An Expository Proof of Bézout’s Theorem

John McGee - Radford University

Two cubics intersect at exactly 9 points, when counted carefully!

Bézout’s Theorem: given
	(Subscript[C, 1])={(x,y)|(Subscript[f, 1])(x,y)=0}, (Subscript[C, 2])={(x,y)|(Subscript[f, 2])(x,y)=0}
      algebraic curves with no common component and
	deg((Subscript[f, 1]))=(Subscript[n, 1]),deg((Subscript[f, 2]))=(Subscript[n, 2]), then #((Subscript[C, 1])⋂(Subscript[C, 2]))≤ (Subscript[n, 1])(Subscript[n, 2])

This theorem was first stated by Isaac Newton in Principia Volume I Section VI Lemma XXVIII.

Hence it is that the intersections of the conic sections with the curves of third order, because they may amount to six, come out together by equations of six dimensions; and the intersections of two curves of the third order, because they may amount to nine, come out together by equations of nine dimensions. ... and so on in infinitum.[1]

Étienne Bézout compiled much of his work in Théorie générale des équations algébraiques published in 1779. This work includes the result known as Bézout’s theorem.

The degree of the final equation resulting from any number of complete equations in the same number
of unknowns, and of any degrees, is equal to the product of the degrees of the equations. [2]

Modern proof based on resultants

This is not hard to do (see exercises), but it seems that a homogeneous formulation of Bezout's theorem, with a rigorous proof that the resultant has degree m n, was not given until the late 1800s. According to Kline (1972), p. 553, the "proper count of multiplicities" was first made by Halphen in 1873. [3]

[1] On the Shoulders of Giants by Stephen Hawking
[2] http://www-history.mcs.st-andrews.ac.uk/Biographies/Bezout.html
[3] Mathematics and its History by John Stillwell

Why is Bézout’s Theorem Important?

(1) Pascal’s Theorem  (Hexagrammum Mysticum Theorem)

(2) Associativity of Elliptic Curve Point Addition

((Subscript[P, 1])+(Subscript[P, 2]))+(Subscript[P, 3])=(Subscript[P, 1])+((Subscript[P, 2])+(Subscript[P, 3]))

(3) Terrance Tao thinks that it is

Algebraic combinatorial geometry: the polynomial method in arithmetic combinatorics,
	incidence combinatorics, and number theory.    arXiv:1310.6482 [math.CO]   Oct, 2013
          
		4. Bezout’s theorem and Stepanov’s method

	“The previous applications of the polynomial method exploited the geometry of
	hypersurfaces through their intersections with lines. Of course, one can also try to
	study such hypersurfaces through their intersection with more complicated varieties.
	One of the most fundamental tools in which to achieve this is Bezout’s theorem.
	This theorem has many different versions; we begin with a classical one.

	Theorem 17 (Bezout’s theorem). Let F be a field, let (Subscript[d, 1]), (Subscript[d, 1]) ≥ 0, and let (Subscript[P, 1]),(Subscript[P, 2])∈F[x,y]
	 be polynomials of degree (Subscript[d, 1]), (Subscript[d, 2]) respectively with no common (factor^([8])). Then
 	𝒵((Subscript[P, 1]),(Subscript[P, 2]))[F]=𝒵((Subscript[P, 1]))[F]⋂ℤ((Subscript[P, 2]))[F] has cardinality at most (Subscript[d, 1]) (Subscript[d, 2]).”

Outline of Proof of Bézout’s Theorem

We followed the outline in Appendix A of Rational Points on Elliptic Curves by Silverman and Tate.
	
(1) For curves that have no intersection at infinity show #((Subscript[C, 1])⋂(Subscript[C, 2])) (Overscript[≤, A]) dim((R/(((Subscript[f, 1]),(Subscript[f, 2]))))) (Overscript[≤, B]) (Subscript[n, 1]) (Subscript[n, 2])
	
(2) Show dim((R/(((Subscript[f, 1]),(Subscript[f, 2])))))≥(Subscript[n, 1]) (Subscript[n, 2]), establishing that (B) is an equality
	
