is tangent to C at P, P′ or P′′.

If the triangle is in good position with respect to C, then it is also in good position with respect to C^{ϕ}.

Indeed, the line Z = 0 is tangent to C^{ϕ} at P′ = (0, 1, 0) iff

I(P′, C^{ϕ} ∩ Z) > m_{P′}(C^{ϕ}),

i.e.,

I(P′, F_{m}(Y, X)X^{n−m−m′′}Y^{n−m−m′} ∩ Z) > n − m − m′′,

i.e.,

I(P′, F_{m}(Y, X) ∩ Z) > 0,

i.e., F_{m}(1, 0) = 0, i.e., Y = 0 is a tangent to C at P = (0, 0, 1), which it is not.

For a triangle in good position we have control over the ‘blowup’ of P: Suppose C^{ϕ} meets the line Z = 0 in the points P_{1}, ..., P_{s} distinct from P′, P′′ (and possibly also in P′, P′′). Then for each i the multiplicity of P_{i} as a point of C^{ϕ} is bounded by m_{P_{i}} ≤ I(P_{i}, C^{ϕ} ∩ Z), and the sum of these intersection numbers equals m.

Indeed ∑_{i} I(P_{i}, C^{ϕ} ∩ Z) = ∑_{i} I(P_{i}, F_{m}(Y, X) ∩ Z) = m. (The powers of X and Y in I(P_{i}, F_{m}(Y, X)X^{n−m−m′′}Y^{n−m−m′} ∩ Z) disappear because P_{i} is distinct from P′ and P′′ and hence does not lie on X = 0 or Y = 0. By the assumption of good position, the points P′ and P′′ do not occur in the sum. Since the field is algebraically closed and F_{m} has degree m, we find m roots.)

So far, everything was symmetric in P, P′, P′′. Introduce asymmetry now. The triangle is in excellent position if it is in good position, and moreover Z = 0 intersects C in n distinct points different from P′, P′′ and the lines X = 0 and Y = 0 intersect C in n−m distinct points different from P, P′, P′′. (That is: we know that P is an m-fold point. All other intersections of C with the triangle must be simple, and C must not meet the other two corners.)

It is clear that given P we can change coordinates in such a way that the triangle is in excellent position (since the field is infinite).

Now C^{ϕ} has the following multiple points:

(i) Outside the triangle, C and C^{ϕ} are isomorphic, and the same multiplicities occur.

(ii) P, P′, P′′ are ordinary multiple points on C^{ϕ} with multiplicities n, n − m, n − m, respectively. (This follows from excellent position.)

(iii) C^{ϕ} has no points on X = 0 or Y = 0 other than P, P′, P′′. (The sum of the intersection multiplicities equals the multiplicity of P′ or P′′ on C, that is, 0.)

Have we made progress? One arbitrary singularity was turned into three ordinary singularities and a number of points on the line Z = 0, that may be bad themselves. Let us introduce a parameter to measure progress.

Let the irreducible plane projective curve C have degree n and multiple points of multiplicity m_{i} = m_{P_{i}}(C). Put

g^{∗}(C) = (n − 1)(n − 2)/2 − ∑ m_{i}(m_{i} − 1)/2.

Then g^{∗} ≥ 0.

We prove this below. But let us assume g^{∗} ≥ 0 for the moment, and compute. We have 2g^{∗}(C^{ϕ}) = (2n−m−1)(2n−m−2)−∑ m_{i}(m_{i} −1)+m(m−1)−n(n− 1) − 2(n − m)(n − m − 1) − σ = 2g^{∗}(C) − σ, where σ = ∑ m_{i}(m_{i} − 1) summed over the points of C^{ϕ} on Z = 0. Thus, either g^{∗} decreases, or it stays the same, but then σ = 0, and hence all points of C^{ϕ} on Z = 0 are simple points. This shows that after finitely many steps all multiple points will be ordinary multiple points.

Remains to show that g^{∗} ≥ 0. This will follow from Bezout.

First apply Bezout to F and its derivative F′ = (∂/∂Z)F. If F has multiple points P_{i} with multiplicities m_{i}, then these same points have multiplicity at least m_{i} − 1 for F′, and Bezout tells us that n(n − 1) ≥ ∑ m_{i}(m_{i} − 1). (Needed: F and F′ do not have a common factor, that is, F does not have a squared factor.)

Next improve this inequality by picking a different curve to intersect F with. Take a polynomial G, homogeneous of degree n − 1 (there are n(n + 1)/2 coefficients to choose), and require for each i that P_{i} has multiplicity at least m_{i} − 1 on G (this imposes m_{i}(m_{i} − 1)/2 linear conditions). We know that n(n − 1)/2 ≥ ∑ m_{i}(m_{i} − 1)/2, so this leaves at least n degrees of freedom, and we can require G to pass through t := n(n + 1)/2 − 1 − ∑ m_{i}(m_{i} − 1)/2 further points of F, and still have a nonzero solution. Now Bezout tells us that n(n − 1) ≥ ∑ m_{i}(m_{i} − 1) + t, i.e., g^{∗} ≥ 0. (Needed: F and G do not have a common factor. Since G is unknown of degree n − 1 we need that F is irreducible.)