Bézout’s Theorem
Jennifer Li
Department of Mathematics University of Massachusetts Amherst, MA ¡ No common components
How many intersection points are there?
Motivation
f (x,y) g(x,y)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 2 / 80 How many intersection points are there?
Motivation
f (x,y) ¡ No common components g(x,y)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 2 / 80 Motivation
f (x,y) ¡ No common components g(x,y)
How many intersection points are there?
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 2 / 80 Algebra Geometry
{ set of solutions } { intersection points of curves }
Motivation
f (x,y) = 0
g(x,y) = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 3 / 80 Geometry
{ intersection points of curves }
Motivation
f (x,y) = 0
g(x,y) = 0
Algebra
{ set of solutions }
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 3 / 80 Motivation
f (x,y) = 0
g(x,y) = 0
Algebra Geometry
{ set of solutions } { intersection points of curves }
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 3 / 80 Motivation
f (x,y) = 0
g(x,y) = 0
How many intersections are there?
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 4 / 80 Motivation
f (x,y) = 0
g(x,y) = 0
How many intersections are there?
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 5 / 80 Motivation
f (x,y) = 0
g(x,y) = 0
How many intersections are there?
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 6 / 80 f (x): polynomial in one variable
2 n f (x) = a0 + a1x + a2x +⋯+ anx
Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of a polynomial f (x) is the largest power of x with nonzero coefficient.
Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of f (x) is 5.
Polynomials: One Variable
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 7 / 80 Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of a polynomial f (x) is the largest power of x with nonzero coefficient.
Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of f (x) is 5.
Polynomials: One Variable
f (x): polynomial in one variable
2 n f (x) = a0 + a1x + a2x +⋯+ anx
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 7 / 80 The degree of a polynomial f (x) is the largest power of x with nonzero coefficient.
Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of f (x) is 5.
Polynomials: One Variable
f (x): polynomial in one variable
2 n f (x) = a0 + a1x + a2x +⋯+ anx
Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 7 / 80 Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of f (x) is 5.
Polynomials: One Variable
f (x): polynomial in one variable
2 n f (x) = a0 + a1x + a2x +⋯+ anx
Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of a polynomial f (x) is the largest power of x with nonzero coefficient.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 7 / 80 Polynomials: One Variable
f (x): polynomial in one variable
2 n f (x) = a0 + a1x + a2x +⋯+ anx
Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of a polynomial f (x) is the largest power of x with nonzero coefficient.
Example. f (x) = 1 + 8x + x2 + 5x3 + 12x5
The degree of f (x) is 5.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 7 / 80 f (x,y): polynomial in two variables
2 2 n n f (x,y) = a00 + a10x + a01y + a20x + a11xy + a02y +⋯+ an0x +⋯+ a0ny
i j a i j x y
Polynomials: Two Variables
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 8 / 80 Polynomials: Two Variables
f (x,y): polynomial in two variables
2 2 n n f (x,y) = a00 + a10x + a01y + a20x + a11xy + a02y +⋯+ an0x +⋯+ a0ny
i j a i j x y
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 8 / 80 Polynomials: Two Variables
f (x,y): polynomial in two variables
2 2 n n f (x,y) = a00 + a10x + a01y + a20x + a11xy + a02y +⋯+ an0x +⋯+ a0ny
i j a ij x y
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 9 / 80 Polynomials: Two Variables
f (x,y): polynomial in two variables
2 2 n n f (x,y) = a00 + a10x + a01y + a20x + a11xy + a02y +⋯+ an0x +⋯+ a0ny
i j a ij x y
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 10 / 80 Polynomials: Two Variables
f (x,y): polynomial in two variables
2 2 n n f (x,y) = a00 + a10x + a01y + a20x + a11xy + a02y +⋯+ an0x +⋯+ a0ny
i + j = 0 i + j = 1 i + j = 2 ⋮ i + j = n
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 11 / 80 Example.
f (x,y) = 5x2 + 8xy + 12y2 − 4x + y + 13
is a polynomial of degree 2.
Degree of a Polynomial
The degree of a polynomial f (x,y) is the largest sum of powers of x and y.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 12 / 80 Degree of a Polynomial
The degree of a polynomial f (x,y) is the largest sum of powers of x and y.
Example.