(3) Show ({{∑}, {P∈X}})I((Subscript[C, 1])⋂(Subscript[C, 1]),P) (Overscript[≤, A+]) dim((R/(((Subscript[f, 1]),(Subscript[f, 2])))));    X=(Subscript[C, 1])⋂(Subscript[C, 2])⋂(𝔸^2)

	 where I((Subscript[C, 1])⋂(Subscript[C, 1]),P)≡dim(((Subscript[𝒪, p])/(Subscript[((Subscript[f, 1]),(Subscript[f, 2])), P])))  is the intersection multiplicity at P
	
(4) Show that (A+) is in fact an equality
	
(5) Show that the definition of I((Subscript[C, 1])⋂(Subscript[C, 1]),P) holds even for points at infinity, i.e. P∈(ℙ^2)\(𝔸^2)

Summer Undergraduate Research Fellowship (SURFS 2013)

Ten weeks of: write, discuss, probe, erase, go to the books, repeat ...

Counting Points of Intersection of Two Curves

Two lines, one point         Parabola and line, two points       Two conics, four points
            
            1×1=1                                    2×1=2                                       2×2=4

We need to count intersection points carefully!

Points in the algebraic closure      Multiplicity of intersection         Second intersection at ∞

Some basic definitions

k is an algebraically closed field, hence non-finite, e.g. ℂ,(OverBar[(Subscript[F, p])])

R=k[x,y] is the ring of polynomials in two unknowns over k (UFD, Notherian)

(Subscript[C, 1]),(Subscript[C, 2]) are two curves in (𝔸^2)=k×k, the solution sets of (Subscript[f, 1])(x,y)=0, (Subscript[f, 2])(x,y)=0

Example

R=ℂ[x,y] is the ring of polynomials with coefficients in ℂ.

        (Subscript[f, 1])(x,y)=((x^2)/(a^2))+((y^2)/(b^2))-1,               (Subscript[C, 1]) =𝒵((Subscript[f, 1])(x,y)), an ellipse
        (Subscript[f, 2])(x,y)=(y^2)+A (x^3)+B x+C,    (Subscript[C, 2])=𝒵((Subscript[f, 2])(x,y)), an elliptic curve

Homogeneous Equations and Projective Space

If f(x,y)∈R with deg(f)=n then F(x,y,z) is the homogeneous polynomial formed as: 

	if (Subscript[a, i])(x^r)(y^s) is a term of f then (Subscript[a, i])(x^r)(y^s)(z^t) is the corresponding term in F where r+s+t=n

	e.g. for f(x,y)=(y^2)-(x^3)+3x-2  we have F(x,y,z)=(z^3)f((x/z),(y/z))=(y^2)z-(x^3)+3x (z^2)-2(z^3)

 Then F(x,y,1)=f(x,y) so the solutions to F(x,y,1)=0 are points in affine space, (𝔸^2)
	 
 If ((Subscript[x, 0]),(Subscript[y, 0]),1) is a solution, then so is (c (Subscript[x, 0]),c (Subscript[y, 0]),c); c≠0.  F(c (Subscript[x, 0]),c (Subscript[y, 0]),c)=(c^n) F((Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0]))
 
 	 [(Subscript[x, 0]),(Subscript[y, 0]),1] is the equivalence class of all multiples of a solution
	 
 The solutions to F(x,y,0)=0 do not correspond to solutions to f(x,y)=0,
      these are points at infinity. [x,y,0]
	 
	 e.g. F(x,y,0)=-(x^3)=0 has y free, so [0,1,0] is the point at infinity on the curve.
	 
(ℙ^2)={[x,y,z]|x,y,z∈k}=(𝔸^2)⋃{[x,y,0]|x,y∈k} note: (0,0,0)∉(ℙ^2).

A Visualization of Projective Space

Intersection of Curves in (ℙ^2)

An algebraic curve C in (ℙ^2) is the zero set of homogeneous polynomial F(x,y,z).
The affine part of the curve is the set of all solutions to F(x,y,1)=f(x,y)=0

Given curves (Subscript[C, 1]),(Subscript[C, 2]) as solutions to (Subscript[F, 1])(x,y,z)=0 and (Subscript[F, 2])(x,y,z)=0, (Subscript[C, 1])⋂(Subscript[C, 2])=𝒵((Subscript[F, 1]))⋂𝒵((Subscript[F, 2])).
We say these curves have no common component when (Subscript[F, 1]),(Subscript[F, 2]) have no common factor.

