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Bezout’s theorem
1 Bezout’s theorem
Let C and D be two plane curves, described by equations f (X, Y ) = 0 and g(X, Y ) = 0, where f and g are nonzero polynomials of degree m and n, respec- tively. Bezout’s theorem says that if all is well, then C and D meet in precisely mn points.
1.1 Too many points in common
What can go wrong? If C = D then both curves have all points in common, probably infinitely many (depending on the field). Also, if the equation factors, then the curve is the union of several compo- nents. For example, if C is given by the equation XY = 0, then it is the union of the two lines given by X = 0 and Y = 0. And if D is given by the equation Y 3 − X^3 Y = 0, then it is the union of the line Y = 0 and the cubic curve Y 2 = X^3. Now C and D have the line Y = 0 in common. But this is the only way there can be too many points in common. So the first statement is: if C and D do not have a common component, then they have at most mn points in common.
1.2 Too few points in common
There are several ways there can be ‘missing’ points of intersection. Missing points can have coordinates in an extension field, they can lie at infinity, and they can coincide with other common points. For example, a straight line and a circle may meet in two points, and all is well. Or in zero points, which means that the quadratic equation one has to solve to find the points of intersection has a discriminant that is not a square in the field, and the two points of intersection have coordinates in a quadratic extension field. Finally, the straight line can be tangent to the circle, and we must count the point of intersection twice. And, for example, two parallel lines have a common point at infinity. So now Bezout’s theorem becomes: If C and D do not have a common component, then they have at most mn points in common. If the field is alge- braically closed, and we also count points at infinity, and we count intersection points with proper multiplicity, then there are precisely mn common points.
1.3 The inequality
Let R = k[X, Y ] be the ring of polynomials in the two variables X, Y with coefficients in the field k.
Proposition 1.1 Let f, g ∈ R be nonzero polynomials of degrees m, n, respec- tively. Let C and D be two plane curves, described by the equations f (X, Y ) = 0 and g(X, Y ) = 0. If f and g do not have a common factor, then |C ∩ D| ≤ dimk R/(f, g) ≤ mn.
Proof: (1) Given distinct points Pi (1 ≤ i ≤ t), there are polynomials hi ∈ R (1 ≤ i ≤ t) such that hi(Pi) 6 = 0 and hi(Pj ) = 0 for all i, j, i 6 = j. (Indeed, if Pi = (xi, yi), put hi(X, Y ) =
xj 6 =xi (X^ −^ xj^ ).^
yj 6 =yi (Y^ −^ yj^ ).) (2) |C ∩ D| ≤ dimk R/(f, g). (Indeed, if C and D have common points Pi, then make polynomials as in (1). If
cihi = uf +vg with u, v ∈ R, then substitute the Pi to find ci = 0. This means that the images hi + (f, g) of the hi in R/(f, g) are linearly independent.)
(3) Let Rd be the k-vectorspace of polynomials p(X, Y ) of total degree at most d. If d ≥ 0 then s(d) := dimk Rd = 1 + · · · + (d + 1) = 12 (d + 1)(d + 2).
(4) We have dimk Rd/(f, g) ≤ mn for all d. (Indeed, consider the sequence of maps
Rd−m × Rd−n →α Rd →π Rd/(f, g) → 0
where α is the map α(u, v) = uf + vg and π is the quotient map. Since f and g do not have a common factor, the kernel of α consists of the pairs (wg, −wf ) with w ∈ Rd−m−n and hence has dimension s(d − m − n) for d ≥ m + n. It follows that the image of α has dimension s(d − m) + s(d − n) − s(d − m − n). Since π is surjective, and πα = 0, we find dimk Rd/(f, g) ≤ s(d) − s(d − m) − s(d − n) + s(d − m − n) = mn.)
(5) We have dimk R/(f, g) ≤ mn. (Indeed, if we can find more than mn linearly independent elements in R/(f, g), then for sufficiently large d they will be in Rd/(f, g), contradicting (4).) 2
1.4 Points at infinity
1.4.1 Affine and projective space
Extend affine space to projective space by adding points at infinity, as follows. A point in n-dimensional affine space has n coordinates (x 1 , ..., xn) in the un- derlying field. A point in n-dimensional projective space has n + 1 coordinates (x 1 , ..., xn+1), not all zero, where only the ratio is significant: if a 6 = 0 then (x 1 , ..., xn+1) = (ax 1 , ..., axn+1).
1.4.2 Equations
An equation like Y = X^2 is meaningful in an affine space, but not in a projective space, since whether equality holds must not change when all cordinates of a point are multiplied by the same nonzero constant. Thus, for a projective space one needs homogeneous equations, equations such that all terms have the same degree, as in Y Z = X^2.