f (x,y) = 5x2 + 8xy + 12y2 − 4x + y + 13
is a polynomial of degree 2.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 12 / 80 Polynomials over a field k
2 n f (x) = a0 + a1x + a2x +⋯+ anx
⊛ In Calculus I: usually over field k = R
⊛ Means ai all belong to R
Polynomials over R
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 13 / 80 ⊛ In Calculus I: usually over field k = R
⊛ Means ai all belong to R
Polynomials over R
Polynomials over a field k
2 n f (x) = a0 + a1x + a2x +⋯+ anx
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 13 / 80 Polynomials over R
Polynomials over a field k
2 n f (x) = a0 + a1x + a2x +⋯+ anx
⊛ In Calculus I: usually over field k = R
⊛ Means ai all belong to R
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 13 / 80 Polynomials over R
Polynomials over a field k
2 n f (x) = a0 + a1x + a2x +⋯+ anx
⊛ In Calculus I: usually over field k = R
⊛ Means ai’s all belong to R
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 14 / 80 2 n f (x) = a0 + a1x + a2x +⋯+ anx
⊛ Means ai’s all belong to C
Polynomials over C
Polynomials over field C
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 15 / 80 Polynomials over C
Polynomials over field C
2 n f (x) = a0 + a1x + a2x +⋯+ anx
⊛ Means ai’s all belong to C
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 15 / 80 Numbers of form
a + bi
where ⊛ numbers a and b are in R ⊛ the symbol i is the imaginary number with property i 2 = −1
a ∶ the real part b ∶ the imaginary part
⊛ every real number is of the form
a + 0 i
Complex Numbers
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 16 / 80 where ⊛ numbers a and b are in R ⊛ the symbol i is the imaginary number with property i 2 = −1
a ∶ the real part b ∶ the imaginary part
⊛ every real number is of the form
a + 0 i
Complex Numbers
Numbers of form
a + bi
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 16 / 80 ⊛ the symbol i is the imaginary number with property i 2 = −1
a ∶ the real part b ∶ the imaginary part
⊛ every real number is of the form
a + 0 i
Complex Numbers
Numbers of form
a + bi
where ⊛ numbers a and b are in R
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 16 / 80 a ∶ the real part b ∶ the imaginary part
⊛ every real number is of the form
a + 0 i
Complex Numbers
Numbers of form
a + bi
where ⊛ numbers a and b are in R ⊛ the symbol i is the imaginary number with property i 2 = −1
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 16 / 80 b ∶ the imaginary part
⊛ every real number is of the form
a + 0 i
Complex Numbers
Numbers of form
a + bi
where ⊛ numbers a and b are in R ⊛ the symbol i is the imaginary number with property i 2 = −1
a ∶ the real part
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 16 / 80 ⊛ every real number is of the form
a + 0 i
Complex Numbers
Numbers of form
a + bi
where ⊛ numbers a and b are in R ⊛ the symbol i is the imaginary number with property i 2 = −1
a ∶ the real part b ∶ the imaginary part
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 16 / 80 Complex Numbers
Numbers of form
a + bi
where ⊛ numbers a and b are in R ⊛ the symbol i is the imaginary number with property i 2 = −1
a ∶ the real part b ∶ the imaginary part
⊛ every real number is of the form
a + 0 i
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 16 / 80 C
R
Algebraic Closure
Algebraically closed?
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 17 / 80 R
Algebraic Closure
Algebraically closed?
C
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 17 / 80 Algebraic Closure
Algebraically closed?
C
R
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 17 / 80 For any nonconstant polynomial f (x) over C, we can find a number r in C
such that f (r) = 0.
Field C is algebraically closed
Algebraically closed? C
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 18 / 80 such that f (r) = 0.
Field C is algebraically closed
Algebraically closed? C
For any nonconstant polynomial f (x) over C, we can find a number r in C
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 18 / 80 such that f (r) = 0.
Field C is algebraically closed
Algebraically closed? C
For any nonconstant polynomial f (x) over C, we can find a number r in C
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 18 / 80 Field C is algebraically closed
Algebraically closed? C
For any nonconstant polynomial f (x) over C, we can find a number r in C
such that f (r) = 0.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 18 / 80 Field R is not algebraically closed
Algebraically closed? R
f (x) = x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 19 / 80 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 f (i ) = i 2 + 1 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
and
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 = (−1)2(i )2 + 1 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 = −1 + 1 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 = 0
Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 Field R is not algebraically closed
Algebraically closed? R
f (x) = 1x2 + 1 polynomial over R
There are only two solutions to x2 + 1 = 0:
f (i ) = i 2 + 1 = −1 + 1 = 0
and
f (−i ) = −i 2 + 1 = (−1)2(i )2 + 1 = −1 + 1 = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 20 / 80 f (x) = x2 + 1
Polynomial over R No solutions in R
Field R is not algebraically closed
But...
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 21 / 80 f (x) = x2 + 1
Polynomial over R No solutions in R
Field R is not algebraically closed
But...
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 21 / 80 Polynomial over R No solutions in R
Field R is not algebraically closed
But...
f (x) = x2 + 1
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 21 / 80 No solutions in R
Field R is not algebraically closed
But...
f (x) = x2 + 1
Polynomial over R
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 21 / 80 Field R is not algebraically closed
But...
f (x) = x2 + 1
Polynomial over R No solutions in R
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 21 / 80 Over R, a polynomial of degree d ≥ 0 has at most d solutions.
Over C, a polynomial of degree d ≥ 0 has exactly d solutions, counted with multiplicity.
The Fundamental Theorem of Algebra
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 22 / 80 Over C, a polynomial of degree d ≥ 0 has exactly d solutions, counted with multiplicity.
The Fundamental Theorem of Algebra
Over R, a polynomial of degree d ≥ 0 has at most d solutions.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 22 / 80 The Fundamental Theorem of Algebra
Over R, a polynomial of degree d ≥ 0 has at most d solutions.
Over C, a polynomial of degree d ≥ 0 has exactly d solutions, counted with multiplicity.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 22 / 80 Tangent at point p Then l and C intersect with multiplicity ≥ 2 at point p.
Intersection Multiplicity
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 23 / 80 Tangent at point p Then l and C intersect with multiplicity ≥ 2 at point p.