(Subscript[f, 1])(x,y)=  2 (x^3)-5 (x^2) y+8 (x^2)-14 x y-6 x-15 (y^2)+15 y   =  (2 x-5 y) ((x^2)+4 x+3 y-3)        		      (Subscript[f, 2])(x,y)=-4 (x^3)+10 (x^2) y-12 (x^2)+34 x y+14 x-10 (y^2)-35 y = -(2 x-5 y) (2 (x^2)+6 x-2 y-7)    
	
	   These curves have the line 2x-5y=0 as a common component

Definitions used in the proof of Bezout’s Theorem

(Subscript[C, 1]),(Subscript[C, 2]) are two curves in (𝔸^2) defined by (Subscript[f, 1])(x,y)=0,(Subscript[f, 2])(x,y)=0

I = ((Subscript[f, 1]),(Subscript[f, 2])) is the ideal in R=k[x,y] generated by (Subscript[f, 1]),(Subscript[f, 2]),  i.e.  I = {(Subscript[g, 1])(Subscript[f, 1])+(Subscript[g, 2])(Subscript[f, 2])|(Subscript[g, 1]),(Subscript[g, 2])∈R}

R/I is the quotient ring R/((Subscript[f, 1]),(Subscript[f, 2])) whose representatives are remainders

K is the function field of R, K={(f/g)|f,g∈R,((Subscript[f, 1])/(Subscript[g, 1]))∼((Subscript[f, 2])/(Subscript[g, 2]))⇔(Subscript[f, 1])(Subscript[g, 2])=(Subscript[f, 2])(Subscript[g, 2])}

(Subscript[𝒪, p]) is the localization of R at the point P, (Subscript[𝒪, p])={(f/g)∈K|g(P)≠0}

(Subscript[M, P])={ϕ∈(Subscript[𝒪, P])|ϕ(P)=0} is the unique maximal ideal in (Subscript[𝒪, P]), hence (Subscript[𝒪, p]) is local.

(Subscript[I, P])=(Subscript[((Subscript[f, 1]),(Subscript[f, 2])), P]) is the ideal in (Subscript[𝒪, P]) generated by (Subscript[f, 1]),(Subscript[f, 2]); (Subscript[I, P])={(Subscript[ϕ, 1]) (Subscript[f, 1])+(Subscript[ϕ, 2])(Subscript[f, 2])|(Subscript[ϕ, 1]),(Subscript[ϕ, 2])∈(Subscript[𝒪, P])}

(Subscript[𝒪, p])/(Subscript[I, p]) is the quotient ring used to define intersection multiplicity at P

(1A) Prove #((Subscript[C, 1])⋂(Subscript[C, 2])⋂𝔸)≤dim(R/((Subscript[f, 1]),(Subscript[f, 2])))

A set of polynomials (Subscript[h, 1]),(Subscript[h, 2]),...,(Subscript[h, m]) is linearly independent mod ((Subscript[f, 1]),(Subscript[f, 2])) iff

	(Subscript[c, 1])(Subscript[h, 1])+(Subscript[c, 2])(Subscript[h, 2])+...+(Subscript[c, m])(Subscript[h, m])∈((Subscript[f, 1]),(Subscript[f, 2])) => (Subscript[c, i])=0 ; i=1...m
	
Let (Subscript[P, 1]),(Subscript[P, 2]),...,(Subscript[P, m]) be m distinct points in (Subscript[C, 1])⋂(Subscript[C, 1])⋂(𝔸^2).  Then (Subscript[f, 1])((Subscript[P, i]))=(Subscript[f, 2])((Subscript[P, i]))=0.

Define (Subscript[h, i])(x,y)= (∏_(i≠j)^m)((λ(x-(Subscript[x, j])) -(y-(Subscript[y, j])))/(λ((Subscript[x, i])-(Subscript[x, j])) -((Subscript[y, i])-(Subscript[y, j])))); λ ≠((((Subscript[y, i])-(Subscript[y, j])))/(((Subscript[x, i])-(Subscript[x, j])) )),  then (Subscript[h, i])((Subscript[P, j]))=({{{, ({{1, i=j}, {0, i≠j}})}})

Then (Subscript[c, 1])=(Subscript[c, 1])(Subscript[h, 1])((Subscript[P, 1]))+(Subscript[c, 2])(Subscript[h, 2])((Subscript[P, 1]))+...+(Subscript[c, m])(Subscript[h, m])((Subscript[P, 1]))=(Subscript[g, 1])(Subscript[f, 1])((Subscript[P, 1]))+(Subscript[g, 2])(Subscript[f, 2])((Subscript[P, 1]))=0
	and similarly for the other (Subscript[P, i]).

