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Science Forum Index » Mathematics Forum » Proof of Bezout Theorem (on Algebraic Geometry)
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| Jose Capco |
 Posted: Fri Dec 26, 2003 7:08 am |
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Dear NG,
Could someone here write the sketch of the proof of the (weaker) Bezout
theorem, i.e. given two curves (say one being
irreducible.. to make matters a bit easier) C1 and C2 defined by functions f1
and f2, the interesection between these
two curves is less or equal deg(f1)*deg(f2). I have not so deep knowledge in
algebraic geometry, my understanding is as far
as the first few chapters of Shafarevich goes. If there is an easy proof of
Bezout theorem I would be glad to see it (I have already
seen the proof of the fact that C1 and C2 interesect finitely without using
Bezout, so maybe that could be implemented).
Thanks in advance.
Sincerely,
Jose Capco
PS: I don't mind if the proof is on P^2 (ie projective plane of three
coordinate, or extended projective plane of the affine plane
of two dimension). In fact I saw an article by Hawthorne in internet on proof
of Bezout theorem on P^2, but unfortunately
it made a lot of assumptions and referred a few books for the detailed proof
(which I dont have). A sketch would suffice and I will
work out the rest. |
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| Timothy Murphy |
| Posted: Fri Dec 26, 2003 10:30 am |
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Jose Capco wrote:
Quote: Dear NG,
Could someone here write the sketch of the proof of the (weaker) Bezout
theorem, i.e. given two curves (say one being
irreducible.. to make matters a bit easier) C1 and C2 defined by functions
f1 and f2, the interesection between these
two curves is less or equal deg(f1)*deg(f2).
I'm not an algebraic geometer (far from it),
but I would prove it by showing that the resolvant (or resultant) R(f1,f2)
of f1(x,y),f2(x,y), regarded as polynomials in y over k[x],
is a polynomial of degree <= mn in x.
There is a minor issue of dealing with the case
where 2 points of intersection have the same x-coordinate,
but this is easily dealt with by applying an affine transformation.
Similarly the projective case can be dealt with
by applying a projective transformation
so that the line at infinity does not contain any point of intersection.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland |
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| Jose Capco |
| Posted: Sat Dec 27, 2003 9:32 am |
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In article <cuYGb.2174$HR.5344@news.indigo.ie>, tim@birdsnest.maths.tcd.ie
says...
Quote: I'm not an algebraic geometer (far from it),
but I would prove it by showing that the resolvant (or resultant) R(f1,f2)
of f1(x,y),f2(x,y), regarded as polynomials in y over k[x],
is a polynomial of degree <= mn in x.
Ok, I am not well versed in the resolvent.. but let me verify if I understand
this correctly. You are telling that if I work with f1 and f2 as functions of
k(y)[x] (we had like k(y) to be a field, thus fraction field), and solve the
resolvent we can arrive to the Bezout theorem. Am I correct that the resolvent
determines the common roots of f1 and f2 (ie if the resolvent is zero at a
point (x,y) then f1 and f2 are zero at those points). I just want to verify if
my understanding of the resolvent was correct. Thanks again
Sincerely,
Jose Capco |
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| Timothy Murphy |
| Posted: Sat Dec 27, 2003 7:36 pm |
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Jose Capco wrote:
Quote: I'm not an algebraic geometer (far from it),
but I would prove it by showing that the resolvant (or resultant) R(f1,f2)
of f1(x,y),f2(x,y), regarded as polynomials in y over k[x],
is a polynomial of degree <= mn in x.
Ok, I am not well versed in the resolvent.. but let me verify if I
understand this correctly. You are telling that if I work with f1 and f2
as functions of k(y)[x] (we had like k(y) to be a field, thus fraction
field), and solve the resolvent we can arrive to the Bezout theorem. Am I
correct that the resolvent determines the common roots of f1 and f2 (ie if
the resolvent is zero at a point (x,y) then f1 and f2 are zero at those
points). I just want to verify if my understanding of the resolvent was
correct. Thanks again
Yes, that is my understanding too.
Suppose f(x), g(x) are monic (leading coefficient 1),
f(x) = x^m + a_1x^{m-1} + ..., g(x) = x^n + b_1x^{n-1} + ...
If we define R(f,g) = prod (a_i - b_j)
where a_i, b_j are the roots of f,g respectively,
then it is relatively straightforward to show that
R(f,g) = det A,
where A is the (m+n)x(m+n) matrix
A = (1,a_1,a_2,.../0,1,a_1,.../.../1.b_1,b_2,.../0,1,b_1,.../...).
Now apply this with f(x) = f1(x,y), g(x) = f2(x,y).
Then a_r, b_r are polynomials of degree r in y,
and R(f,g) is a polynomial of degree mn in y.