Intersection Multiplicity
l ∶ Line C ∶ Curve
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 23 / 80 Then l and C intersect with multiplicity ≥ 2 at point p.
Intersection Multiplicity
l ∶ Line C ∶ Curve
Tangent at point p
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 23 / 80 Intersection Multiplicity
l ∶ Line C ∶ Curve
Tangent at point p Then l and C intersect with multiplicity ≥ 2 at point p.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 23 / 80 Intersection Multiplicity
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 24 / 80 Intersection Multiplicity
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 24 / 80 ⊛ All polynomials over algebraically closed field C
⊛ Count with multiplicity
Warning: In pictures, we can only see the real (R) part!
So far . . .
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 25 / 80 ⊛ Count with multiplicity
Warning: In pictures, we can only see the real (R) part!
So far . . .
⊛ All polynomials over algebraically closed field C
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 25 / 80 Warning: In pictures, we can only see the real (R) part!
So far . . .
⊛ All polynomials over algebraically closed field C
⊛ Count with multiplicity
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 25 / 80 So far . . .
⊛ All polynomials over algebraically closed field C
⊛ Count with multiplicity
Warning: In pictures, we can only see the real (R) part!
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 25 / 80 The points (x,y) that satisfy f (x,y) = 0 cut out an affine plane curve.
The polynomial f (x,y) is called the defining polynomial of the curve.
Affine Plane Curves
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 26 / 80 The polynomial f (x,y) is called the defining polynomial of the curve.
Affine Plane Curves
The points (x,y) that satisfy f (x,y) = 0 cut out an affine plane curve.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 26 / 80 Affine Plane Curves
The points (x,y) that satisfy f (x,y) = 0 cut out an affine plane curve.
The polynomial f (x,y) is called the defining polynomial of the curve.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 26 / 80 Example.
x2 + y2 = 9 x2 + y2 − 9 = 0
Defining equation: f (x,y) = x2 + y2 − 9 degree 2
∴ the curve is of degree 2, called a conic.
Degree of a Curve
The degree of a curve is the degree of its defining polynomial f (x,y).
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 27 / 80 Defining equation: f (x,y) = x2 + y2 − 9 degree 2
∴ the curve is of degree 2, called a conic.
Degree of a Curve
The degree of a curve is the degree of its defining polynomial f (x,y).
Example.
x2 + y2 = 9 x2 + y2 − 9 = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 27 / 80 f (x,y) = x2 + y2 − 9 degree 2
∴ the curve is of degree 2, called a conic.
Degree of a Curve
The degree of a curve is the degree of its defining polynomial f (x,y).
Example.
x2 + y2 = 9 x2 + y2 − 9 = 0
Defining equation:
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 27 / 80 degree 2
∴ the curve is of degree 2, called a conic.
Degree of a Curve
The degree of a curve is the degree of its defining polynomial f (x,y).
Example.
x2 + y2 = 9 x2 + y2 − 9 = 0
Defining equation: f (x,y) = x2 + y2 − 9
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 27 / 80 ∴ the curve is of degree 2, called a conic.
Degree of a Curve
The degree of a curve is the degree of its defining polynomial f (x,y).
Example.
x2 + y2 = 9 x2 + y2 − 9 = 0
Defining equation: f (x,y) = x2 + y2 − 9 degree 2
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 27 / 80 Degree of a Curve
The degree of a curve is the degree of its defining polynomial f (x,y).
Example.
x2 + y2 = 9 x2 + y2 − 9 = 0
Defining equation: f (x,y) = x2 + y2 − 9 degree 2
∴ the curve is of degree 2, called a conic.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 27 / 80 Defining equation: g(x,y) = y2 − x2 − x3 degree 3
∴ the curve is of degree 3, called a cubic.
Degree of a Curve
Example.
y2 = x2 + x3 y2 − x2 − x3 = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 28 / 80 g(x,y) = y2 − x2 − x3 degree 3
∴ the curve is of degree 3, called a cubic.
Degree of a Curve
Example.
y2 = x2 + x3 y2 − x2 − x3 = 0
Defining equation:
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 28 / 80 degree 3
∴ the curve is of degree 3, called a cubic.
Degree of a Curve
Example.
y2 = x2 + x3 y2 − x2 − x3 = 0
Defining equation: g(x,y) = y2 − x2 − x3
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 28 / 80 ∴ the curve is of degree 3, called a cubic.
Degree of a Curve
Example.
y2 = x2 + x3 y2 − x2 − x3 = 0
Defining equation: g(x,y) = y2 − x2 − x3 degree 3
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 28 / 80 Degree of a Curve
Example.
y2 = x2 + x3 y2 − x2 − x3 = 0
Defining equation: g(x,y) = y2 − x2 − x3 degree 3
∴ the curve is of degree 3, called a cubic.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 28 / 80 Example.
Intersection of Curves
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 29 / 80 Intersection of Curves
Example.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 29 / 80 Intersection of Curves
Circle ∩ line
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 30 / 80 Intersection of Curves
Circle ∩ line
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 30 / 80 Example.
Intersection of Curves
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 31 / 80 Intersection of Curves
Example.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 31 / 80 Intersection of Curves
Parabola ∩ line
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 32 / 80 Example.