Since (Subscript[c, i])=0, we have m linearly independent polynomials (Subscript[h, i])+((Subscript[f, 1]),(Subscript[f, 2]))∈R/((Subscript[f, 1]),(Subscript[f, 2]))

	hence dim(R/((Subscript[f, 1]),(Subscript[f, 2])))≥ #((Subscript[C, 1])⋂(Subscript[C, 2])⋂𝔸)

(1A) Example of such (Subscript[h, 1]),(Subscript[h, 2]),(Subscript[h, 3]),(Subscript[h, 4]) for four points in (ℝ^2)

(Subscript[P, 1])=(9,-5),(Subscript[P, 2])=(-2,4),(Subscript[P, 3])=(3,-8),(Subscript[P, 4])=(-6,-3)
	(Subscript[h, 1])(x,y)=(32/10971) (((3 )/4)(x-3)-y-8) ((3/4) (x+2)-y+4) ((3/4) (x+6)-y-3)
	(Subscript[h, 2])(x,y)= (1/1020)(-x+y-3) (-x+y+11) (-x+y+14)
	(Subscript[h, 3])(x,y)=((-32)/42705)((5/4)(x-9)-y-5) ((5/4)(x+2)-y+4) ((5/4)(x+6)-y-3)
	(Subscript[h, 4])(x,y)=(4/1813)((3/2)(x-9)-y-5) ((3/2)(x-3)-y-8) ((3/2)(x+2)-y+4)

(1B) Prove dim(R/((Subscript[f, 1]),(Subscript[f, 2])))≤ (Subscript[n, 1]) (Subscript[n, 2])

Define: (Subscript[R, d])={f(x,y)∈R|deg(f)≤ d},  (Subscript[W, d])=(Subscript[R, d-(Subscript[n, 1])])(Subscript[f, 1])+(Subscript[R, d-(Subscript[n, 2])])(Subscript[f, 2]),   ϕ(d)=(1/2)(d+1)(d+2)

We have ϕ(d)-ϕ(d-1)=(1/2)(d+1)(d+2)-(1/2)d(d+1)=d+1
	(number of monomials (x^i)(y^j)|i+j=d) hence, by induction, dim((Subscript[R, d]))=ϕ(d)
	
We were able to show that (Subscript[R, d-(Subscript[n, 1])])(Subscript[f, 1])⋂(Subscript[R, d-(Subscript[n, 2])])(Subscript[f, 2])=(Subscript[R, d-(Subscript[n, 1])-(Subscript[n, 2])])(Subscript[f, 1])(Subscript[f, 2]) so that

dim (Subscript[W, d])=dim (Subscript[R, d-(Subscript[n, 1])])(Subscript[f, 1])+dim (Subscript[R, d-(Subscript[n, 2])])(Subscript[f, 2]) -dim ((Subscript[R, d-(Subscript[n, 1])])(Subscript[f, 1])⋂(Subscript[R, d-(Subscript[n, 2])])(Subscript[f, 2]) )
	=ϕ(d-(Subscript[n, 1]))+ϕ(d-(Subscript[n, 2]))-ϕ(d-(Subscript[n, 1])-(Subscript[n, 2]))

Then for d≥ (Subscript[n, 1])+(Subscript[n, 2]) we have
	dim((Subscript[R, d]))-dim ((Subscript[W, d]))=ϕ(d) -ϕ(d-(Subscript[n, 1]))-ϕ(d-(Subscript[n, 2]))+ϕ(d-(Subscript[n, 1])-(Subscript[n, 2])) =(Subscript[n, 1]) (Subscript[n, 2]).