1.5 Intersection multiplicity
Let C and D be two plane curves defined by f (X, Y ) = 0 and g(X, Y ) = 0 and let P be a point. We want to define the intersection multiplicity IP (f, g) of C and D at the point P. It should be a nonnegative integer, or ∞ in case C and D have a common component that passes through P. First an operational definition, a series of rules that suffice to compute IP (f, g). We have IP (f, g) = IP (g, f ), and IP (f, g + f h) = IP (f, g), and IP (f, gh) = IP (f, g) + IP (f, h) and IP (f, g) = 0 if P is not a common point of C and D, and IP (f, g) = 1 if C and D are nonsingular at P with distinct tangents, and IP (f, g) = ∞ iff f and g have a common factor.
Example Consider the two circles X^2 + Y 2 = 1 and X^2 + Y 2 = 2. Clearly, any common points must lie at infinity. The homogeneous equations are X^2 + Y 2 − Z^2 = 0 and X^2 + Y 2 − 2 Z^2 = 0, and the common points are the two points (±i, 1 , 0). Now let P = (i, 1 , 0) and consider the intersection multiplicity at P. We have IP (X^2 + Y 2 − Z^2 , X^2 + Y 2 − 2 Z^2 ) = IP (X^2 + Y 2 − Z^2 , Z^2 ) = IP (X^2 + Y 2 , Z^2 ) = 2IP (X^2 + Y 2 , Z) = 2IP (X + iY, Z) + 2IP (X − iY, Z) = 0 + 2 = 2. So, the four common points are the two points (±i, 1 , 0), each counted twice.
Example Consider the two curves Y = X^3 and Y = X^5. The homogeneous equations are Y Z^2 = X^3 and Y Z^4 = X^5 , and the common points are the points (0, 0 , 1), (1, 1 , 1), (− 1 , − 1 , 1), (0, 1 , 0). Since (1, 1 , 1) and (− 1 , − 1 , 1) are ordinary points on the curves, and the curves have different tangents at each of these points, the intersection multiplicity at (1, 1 , 1) and (− 1 , − 1 , 1) is 1. Let P = (0, 0). Then IP (Y − X^3 , Y − X^5 ) = IP (Y − X^3 , X^3 − X^5 ) = IP (Y − X^3 , X^3 ) + IP (Y − X^3 , 1 − X^2 ) = 3IP (Y, X) + 0 = 3, so that the origin is a point with intersection multiplicity 3. Let Q = (0, 1 , 0) and make this the origin by choosing affine coordinates (X/Y, Z/Y ), that is, by putting Y = 1. Then IQ(X^3 −Z^2 , X^5 −Z^4 ) = IQ(X^3 −Z^2 , X^5 −X^3 Z^2 ) = IQ(X^3 −Z^2 , X^3 )+IQ(X^3 − Z^2 , X^2 − Z^2 ) = IQ(Z^2 , X^3 ) + IQ(X^3 − X^2 , X^2 − Z^2 ) = 6IQ(Z, X) + 4IQ(X, Z) + 0 = 10. So, the fifteen common points are the two points (1, 1) and (− 1 , −1) each counted once, the point (0, 0) counted three times, and the point at infinity of the Y -axis (0, 1 , 0), counted ten times.
Algorithm Note that the operational definition will always compute some answer. We may assume that f and g do not have a common factor and that f (P ) = g(P ) = 0. If P = (x, y) then consider the polynomials f ∗(X) := f (X, y) and g∗(X) := g(X, y). We may suppose f ∗^ has degree not larger than that of g∗. If f ∗^ is the zero polynomial, then f has a factor (Y − y), and IP (f, g) = IP (Y − y, g∗) + IP (f 0 , g), where f = (Y − y)f 0. Since f and g do not have a common factor, g∗^ is not the zero polynomial, and g∗(X) = (X − x)ig 2 (X) with i ≥ 1 and g 2 (x) 6 = 0, and now IP (Y − y, g∗) = i. So, finding IP (f, g) has been reduced to finding IP (f 0 , g). Since f 6 = 0, we arrive after finitely many steps in the situation where f ∗^6 =
- Let f ∗^ have leading term axd, and let g∗^ have leading term bXe. Put g 0 = g − ba (X − x)e−df. Now IP (f, g) = IP (f, g 0 ) and g 0 ∗ has smaller degree than g∗^ so by induction we are done. Induction on what? The degrees of f ∗^ and g∗^ go down until one of them is zero, then we divide f or g by (Y − y), and afterwards the degree of f ∗^ or g∗ may be very large again. But we have the promise that IP (f, g) is going to be
finite, and each time that f ∗^ = 0 or g∗^ = 0 we get a contribution of at least 1, so this can happen only finitely many times, and the algorithm terminates.
1.5.1 The local ring at P
A local ring is a ring with a unique maximal ideal. Given a field k and a point P ∈ k^2 , let OP be the ring of rational functions u v with^ u, v^ ∈^ R^ and^ v(P^ )^6 = 0.^ This ring has a unique maximal ideal^ MP^ = { uv ∈ OP | u(P ) = 0} and is called the local ring at P.