Intersection of Curves
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 33 / 80 Intersection of Curves
Example.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 33 / 80 Intersection of Curves
Ellipse ∩ Ellipse
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 34 / 80 Example.
Intersection of Curves
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 35 / 80 Intersection of Curves
Example.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 35 / 80 Intersection of Curves
Cubic ∩ Line
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 36 / 80 C : curve of degree m D : curve of degree n
Let S C ∩ D S = number of intersection points
Then
S C ∩ D S ≤ m ⋅ n ↑
Observation
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 37 / 80 D : curve of degree n
Let S C ∩ D S = number of intersection points
Then
S C ∩ D S ≤ m ⋅ n ↑
Observation
C : curve of degree m
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 37 / 80 Let S C ∩ D S = number of intersection points
Then
S C ∩ D S ≤ m ⋅ n ↑
Observation
C : curve of degree m D : curve of degree n
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 37 / 80 Then
S C ∩ D S ≤ m ⋅ n ↑
Observation
C : curve of degree m D : curve of degree n
Let S C ∩ D S = number of intersection points
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 37 / 80 Observation
C : curve of degree m D : curve of degree n
Let S C ∩ D S = number of intersection points
Then
S C ∩ D S ≤ m ⋅ n ↑
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 37 / 80 Observation
C : curve of degree m D : curve of degree n
Let S C ∩ D S = number of intersection points
Then
S C ∩ D S ≤ m ⋅ n ↑
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 38 / 80 Intersection of Two Circles
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 39 / 80 Intersection of Two Circles
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 40 / 80 False! They intersect in at most two points.
Intersection of Two Circles
Expected: Circles intersect at2 × 2 = 4 points.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 41 / 80 Intersection of Two Circles
Expected: Circles intersect at2 × 2 = 4 points.
False! They intersect in at most two points.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 41 / 80 Intersection of Two Circles
Expected: Circles intersect at2 × 2 = 4 points.
False! They intersect in at most two points.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 42 / 80 Intersection of Two Circles
Expected: Circles intersect at2 × 2 = 4 points.
False! They intersect in at most two points.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 43 / 80 Intersection of Two Circles
Expected: Circles intersect at2 × 2 = 4 points.
False! They intersect in at most two points.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 44 / 80 Intersection of Two Circles
Expected: Circles intersect at2 × 2 = 4 points.
False! They intersect in at most two points.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 45 / 80 Intersection of Two Circles
Expected: Circles intersect at2 × 2 = 4 points.
False! They intersect in at most two points.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 46 / 80 2 2 2 (x − a) +(y − b) = r1
2 2 2 − (x − c) +(y − d) = r2
r2 − r2 + c 2 + d 2 − a 2 − b 2 (c − a)x +(d − b)y = 1 2 2 A linear equation
Intersection of Two Circles
Reason:
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 47 / 80 Intersection of Two Circles
Reason:
2 2 2 (x − a) +(y − b) = r1
2 2 2 − (x − c) +(y − d) = r2
r2 − r2 + c 2 + d 2 − a 2 − b 2 (c − a)x +(d − b)y = 1 2 2 A linear equation
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 47 / 80 r2 − r2 + c 2 + d 2 − a 2 − b 2 (c − a)x +(d − b)y = 1 2 2 A linear equation
Intersection of Two Circles
Reason:
2 2 2 (x − a) +(y − b) = r1
2 2 2 − (x − c) +(y − d) = r2
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 48 / 80 Intersection of Two Circles
Reason:
2 2 2 (x − a) +(y − b) = r1
2 2 2 − (x − c) +(y − d) = r2
r2 − r2 + c 2 + d 2 − a 2 − b 2 (c − a)x +(d − b)y = 1 2 2 A linear equation
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 48 / 80 circle ∩ circle ≡ circle ∩ line
Intersection of Two Circles
Reason:
circle- circle= linear equation
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 49 / 80 Intersection of Two Circles
Reason:
circle- circle= linear equation
circle ∩ circle ≡ circle ∩ line
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 49 / 80 Intersection of Two Lines
An even simpler example:
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 50 / 80 Intersection of Two Lines
An even simpler example:
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 50 / 80 Not if they’re parallel!
Intersection of Two Lines
Expected: Two lines intersect at1 × 1 = 1 point.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 51 / 80 Intersection of Two Lines
Expected: Two lines intersect at1 × 1 = 1 point.
Not if they’re parallel!
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 51 / 80 Off to Projective Space!
Motivation: affine space Ð→ projective space.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 52 / 80 Projective Space: The Idea
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 53 / 80 Projective Space: The Idea
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 54 / 80 Projective space Pn: Equivalence class of nonzero vectors
(A0,...,An)
where equivalence is defined as follows:
(A0,...,An) ∼ (B0,...,Bn) if and only if (A0,...,An) = (λB0,...,λBn)
Note: In projective space, there is no point (0,...,0)!
The Projective Space
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 55 / 80 where equivalence is defined as follows:
(A0,...,An) ∼ (B0,...,Bn) if and only if (A0,...,An) = (λB0,...,λBn)
Note: In projective space, there is no point (0,...,0)!
The Projective Space
Projective space Pn: Equivalence class of nonzero vectors
(A0,...,An)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 55 / 80 Note: In projective space, there is no point (0,...,0)!