If we chose (Subscript[n, 1])(Subscript[n, 2])+1 elements of (Subscript[g, i])∈R must be dependent mod I else dim(R/I)>dim((Subscript[R, d])/(Subscript[W, d]))

Computations

ϕ[d_]:=(1/2)(d+1)(d+2);

dimrdw=ϕ[d]-ϕ[d-(Subscript[n, 1])]-ϕ[d-(Subscript[n, 2])]+ϕ[d-(Subscript[n, 1])-(Subscript[n, 2])]

dimrdw//Simplify

(2) Show dim(R/I)≥(Subscript[n, 1]) (Subscript[n, 2]), when (Subscript[C, 1]),(Subscript[C, 2]) do not intersect at ∞

(SuperStar[f])(x,y)=(∏_(i=1)^n)((Subscript[a, i])x+(Subscript[b, i])y) the homogeneous part of f with deg f=deg (SuperStar[f])=n

	e.g., f(x,y) = a c (x^2)+a d x y+b c x y+b d (y^2)+ e x+ f y+g
	 ⟹ (SuperStar[f])(x,y) = a c (x^2)+a d x y+b c x y+b d (y^2) =(a x+b y)(c x +d y)

With f∈I⋂(Subscript[R, d]), f=(Subscript[g, 1]) (Subscript[f, 1])+(Subscript[g, 2]) (Subscript[f, 2]), for d≥(Subscript[n, 1])+(Subscript[n, 2]) then (Subscript[g, 1]^( * ))(Subscript[f, 1]^( * ))+(Subscript[g, 2]^( * ))(Subscript[f, 2]^( * ))=0 else deg(f)>d.

Since gcd((Subscript[f, 1]^( * )),(Subscript[f, 2]^( * )))=1, ∃ (Subscript[h, 1]),(Subscript[h, 2])∈R ∍ (Subscript[h, 1])(Subscript[f, 1]^( * ))+(Subscript[h, 2])(Subscript[f, 2]^( * ))=1

then for h=(Subscript[g, 2]^( * ))(Subscript[h, 1])-(Subscript[g, 1]^( * ))(Subscript[h, 2]) we have 0=deg((Subscript[g, 1])+h (Subscript[f, 2]))<deg((Subscript[g, 1])),0=deg((Subscript[g, 2])-h (Subscript[f, 1]))<deg((Subscript[g, 2]))

⟹ deg((Subscript[g, i]))≤d-(Subscript[n, i])=>deg((Subscript[g, 1])(Subscript[f, 1])+(Subscript[g, 2])(Subscript[f, 2]))≤d⟹f∈(Subscript[W, d]) ∴(Subscript[W, d])=I⋂(Subscript[R, d])

Since dim((Subscript[R, d])/(Subscript[W, d]))=(Subscript[n, 1])(Subscript[n, 2]) we have at least (Subscript[n, 1])(Subscript[n, 2]) L.I. elements in R/I hence dim(R/I)≥(Subscript[n, 1])(Subscript[n, 2])

(3) Show ({{∑}, {P∈X}})I((Subscript[C, 1])⋂(Subscript[C, 1]),P) ≤ dim(R/I)

Define I((Subscript[C, 1])⋂(Subscript[C, 2]),P)=dim((Subscript[𝒪, p])/(Subscript[I, P]))  (proof given in part (6))
The map α:R->({{∏}, {P∈X}})(Subscript[𝒪, P])/(Subscript[I, P]) with f↦(f mod (Subscript[I, (Subscript[P, 1])]),f mod (Subscript[I, (Subscript[P, 2])]),...,f mod (Subscript[I, (Subscript[P, m])]))
	with J=ker α we have 	
We show that α is onto with α((Subscript[b, 1]) (Subscript[h, 1]^r)+(Subscript[b, 2])(Subscript[h, 2]^r)+...(Subscript[b, m])(Subscript[h, m]^r))=((Subscript[ϕ, 1]),(Subscript[ϕ, 2]),...,(Subscript[ϕ, m]))
for (Subscript[h, i]) as defined in part (1), (Subscript[b, i])≡(Subscript[ϕ, i])(Subscript[h, i]^(-r))(mod (Subscript[I, (Subscript[P, i])])) and r≥ max { dim((Subscript[𝒪, (Subscript[p, i])])/(Subscript[I, (Subscript[p, i])])) }

Therefore dim(R/J)=({{∑}, {P∈X}})dim((Subscript[𝒪, P])/(Subscript[I, P]))

Then I⊆J so R/J⊆R/I hence dim(R/J)≤dim(R/I)