1.5.2 Definition of the intersection multiplicity
Let C and D be curves in the plane given by equations f (X, Y ) = 0 and g(X, Y ) = 0. Let P be a point. Let (f, g)P be the ideal OP f + OP g in OP generated by f and g.
Definition IP (C, D) = IP (f, g) = dimk OP /(f, g)P.
Proposition 1.3 If f, g do not have a common factor, then OP = R + (f, g)P (that is, elements of OP have polynomial representatives), and we have IP (f, g) = dimk OP /(f, g)P ≤ dimk R/(f, g).
Proof: Given finitely many elements of OP , we can write them with common denominator. If the images of u v^1 , ..., u vt are linearly independent in OP /(f, g)P , then u 1 , ..., ut are linearly independent in R/(f, g) since (^1) v ∈ OP. This proves the statement about dimensions. Since f, g do not have a common factor, we have dimk R/(f, g) ≤ mn, so dimk OP /(f, g)P is finite. If u v^1 , ..., u vt is a basis of OP /(f, g)P , then (since v, (^1) v ∈ OP so multiplication by v is invertible) also u 1 , ..., ut is a basis. 2
Example Let f (X, Y ) = Y and g(X, Y ) = Y − X^3. The intersection multi- plicity of the cubic Y = X^3 and the line Y = 0 at P = (0, 0) should be 3. The quotient ring R/(f, g) is a vector space over k, and the images of 1, X, X^2 form a basis, so dimk R/(f, g) = 3 and also dimk OP /(f, g)P = 3.
Example Let f (X, Y ) = Y 2 − X^3 and g(X, Y ) = Y 3 − X^4. Then IP (f, g) = 8 for P = (0, 0). A Gr¨obner basis for (f, g) is given by {X^3 −Y 2 , XY 2 −Y 3 , Y 5 − Y 4 } so that R/(f, g) has basis with representatives X^2 Y , X^2 , XY , X, Y 4 , Y 3 , Y 2 , Y , 1, and dimk R/(f, g) = 9. But Y − 1 is nonzero in P , so (f, g)P also contains (Y 5 − Y 4 )/(Y − 1) = Y 4 , and dimk OP /(f, g)P = 8.
In these examples it was clear that dimk OP /(f, g)P had at most the given value. That it has precisely the claimed value will follow if we show that IP (f, g) defined in this way satisfies the rules given earlier.
Proposition 1.4 The algorithm given earlier computes IP (f, g) as defined above. The rules given earlier are valid.
Proof: The rules IP (f, g) = IP (g, f ) and IP (f, g + f h) = IP (f, g) are obvious, since the ideal (f, g)P does not change.
If f and g have a common factor h, dimk OP /(f, g)P ≥ dimk OP /(h)P = ∞. Conversely, if f and g do not have a common factor, then dimk OP /(f, g)P ≤ dimk R/(f, g) ≤ mn < ∞ if f and g have degrees m and n, respectively.
Remains to show the existence of N. It suffices to show: Let p be a polynomial with p(Q) = 0. Let N ≥ d := dimk OQ/(f, g)Q. Then pN^ ∈ (f, g)Q. Indeed, let Ji := piOQ + (f, g)Q. The sequence (Ji)i≥ 0 of ideals is decreasing, but can have at most d + 1 different members, so there is an i with 0 ≤ i ≤ d such that Ji = Ji+1. This means that pi^ = pi+1u + v with u ∈ OQ and v ∈ (f, g)Q. Since 1 1 −pu ∈ OQ^ it follows that^ p
i (^) = v 1 −pu ∈^ (f, g)Q, as desired.^2
1.5.4 Equality
Proposition 1.6 Assume that the field k is algebraically closed. Then
∑
P
dimk OP /(f, g)P = dimk R/(f, g).
Proof:∏ We have to show that (f, g) is the full kernel of the map π : R →
P OP^ /(f, g)P^.^ Pick^ h^ in this kernel.^ Consider^ L^ :=^ {p^ ∈^ R^ |^ ph^ ∈^ (f, g)}. This is an ideal in R. If (x, y) ∈ V (L) then P := (x, y) ∈ C ∩ D since f, g ∈ L. Since π(h)P = 0 we have h = uf + vg for certain u, v ∈ OP. Write u, v with common denominator p, then p ∈ L and p(P ) 6 = 0, contradiction. Hence V (L) = ∅. Since k is algebraically closed we can apply the Nullstellensatz and conclude that 1 ∈ L, that is, h ∈ (f, g). 2
1.6 Conclusion
We did all the work required. Assume that k is algebraically closed. The curves C and D have only finitely many points (affine or at infinity) in common, and there is a line that misses all these points (since k is infinite). Choose such a line as line at infinity to find that C and D have precisely mn points in common, counting multiplicities.
(A detail to check: is the definition of intersection multiplicity invariant for change of coordinates? But it is.)