The Projective Space
Projective space Pn: Equivalence class of nonzero vectors
(A0,...,An)
where equivalence is defined as follows:
(A0,...,An) ∼ (B0,...,Bn) if and only if (A0,...,An) = (λB0,...,λBn)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 55 / 80 The Projective Space
Projective space Pn: Equivalence class of nonzero vectors
(A0,...,An)
where equivalence is defined as follows:
(A0,...,An) ∼ (B0,...,Bn) if and only if (A0,...,An) = (λB0,...,λBn)
Note: In projective space, there is no point (0,...,0)!
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 55 / 80 P1: Projective line
Points look like (A0,A1)
P2: Projective plane
Points look like (A0,A1,A2)
The Projective Space
Note: In projective space, there is no point (0,...,0)!
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 56 / 80 P2: Projective plane
Points look like (A0,A1,A2)
The Projective Space
Note: In projective space, there is no point (0,...,0)!
P1: Projective line
Points look like (A0,A1)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 56 / 80 The Projective Space
Note: In projective space, there is no point (0,...,0)!
P1: Projective line
Points look like (A0,A1)
P2: Projective plane
Points look like (A0,A1,A2)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 56 / 80 Projective plane P2: Equivalence class of nonzero vectors
(X,Y ,Z)
where equivalence is defined as follows:
(X,Y ,Z) ∼ (X′,Y ′,Z′) if and only if (X,Y ,Z) = (λX′,λY ′,λZ′)
Complex Projective Plane CP2
Note: In CP2, there is no point (0,0,0)!
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 57 / 80 where equivalence is defined as follows:
(X,Y ,Z) ∼ (X′,Y ′,Z′) if and only if (X,Y ,Z) = (λX′,λY ′,λZ′)
Complex Projective Plane CP2
Note: In CP2, there is no point (0,0,0)!
Projective plane P2: Equivalence class of nonzero vectors
(X,Y ,Z)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 57 / 80 Complex Projective Plane CP2
Note: In CP2, there is no point (0,0,0)!
Projective plane P2: Equivalence class of nonzero vectors
(X,Y ,Z)
where equivalence is defined as follows:
(X,Y ,Z) ∼ (X′,Y ′,Z′) if and only if (X,Y ,Z) = (λX′,λY ′,λZ′)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 57 / 80 A polynomial is homogeneous of degree d if all monomials with nonzero coefficient have degree d.
Example.
f (x,y) = 3x4y + x2y3 + 2x3y2 + 8y5
3x4 y1 ←→ 1 + 4 = 5 x2 y3 ←→ 2 + 3 = 5 2x3 y2 ←→ 3 + 2 = 5 x0 8y5 ←→ 0 + 5 = 5
Polynomial f (x,y) is homogeneous of degree 5
Homogeneous Polynomials
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 58 / 80 Example.
f (x,y) = 3x4y + x2y3 + 2x3y2 + 8y5
3x4 y1 ←→ 1 + 4 = 5 x2 y3 ←→ 2 + 3 = 5 2x3 y2 ←→ 3 + 2 = 5 x0 8y5 ←→ 0 + 5 = 5
Polynomial f (x,y) is homogeneous of degree 5
Homogeneous Polynomials
A polynomial is homogeneous of degree d if all monomials with nonzero coefficient have degree d.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 58 / 80 3x4 y1 ←→ 1 + 4 = 5 x2 y3 ←→ 2 + 3 = 5 2x3 y2 ←→ 3 + 2 = 5 x0 8y5 ←→ 0 + 5 = 5
Polynomial f (x,y) is homogeneous of degree 5
Homogeneous Polynomials
A polynomial is homogeneous of degree d if all monomials with nonzero coefficient have degree d.
Example.
f (x,y) = 3x4y + x2y3 + 2x3y2 + 8y5
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 58 / 80 x2 y3 ←→ 2 + 3 = 5 2x3 y2 ←→ 3 + 2 = 5 x0 8y5 ←→ 0 + 5 = 5
Polynomial f (x,y) is homogeneous of degree 5
Homogeneous Polynomials
A polynomial is homogeneous of degree d if all monomials with nonzero coefficient have degree d.
Example.
f (x,y) = 3x4y + x2y3 + 2x3y2 + 8y5
3x4 y1 ←→ 1 + 4 = 5
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 58 / 80 2x3 y2 ←→ 3 + 2 = 5 x0 8y5 ←→ 0 + 5 = 5
Polynomial f (x,y) is homogeneous of degree 5
Homogeneous Polynomials
A polynomial is homogeneous of degree d if all monomials with nonzero coefficient have degree d.
Example.
f (x,y) = 3x4y + x2y3 + 2x3y2 + 8y5
3x4 y1 ←→ 1 + 4 = 5 x2 y3 ←→ 2 + 3 = 5
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 58 / 80 Polynomial f (x,y) is homogeneous of degree 5
Homogeneous Polynomials
A polynomial is homogeneous of degree d if all monomials with nonzero coefficient have degree d.