Example of why  I((Subscript[C, 1])⋂(Subscript[C, 1]),P) = dim((Subscript[𝒪, p])/(Subscript[I, p])) works

f1[x_,y_]:=(x^2)+2 x y+3 (y^2)-15;
f2[x_,y_]:=((Sqrt[43])+2)x+3 (Sqrt[43])y-45;
f2[x_,y_,e_]:=f2[x,y+e];
tp={1,(1/3)((Sqrt[43])-1)};

Manipulate[ContourPlot[{f1[x,y]==0,f2[x,y,e]==0},{x,-5,5},{y,-5,5},
Epilog->{PointSize[0.02],Point[tp],Red,
Point[{x,y}]/.Solve[f1[x,y]==0∧f2[x,y,e]==0,{x,y}]}],{e,0,1/8,1/128}]

{grb1,grb2}=GroebnerBasis[{f1[x,y],f2[x,y]},{x,y}]//FullSimplify

sol1=Collect[Solve[grb2==0,x],y]//Flatten//First

sol2=Collect[Solve[(Collect[grb1,y]/.(y^2)->z)==0,z],y]/.z->(y^2)//Flatten//First

We can use these rules to transform any polynomial in f(x,y) 
	to the form a y+b+q(x,y)I
    	thus dim R/I=2

A Mathematica-assisted proof for the degree of intersection
	of a conic with tangent line.

General conic with tangent line, prove that dim(R/((Subscript[f, 1]),(Subscript[f, 2])))=2

Define a generic conic and a general linear equation in ℂ[x,y]

f3[x_,y_]:=a1 (x^2)+a2 x y+a3 (y^2)+a4 x +a5 y+a6;

f4[x_,y_]:=c1 x+c2 y+c3;

Notice that the Groebner basis has 4 elements, because the line is not specifically a tangent line.

gb=GroebnerBasis[{f3[x,y],f4[x,y]},{x,y}]//FullSimplify;

Length[gb]

Collect[gb[[1]],{x,y}]

Collect[gb[[2]],{x,y}]

Form the general equation of the tangent line to (Subscript[f, 3])(x,y) at ((Subscript[x, 1]),(Subscript[x, 2]))

df3x=D[f3[x,y],x]

df3y=D[f3[x,y],y]

mx=df3x/.{x->x1,y->y1}

my=df3y/.{x->x1,y->y1}

tngln=Collect[f3[x1,y1]+mx (x-x1)+my (y-y1)//FullSimplify,{x,y}]

Determine a rule to transform x↦f(y) using the tangent line (Subscript[f, 4])
This is equivalent to determining R/((Subscript[f, 4]))

sol=Solve[tngln==0,x]//First

Apply this rule to (Subscript[f, 3]) to obtain a rule (y^2)→a y +b.
The combination of these rules is equivalent to determining R/((Subscript[f, 3]),(Subscript[f, 4]))

rem1=Collect[f3[x,y]/.sol//FullSimplify,y];

Length[CoefficientList[rem1,y]]

r2=Coefficient[rem1,y,2]//FullSimplify

r1=Coefficient[rem1,y,1]//FullSimplify

r0=Coefficient[rem1,y,0]//FullSimplify

s1=r1/r2//FullSimplify

s0=r0/r2//FullSimplify

s3=Coefficient[x/.sol,y,1]

s4=Coefficient[x/.sol,y,0]

Given f(x,y)∈R then (Subscript[f, 2])(y)=f(x, y) : x ↦(Subscript[s, 3])y+(Subscript[s, 4]) is a polynomial in y only
then (Subscript[f, 3])(y)=(Subscript[f, 2])(y):(y^2)→(Subscript[s, 1]) y+(Subscript[s, 2]) is a linear polynomial of the form a y+b.
Hence dim(R/((Subscript[f, 1]),(Subscript[f, 2])))=2

(4) Show ({{∑}, {P∈X}})I((Subscript[C, 1])⋂(Subscript[C, 2]),P) = dim(R/I)

Define L={g∈R|g f∈((Subscript[f, 1]),(Subscript[f, 2]))}for f∈J=ker α.