Basically, given polynomials of degrees m,n in x,y
this shows that one of the variables, x say,
can be eliminated to give a polynomial of degree mn
in the other variable.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland |
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| KRamsay |
| Posted: Sat Dec 27, 2003 7:54 pm |
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In article <bsk56c$idj$1@wsc10.lrz-muenchen.de>, you@somehost.somedomain (Jose
Capco) writes:
|Ok, I am not well versed in the resolvent.. but let me verify if I understand
|this correctly.
If two nonzero polynomials P and Q of degrees m and n have a nonconstant
common factor R, then P*(Q/R)-Q*(P/R)=0, and S=Q/R and T=-P/R are
polynomials of degrees <n and <m. Conversely, if there are nonzero
polynomials S and T of degrees <n and <m such that P*S+Q*T=0,
then P and Q must have a nonconstant common factor.
The resolvant is the determinant of the linear transformation from
the n+m-dimensional space of pairs (S,T) of polynomials of degrees
<n and <m respectively to the n+m-dimensional space of polynomials
of degree <n+m, given by f(S,T)=P*S+Q*T. It's not hard to write out
the entries of the matrix of this transformation; they're coefficients
of P and Q, arranged sort of like a band matrix.
So when the resolvant is zero, the transformation is singular, there
exist nonzero S and T as above satisfying P*S+Q*T=0, and the
polynomials P and Q have a common factor (which is the same as
having a common root). When the resolvant isn't zero, the
transformation is nonsingular, the only S and T of the given degrees
satisfying P*S+Q*T=0 are S=T=0, and P and Q have no common
factor.
Keith Ramsay |
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| Timothy Murphy |
| Posted: Sun Dec 28, 2003 8:22 am |
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Timothy Murphy wrote:
Quote: Could someone here write the sketch of the proof of the (weaker) Bezout
theorem, i.e. given two curves (say one being
irreducible.. to make matters a bit easier) C1 and C2 defined by
functions f1 and f2, the interesection between these
two curves is less or equal deg(f1)*deg(f2).
I'm not an algebraic geometer (far from it),
but I would prove it by showing that the resolvant (or resultant) R(f1,f2)
of f1(x,y),f2(x,y), regarded as polynomials in y over k[x],
is a polynomial of degree <= mn in x.
I guess if one doesn't want to go into the theory of the resolvent,
one could regard y as fixed and consider the m+n polynomails
f1(x,y), x*f1(x,y), x^2*f1(x,y), ..., x^{n-1}*f1(x,y),
f2(x,y), x*f2(x,y), x^2*f2(x,y), ..., x^{m-1}*f2(x,y).
One can regard these as m+n linear equations in 1,x,x^2,...,x^{m+n-1}.
If f1(x,y),f2(x,y) have a common root in x for a given value of y,
then det(A) = 0, where A is the matrix of this set of equations.
This is an equation of degree <= mn in y.
This is exactly the same argument --
I'm just pointing out that for your purpose
one only needs to establish the property of the resolvent in one direction.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland |
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| JHS |
| Posted: Sun Dec 28, 2003 10:29 am |
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Using the resultant in this way should be okay to prove Bezout's
theorem _if_ the curves intersect in distinct points, but things get
subtler when there are higher multiplicities, i.e., when the curves
intersect tangentially and/or intersect at singular points.
The way I've generally seen it done is that first one defines the
intersection index (or multiplicity) of the two curves at each point
of intersection. For example, if the curves are f(x,y)=0 and g(x,y)=0
and if they intersect at the point (0,0), then the intersection index
at (0,0) is the dimension of the complex vector space
C[x,y]_{0,0}/(f(x,y),g(x,y)).
Here C[x,y]_(0,0) is the ring of polynomials localized at (0,0), which
means the ring of all rational functions p(x,y)/q(x,y) with the
property that q(0,0) is not 0.
Of course, this definition works anywhere, since you can always do a
translation X=x-a, Y=y-b to move the intersection point (a,b) to
(0,0). And for points on the line at infinity in projective space, you
can dehomogenize with respect to one of the other coordinates.
Bezout's theorem then says that if you add up all of the intersection
indices for all of the intersection points, the total is equal to
deg(f)*deg(g). The beauty and strength of Bezout's theorem lies in the
fact that the intersection indices are computed locally in a
neighborhood of each point, but that their total is equal to a global
quantity, namely the product of the degrees of the curves.
The proof of Bezout's theorem is in most basic algebraic geometry
books, using varying amounts of fancy machinery. John Tate and I put a
fairly elementary proof into the appendix of our book "Rational Points
on Elliptic Curves" (Springer-Verlag).
Joe Silverman
you@somehost.somedomain (Jose Capco) wrote in message news:<bsk56c$idj$1@wsc10.lrz-muenchen.de>...
Quote: In article <cuYGb.2174$HR.5344@news.indigo.ie>, tim@birdsnest.maths.tcd.ie
says...