Example.
f (x,y) = 3x4y + x2y3 + 2x3y2 + 8y5
3x4 y1 ←→ 1 + 4 = 5 x2 y3 ←→ 2 + 3 = 5 2x3 y2 ←→ 3 + 2 = 5 x0 8y5 ←→ 0 + 5 = 5
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 58 / 80 Homogeneous Polynomials
A polynomial is homogeneous of degree d if all monomials with nonzero coefficient have degree d.
Example.
f (x,y) = 3x4y + x2y3 + 2x3y2 + 8y5
3x4 y1 ←→ 1 + 4 = 5 x2 y3 ←→ 2 + 3 = 5 2x3 y2 ←→ 3 + 2 = 5 x0 8y5 ←→ 0 + 5 = 5
Polynomial f (x,y) is homogeneous of degree 5
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 58 / 80 Projective Plane ←→ Affine Plane
Points in affine plane:
(x,y) lower case
Points in the projective plane:
(X,Y ,Z) upper case
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 59 / 80 Note: There is no point (0,0,0)! Suppose Z ≠ 0
X Y Z X Y (X,Y ,Z)∼ , , ∼ , ,1 Z Z Z Z Z
Then we have:
(X,Y ,Z) ∼ x,y,1 ←→ (x,y)
Projective Plane ←→ Affine Plane
Let (X,Y ,Z) be a point in CP2
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 60 / 80 X Y Z X Y (X,Y ,Z)∼ , , ∼ , ,1 Z Z Z Z Z
Then we have:
(X,Y ,Z) ∼ x,y,1 ←→ (x,y)
Projective Plane ←→ Affine Plane
Let (X,Y ,Z) be a point in CP2 Note: There is no point (0,0,0)! Suppose Z ≠ 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 60 / 80 Projective Plane ←→ Affine Plane
Let (X,Y ,Z) be a point in CP2 Note: There is no point (0,0,0)! Suppose Z ≠ 0
X Y Z X Y (X,Y ,Z)∼ , , ∼ , ,1 Z Z Z Z Z
Then we have:
(X,Y ,Z) ∼ x,y,1 ←→ (x,y)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 60 / 80 X Y ∼ , ,1 Z Z
Then we have:
(X,Y ,Z) ∼ x,y,1 ←→ (x,y)
Projective Plane ←→ Affine Plane
Let (X,Y ,Z) be a point in CP2 Note: There is no point (0,0,0)! Suppose Z ≠ 0
X Y Z (X,Y ,Z) ∼ , , Z Z Z
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 61 / 80 Projective Plane ←→ Affine Plane
Let (X,Y ,Z) be a point in CP2 Note: There is no point (0,0,0)! Suppose Z ≠ 0
X Y Z X Y (X,Y ,Z) ∼ , , ∼ , ,1 Z Z Z Z Z
Then we have:
(X,Y ,Z) ∼ x,y,1 ←→ (x,y)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 61 / 80 Projective Plane ←→ Affine Plane
Let (X,Y ,Z) be a point in CP2 Note: There is no point (0,0,0)! Suppose Z ≠ 0
X Y Z X Y (X,Y ,Z) ∼ , , ∼ , ,1 Z Z Z Z Z
Then we have:
(X,Y ,Z) ∼ x,y,1 ←→ (x,y)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 62 / 80 2 Then P = UZ ∪ l Z
Projective Plane ←→ Affine Plane
Points at Finite Distance Points Infinitely Far Away (Affine Part) (Projective Part)
UZ = {(X,Y ,Z)S Z ≠ 0} l Z = {(X,Y ,0)}
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 63 / 80 Projective Plane ←→ Affine Plane
Points at Finite Distance Points Infinitely Far Away (Affine Part) (Projective Part)
UZ = {(X,Y ,Z)S Z ≠ 0} l Z = {(X,Y ,0)}
2 Then P = UZ ∪ l Z
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 63 / 80 Points at Finite Distance Points Infinitely Far Away (Affine Part) (Projective Part)
UX = {(X,Y ,Z)S X ≠ 0} l X = {(0,Y ,Z)} UY = {(X,Y ,Z)S Y ≠ 0} l Y = {(X,0, Z)} UZ = {(X,Y ,Z)S Z ≠ 0} l Z = {(X,Y ,0)}
Projective Plane ←→ Affine Plane
Analogously,
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 64 / 80 Projective Plane ←→ Affine Plane
Analogously,
Points at Finite Distance Points Infinitely Far Away (Affine Part) (Projective Part)
UX = {(X,Y ,Z)S X ≠ 0} l X = {(0,Y ,Z)} UY = {(X,Y ,Z)S Y ≠ 0} l Y = {(X,0, Z)} UZ = {(X,Y ,Z)S Z ≠ 0} l Z = {(X,Y ,0)}
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 64 / 80 Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + cZ = 0Z set Z = 0 aX + bY + dZ = 0Z set Z = 0
Then X = b and Y = −a
∴ the parallel lines intersect at the point (b,−a,0) which lies on lZ
Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 65 / 80 ⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + cZ = 0Z set Z = 0 aX + bY + dZ = 0Z set Z = 0
Then X = b and Y = −a
∴ the parallel lines intersect at the point (b,−a,0) which lies on lZ
Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 65 / 80 aX + bY + cZ = 0Z set Z = 0 aX + bY + dZ = 0Z set Z = 0
Then X = b and Y = −a
∴ the parallel lines intersect at the point (b,−a,0) which lies on lZ
Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 65 / 80 Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + cZ = 0Z set Z = 0 aX + bY + dZ = 0Z set Z = 0
Then X = b and Y = −a
∴ the parallel lines intersect at the point (b,−a,0) which lies on lZ
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 65 / 80 Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + cZ = 0Z homogeneous of degree 1 aX + bY + dZ = 0Z homogeneous of degree 1
Then X = b and Y = −a
∴ the parallel lines intersect at the point (b,−a,0) which lies on lZ
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 66 / 80 Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + c 0 = 0 setZ=0 aX + bY + d 0 = 0 setZ=0