We show that ∀P∈(𝔸^2),∃g∈L∍g(P)≠0

	Note: f ∈(Subscript[I, (Subscript[P, i])]) for all (Subscript[P, i])∈X so f=((Subscript[r, 1])/(Subscript[s, 1]))(Subscript[f, 1])+((Subscript[r, 2])/(Subscript[s, 2]))(Subscript[f, 2]) and (Subscript[f, 1])((Subscript[P, i]))=(Subscript[f, 2])((Subscript[P, i]))=0  
	
	If P∈X then g=(Subscript[s, 1])(Subscript[s, 2]) gives g f=((Subscript[s, 2])(Subscript[r, 1])) (Subscript[f, 1])+((Subscript[s, 1])(Subscript[r, 2]))(Subscript[f, 2])∈((Subscript[f, 1]),(Subscript[f, 2])) and (Subscript[s, 1])(Subscript[s, 2])(P)≠0
	
	otherwise one of g=(Subscript[f, 1])+(Subscript[f, 2]) or (Subscript[f, 1])-(Subscript[f, 2]) is not zero at P then f((Subscript[f, 1])±(Subscript[f, 2]))∈((Subscript[f, 1]),(Subscript[f, 2]))
		
Since dim(R/L) is finite ∃r(x)∈L with r(x)=(∑_(i=1)^n)(x-(Subscript[a, i])). If 1∈L+R(x-(Subscript[a, i]))∀i then 1∈L \ NOT

So ∃a,b∈k ∍1∉L+R(x-a)+R(y-b) and g(x,y)=g(a,b)+(Subscript[g, 1])(x-a)+(Subscript[g, 2])(y-b)
	=>g(a,b)∈L so g(a,b)=0,else 1∈L

So 1∈L=>f∈((Subscript[f, 1]),(Subscript[f, 2])) => J⊆I => J=I => R/J=R/I hence dim(R/I)=dim(R/J)

(5) Show ({{∑}, {P∈X}})I((Subscript[C, 1])⋂(Subscript[C, 1]),P) = (Subscript[n, 1])(Subscript[n, 2]) for points at infinity also

Extend the definitions of (Subscript[𝒪, p]),(Subscript[M, p]),(Subscript[I, p]) to K={((F(X,Y,Z))/(G(X,Y,Z)))|deg(F)=deg(G)}
	this amounts to showing the definitions consistent with the affine version for P∈(𝔸^2)
	
	if ϕ=(f/g)∈(Subscript[𝒪, P]) then g(P)≠0⇔t g(P)=g(t P)≠0=>(Subscript[𝒪, P])=(Subscript[𝒪, t P]) => dim(((Subscript[𝒪, p])/(Subscript[((Subscript[F, 1]),(Subscript[F, 2])), P]))) well-defined

Choose a line L such that L⋂(Subscript[C, 1])⋂(Subscript[C, 2])=∅  with L:a x+b y+c z=0  (projective transformation)

	Put w=<a,b,c>, choose u,v∈(k^3) such that {u,v,w} is linearly independent

	then the linear transformation T(x)=A x, where A = (({{u}, {v}, {w}})), is an isomorphism
	of vector spaces that transforms all points on L to infinity and all points in (Subscript[C, 1])⋂(Subscript[C, 2]) to (𝔸^2)

Acknowledgments

Project funded in part by a Summer Undergraduate Research Fellowship (SURFs)
provided through the Office of Undegraduate Research and Scholarship (OURS)

	 James Grenier - for taking on the project  (presented at MathFest 2013)
	* Camron Withrow - for reviewing our proofs
	* Ezra Brown - for introducing me to the beauty of algebraic curves
	
	

Primary References
(1) Kendig, K. A Guide to Plane Algebraic Curves
(2) Nicholsen, K. Introduction to Abstract Algebra
(3) Silverman and Tate Rational Points on Elliptic Curves
(4) Watkins, J. Topics in Commutative Ring Theory

Thanks

Bézout’s Theorem:
	Given (Subscript[C, 1])=𝒵((Subscript[F, 1])(X,Y,Z)), (Subscript[C, 2])=𝒵((Subscript[F, 2])(X,Y,Z)), with
	deg((Subscript[F, 1]))=(Subscript[n, 1]),deg((Subscript[F, 2]))=(Subscript[n, 2]), gcd((Subscript[F, 1]),(Subscript[F, 2]))=1, then ({{∑}, {P∈(Subscript[C, 1])⋂(Subscript[C, 1])}})I((Subscript[C, 1])⋂(Subscript[C, 2]),(Subscript[P, i]))=(Subscript[n, 1])(Subscript[n, 2])
	
				  Questions?