I'm not an algebraic geometer (far from it),
but I would prove it by showing that the resolvant (or resultant) R(f1,f2)
of f1(x,y),f2(x,y), regarded as polynomials in y over k[x],
is a polynomial of degree <= mn in x.
Ok, I am not well versed in the resolvent.. but let me verify if I understand
this correctly. You are telling that if I work with f1 and f2 as functions of
k(y)[x] (we had like k(y) to be a field, thus fraction field), and solve the
resolvent we can arrive to the Bezout theorem. Am I correct that the resolvent
determines the common roots of f1 and f2 (ie if the resolvent is zero at a
point (x,y) then f1 and f2 are zero at those points). I just want to verify if
my understanding of the resolvent was correct. Thanks again
Sincerely,
Jose Capco |
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| Timothy Murphy |
| Posted: Sun Dec 28, 2003 4:09 pm |
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JHS wrote:
Quote: Using the resultant in this way should be okay to prove Bezout's
theorem _if_ the curves intersect in distinct points, but things get
subtler when there are higher multiplicities, i.e., when the curves
intersect tangentially and/or intersect at singular points.
Actually the original poster asked for a proof of what he/she called
the "weak Bezout's Theorem", that the number of intersections is <= mn.
Quote: The proof of Bezout's theorem is in most basic algebraic geometry
books, using varying amounts of fancy machinery. John Tate and I put a
fairly elementary proof into the appendix of our book "Rational Points
on Elliptic Curves" (Springer-Verlag).
I must admit I found this to be one of the weaker sections
of what I rate as one of the best popular mathematical books I know --
one of the very few worthy to stand on the bookshelf next to Hardy & Wright.
Anyone with the slightest interest in elliptic curves
should start with "Silverman & Tate", IMHO.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland |
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| Steven E. Landsburg |
| Posted: Sun Dec 28, 2003 5:42 pm |
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In article <KDHHb.2350$HR.5705@news.indigo.ie>,
Timothy Murphy <tim@birdsnest.maths.tcd.ie> wrote:
Quote: Anyone with the slightest interest in elliptic curves
should start with "Silverman & Tate", IMHO.
I just want to second this. I absolutely love this book.
Steven E. Landsburg
www.landsburg.com/about2.html
-- |
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| JHS |
| Posted: Sun Dec 28, 2003 10:00 pm |
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Timothy Murphy <tim@birdsnest.maths.tcd.ie> wrote in message news:<KDHHb.2350$HR.5705@news.indigo.ie>...
Quote: JHS wrote:
Using the resultant in this way should be okay to prove Bezout's
theorem _if_ the curves intersect in distinct points, but things get
subtler when there are higher multiplicities, i.e., when the curves
intersect tangentially and/or intersect at singular points.
Actually the original poster asked for a proof of what he/she called
the "weak Bezout's Theorem", that the number of intersections is <= mn.
True. I was just expanding the discusssion a bit to the general case.
For the weak form, as other posters noted, the resultant argument will
give the upper bound of mn, and it is certainly very elementary.
Quote: The proof of Bezout's theorem is in most basic algebraic geometry
books, using varying amounts of fancy machinery. John Tate and I put a
fairly elementary proof into the appendix of our book "Rational Points
on Elliptic Curves" (Springer-Verlag).
I must admit I found this to be one of the weaker sections
of what I rate as one of the best popular mathematical books I know --
one of the very few worthy to stand on the bookshelf next to Hardy & Wright.
Anyone with the slightest interest in elliptic curves
should start with "Silverman & Tate", IMHO.
Sorry you didn't like that section, but in any case, thank you for the
kind words regarding RPEC. It's an honor to be mentioned in the same
sentence as the H&W classic.
Joe Silverman |
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| Jose Capco |
| Posted: Mon Dec 29, 2003 6:54 am |
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In article <KDHHb.2350$HR.5705@news.indigo.ie>, tim@birdsnest.maths.tcd.ie
says...
Quote:
JHS wrote:
Actually the original poster asked for a proof of what he/she called
the "weak Bezout's Theorem", that the number of intersections is <= mn.
The proof of Bezout's theorem is in most basic algebraic geometry
books, using varying amounts of fancy machinery. John Tate and I put a
fairly elementary proof into the appendix of our book "Rational Points
on Elliptic Curves" (Springer-Verlag).
I must admit I found this to be one of the weaker sections
of what I rate as one of the best popular mathematical books I know --
one of the very few worthy to stand on the bookshelf next to Hardy & Wright.
Anyone with the slightest interest in elliptic curves
should start with "Silverman & Tate", IMHO.
Ok thanks all  .. I didn't know that this post could bother Mr. Silverman too
.. anyway, you might be right (I never looked at that book but I certainly
heard a lot about it). My original purpose was actually to prove the
associativity law of the elliptic curve and for that I believe "weak Bezout
theorem" suffices. The previous threads gave me enough ideas to start with and
I dont suppose I need to go very much into resolvent theory.
Sincerely,
Jose Capco |
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