Then X = b and Y = −a
∴ the parallel lines intersect at the point (b,−a,0) which lies on lZ
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 67 / 80 Then X = b and Y = −a
∴ the parallel lines intersect at the point (b,−a,0) a point on lZ
Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + 0 = 0 aX + bY + 0 = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 68 / 80 ∴ the parallel lines intersect at the point (b,−a,0) a point on lZ
Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + 0 = 0 aX + bY + 0 = 0
Then X = b and Y = −a
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 68 / 80 a point on lZ
Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + 0 = 0 aX + bY + 0 = 0
Then X = b and Y = −a ∴ the parallel lines intersect at the point (b,−a,0)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 68 / 80 Intersection of Parallel Lines
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider parallel lines
ax + by + c = 0 ax + by + d = 0
⋆⋆⋆ Homogenize! ⋆⋆⋆
aX + bY + 0 = 0 aX + bY + 0 = 0
Then X = b and Y = −a
∴ the parallel lines intersect at the point (b,−a,0) a point on lZ
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 68 / 80 If they are not parallel . . . If they are parallel . . .
Two lines intersect at one point.
Intersection of Two Lines
Expected: Two lines intersect at1 × 1 = 1 point.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 69 / 80 If they are parallel . . .
Two lines intersect at one point.
Intersection of Two Lines
Expected: Two lines intersect at1 × 1 = 1 point.
If they are not parallel . . .
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 69 / 80 Two lines intersect at one point.
Intersection of Two Lines
Expected: Two lines intersect at1 × 1 = 1 point.
If they are not parallel . . . If they are parallel . . .
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 69 / 80 Intersection of Two Lines
Expected: Two lines intersect at1 × 1 = 1 point.
If they are not parallel . . . If they are parallel . . .
Two lines intersect at one point.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 69 / 80 Consider circle
(x − a)2 +(y − b)2 = r2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aXZ + a2Z2)+(Y 2 − 2bYZ + b2Z2) = r2Z2 Set Z = 0 homogeneous of degree 2
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 70 / 80 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aXZ + a2Z2)+(Y 2 − 2bYZ + b2Z2) = r2Z2 Set Z = 0 homogeneous of degree 2
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 = r2
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 70 / 80 ⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aXZ + a2Z2)+(Y 2 − 2bYZ + b2Z2) = r2Z2 Set Z = 0 homogeneous of degree 2
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 = r2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 70 / 80 (X 2 − 2aXZ + a2Z2)+(Y 2 − 2bYZ + b2Z2) = r2Z2 Set Z = 0 homogeneous of degree 2
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 = r2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 70 / 80 homogeneous of degree 2
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 = r2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aXZ + a2Z2)+(Y 2 − 2bYZ + b2Z2) = r2Z2 Set Z = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 70 / 80 Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 = r2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aXZ + a2Z2)+(Y 2 − 2bYZ + b2Z2) = r2Z2 Set Z = 0 homogeneous of degree 2
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 70 / 80 X 2 + Y 2 = 0
Then X = 1 and Y = ±i.
∴ any circle intersects the line at infinity lZ at points (1,i,0) and (1,−i,0) ∴ any two circles intersect at (1,i,0) and (1,−i,0)
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aX 0 + a2 02)+(Y 2 − 2bY 0 + b2 02) = r2 02 Set Z = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 71 / 80 Then X = 1 and Y = ±i.
∴ any circle intersects the line at infinity lZ at points (1,i,0) and (1,−i,0) ∴ any two circles intersect at (1,i,0) and (1,−i,0)
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aX 0 + a2 02)+(Y 2 − 2bY 0 + b2 02) = r2 02 Set Z = 0 X 2 + Y 2 = 0
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 71 / 80 ∴ any circle intersects the line at infinity lZ at points (1,i,0) and (1,−i,0) ∴ any two circles intersect at (1,i,0) and (1,−i,0)
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aX 0 + a2 02)+(Y 2 − 2bY 0 + b2 02) = r2 02 Set Z = 0 X 2 + Y 2 = 0
Then X = 1 and Y = ±i.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 71 / 80 ∴ any two circles intersect at (1,i,0) and (1,−i,0)
Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aX 0 + a2 02)+(Y 2 − 2bY 0 + b2 02) = r2 02 Set Z = 0 X 2 + Y 2 = 0
Then X = 1 and Y = ±i.
∴ any circle intersects the line at infinity lZ at points (1,i,0) and (1,−i,0)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 71 / 80 Intersection of Two Circles
Choose UZ = {(X,Y ,Z)S Z ≠ 0} and l Z = {(X,Y ,0)}. Consider circle
(x − a)2 +(y − b)2 (x2 − 2ax + a2)+(y2 − 2by + b2) = r2
⋆⋆⋆ Homogenize! ⋆⋆⋆
(X 2 − 2aX 0 + a2 02)+(Y 2 − 2bY 0 + b2 02) = r2 02 Set Z = 0 X 2 + Y 2 = 0
Then X = 1 and Y = ±i.
∴ any circle intersects the line at infinity lZ at points (1,i,0) and (1,−i,0) ∴ any two circles intersect at (1,i,0) and (1,−i,0)
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 71 / 80 If they are tangent . . . If they are not tangent . . .
Two circles intersect at four points (counting with multiplicity).
Intersection of Two Circles
Expected: Two circles intersect at2 × 2 = 4 points (counting with multiplicity).
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 72 / 80 If they are not tangent . . .
Two circles intersect at four points (counting with multiplicity).
Intersection of Two Circles
Expected: Two circles intersect at2 × 2 = 4 points (counting with multiplicity).
If they are tangent . . .
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 72 / 80 Two circles intersect at four points (counting with multiplicity).
Intersection of Two Circles
Expected: Two circles intersect at2 × 2 = 4 points (counting with multiplicity).
If they are tangent . . . If they are not tangent . . .
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 72 / 80 Intersection of Two Circles
Expected: Two circles intersect at2 × 2 = 4 points (counting with multiplicity).
If they are tangent . . . If they are not tangent . . .
Two circles intersect at four points (counting with multiplicity).
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 72 / 80 Bézout’s Theorem
Bézout’s Theorem. Suppose we have complex projective plane curves C : plane curve of degree m D : plane curve of degree n and C and D do not share any components.
Then the number of intersections of C and D, counting multiplicities, is mn.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 73 / 80 Bézout’s Theorem
Bézout’s Theorem. Suppose we have complex projective plane curves C : plane curve of degree m D : plane curve of degree n and C and D do not share any components.
Then the number of intersections of C and D, counting multiplicities, is mn.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 74 / 80 Bézout’s Theorem
Bézout’s Theorem. Suppose we have complex projective plane curves C : plane curve of degree m D : plane curve of degree n and C and D do not share any components.
Then the number of intersections of C and D, counting multiplicities, is mn.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 75 / 80 Bézout’s Theorem
Bézout’s Theorem. Suppose we have complex projective plane curves C : plane curve of degree m D : plane curve of degree n and C and D do not share any components.
Then the number of intersections of C and D, counting multiplicities, is mn.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 76 / 80 Fundamental Theorem of Algebra
A polynomial of degree d over R has less than or equal to d solutions. Bézout’s Theorem A polynomial of degree d over C has S C ∩ D S ≤ n ⋅ m exactly d solutions, counting multiplicities. S C ∩ D S = n ⋅ m This is just a spacefiller because the two columns I’ve created here
Ending Note
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 77 / 80 Bézout’s Theorem
S C ∩ D S ≤ n ⋅ m
S C ∩ D S = n ⋅ m This is just a spacefiller because the two columns I’ve created here
Ending Note
Fundamental Theorem of Algebra
A polynomial of degree d over R has less than or equal to d solutions.
A polynomial of degree d over C has exactly d solutions, counting multiplicities.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 77 / 80 Ending Note
Fundamental Theorem of Algebra
A polynomial of degree d over R has less than or equal to d solutions. Bézout’s Theorem A polynomial of degree d over C has S C ∩ D S ≤ n ⋅ m exactly d solutions, counting multiplicities. S C ∩ D S = n ⋅ m This is just a spacefiller because the two columns I’ve created here
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 77 / 80 Bézout’s Theorem
S C ∩ D S ≤ n ⋅ m
S C ∩ D S = n ⋅ m This is just a spacefiller because the two columns I’ve created here
Ending Note
Fundamental Theorem of Algebra
A polynomial of degree d over R has less than or equal to d solutions.
A polynomial of degree d over C has exactly d solutions, counting multiplicities.
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 78 / 80 Ending Note
Fundamental Theorem of Algebra Bézout’s Theorem
A polynomial of degree d over R has S C ∩ D S ≤ n ⋅ m less than or equal to d solutions. S C ∩ D S = n ⋅ m A polynomial of degree d over C has This is just a spacefiller exactly d solutions, counting because the two columns I’ve created multiplicities. here
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 78 / 80 Ending Note
Fundamental Theorem of Algebra Bézout’s Theorem
A polynomial of degree d over R has S C ∩ D S ≤ n ⋅ m less than or equal to d solutions. S C ∩ D S = n ⋅ m A polynomial of degree d over C has This is just a spacefiller exactly d solutions, counting because the two columns I’ve created multiplicities. here
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 78 / 80 Ending Note
Fundamental Theorem of Algebra Bézout’s Theorem
A polynomial of degree d over R has S C ∩ D S ≤ n ⋅ m less than or equal to d solutions. S C ∩ D S = n ⋅ m A polynomial of degree d over C has This is just a spacefiller exactly d solutions, counting because the two columns I’ve created multiplicities. here
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 79 / 80 References
1. Basic Algebraic Geometry I, by Igor Shafarevich
Jennifer Li (University of Massachusetts) Bézout’s Theorem November 30, 2016 80 / 80