CLASSICAL
ALGEBRAIC GEOMETRY
Daniel Plaumann
Universität Konstanz
Summer ����
A brief inaccurate history of algebraic geometry
���� - ����
���� - ����
���� - ����
���� - ����
from ����
from ����
Projective geometry. Emergence of ’analytic’ geometry with cartesian
coordinates, as opposed to ’synthetic’ (axiomatic) geometry in the style
of Euclid. (Celebrities: Plücker, Hesse, Cayley)
Complex analytic geometry. Powerful new tools for the study of geometric problems over C. (Celebrities: Abel, Jacobi, Riemann)
Classical school. Perfected the use of existing tools without any ’dogmatic’ approach. (Celebrities: Castelnuovo, Segre, Severi, M. Noether)
Algebraization. Development of modern algebraic foundations (’commutative ring theory’) for algebraic geometry. (Celebrities: Hilbert,
E. Noether, Zariski)
Modern algebraic geometry. All-encompassing abstract frameworks
(schemes, stacks), greatly widening the scope of algebraic geometry.
(Celebrities: Weil, Serre, Grothendieck, Deligne, Mumford)
Computational algebraic geometry Symbolic computation and discrete methods, many new applications. (Celebrities: Buchberger)
Literature
Primary source
[Ha]
J. Harris, Algebraic Geometry: A first course. Springer GTM ��� (����)
Classical algebraic geometry
[BCGB] M. C. Beltrametti, E. Carletti, D. Gallarati, G. Monti Bragadin. Lectures on Curves, Surfaces and Projective Varieties. A classical view of algebraic geometry. EMS Textbooks
(translated from Italian) (����)
[Do]
I. Dolgachev. Classical Algebraic Geometry. A modern view. Cambridge UP (����)
Algorithmic algebraic geometry
[CLO]
D. Cox, J. Little, D. O’Shea. Ideals, Varieties, and Algorithms. Springer UTM (����)
[EGSS]
D. Eisenbud, D. R. Grayson, M. Stillman, B. Sturmfels. Computations in Algebraic
Geometry with Macaulay �. Springer (����).
’Big books’
[GH]
P. Griffiths, J. Harris. Principles of Algebraic Geometry. John Wiley & Sons (����)
[Hs]
R. Hartshorne. Algebraic Geometry. Springer GTM �� (����)
[Sh]
I. Shafarevich. Basic Algebraic Geometry I: Varieties in projective space. Springer
(translated from Russian) (����)
§�
Projective Varieties
Affine varieties
K
algebraically closed field
An = K n affine space
V ⊂ An is an affine variety if there is a set of polynomials M ⊂ K[x� , . . . , x n ] such that
V = V(M) = {p ∈ An ∶ f (p) = � for all f ∈ M}.
There is a finite subset M ′ ⊂ M such that V(M) = V(M ′) (Hilbert Basis Theorem).
Projective space
Let V be a K -vector space.
P(V ) = {one-dimensional subspaces of V }, the projective space of V
Pn = PK n+� = �K n+� � {�}��∼
where v ∼ w ⇐⇒ ∃λ ∈ K ×∶ v = λw.
Points of Pn are denoted in homogeneous coordinates [Z� , . . . , Z n ] where
[Z� , . . . , Z n ] = [λZ� , . . . , λZ n ] for λ ∈ K × .
Affine varieties
Affine varieties
K
Kn
n
A
=
K
n
A = Kn
algebraically
algebraically closed
closed field
field
affine
affine space
space
n is an affine variety if there is a set of polynomials M ⊂ K[x , . . . , x ] such that
V
⊂
A
V ⊂ An is an affine variety if there is a set of polynomials M ⊂ K[x�� , . . . , x nn ] such that
n ∶ f (p) = � for all f ∈ M}.
V
=
V(M)
=
{p
∈
A
V = V(M) = {p ∈ An ∶ f (p) = � for all f ∈ M}.
′ ⊂ M such that V(M) = V(M ′) (Hilbert Basis Theorem).
There
is
a
finite
subset
M
If I is the ideal generated by M , then V(M) = V(I). By the Hilbert Basis Theorem, there is a
finite subset M ′ ⊂ M that also generates I , so in particular V(M ′) = V(M).
space
If I and J are two ideals in K[x� , . . . , xProjective
]
,
then
n
Let VVbe
-vector
(I)a ∪KV
(J) = space.
V (IJ) = V(I ∩ J)
V (I))∩=V{(J)
= V (I + J) subspaces of V }, the projective space of V
P(V
one-dimensional
where IJ is
the ideal generated by all products f g , f ∈ I , g ∈ J .
Pn = PK n+� = �K n+� � {�}��∼
where v ∼ w ⇐⇒ ∃λ ∈ K ×∶ v = λw.
Points of Pn are denoted in homogeneous coordinates [Z� , . . . , Z n ] where
Projective space
×
Let V[Z
be� ,a. .K. -vector
space.
, Z n ] = [λZ
� , . . . , λZ n ] for λ ∈ K .
P(V ) = {one-dimensional subspaces of V }, the projective space of V
Pn = PK n+� = �K n+� � {�}��∼
where v ∼ w ⇐⇒ ∃λ ∈ K ×∶ v = λw.
K
Kn
n
A
=
K
n
K = Kn
A
algebraically
algebraically closed
closed field
field
affine
space
algebraically
affine
space closed field
Affine varieties
Affine varieties
Affine varieties
n affine space
An = K
n
V
V⊂
⊂A
An is
is an
an affine
affine variety
variety if
if there
there is
is aa set
set of
of polynomials
polynomials M
M⊂
⊂ K[x
K[x�� ,, .. .. .. ,, xx nn ]] such
such that
that
n is an affine variety
n ∶ iff (p)
V ⊂A
there= is
a set
off polynomials
M ⊂ K[x� , . . . , x n ] such that
V
=
V(M)
=
{p
∈
A
�
for
all
∈
M}.
n
V = V(M) = {p ∈ A ∶ f (p) = � for all f ∈ M}.
n′ ∶ f (p) = � for all f ∈ M}.
′) (Hilbert Basis Theorem).
V
=
V(M)
=
{p
∈
A
There
is
a
finite
subset
M
⊂
M
such
that
V(M)
=
V(M
If I is the ideal generated by M , then V(M) = V(I). By the Hilbert Basis Theorem, there is a
′ ⊂ M that by
If I is the
ideal
Mgenerates
, then V(M)
V(I)
. Furthermore,
finitely
finite
subset
Mgenerated
also
I , so= in
particular
V(M ′) I= isV(M)
. generated by the
Hilbert Basis Theorem. In particular, there is a finite subset M ′ ⊂ M such that V(M) = V(M ′).
space
If I and J are two ideals in K[x� , . . . , xProjective
]
,
then
n
Let VVbe
-vector
(I)a ∪KV
(J) = space.
V (IJ) = V(I ∩ J)
V (I))∩=V{(J)
= V (I + J) subspaces of V }, the projective space of V
P(V
one-dimensional
where IJ is
the ideal generated by all products f g , f ∈ I , g ∈ J .
Pn = PK n+� = �K n+� � {�}��∼
where v ∼ w ⇐⇒ ∃λ ∈ K ×∶ v = λw.
Projective
space[Z� , . . . , Z n ] where
Points of Pn are denoted in homogeneous
coordinates
Projective space
×.
Let V[Z
be� ,a. .K. -vector
space.
,
Z
]
=
[λZ
,
.
.
.
,
λZ
]
for
λ
∈
K
n
n
�
Let V be a K -vector space.
P(V ) = {one-dimensional subspaces of V }, the projective space of V
P(V ) = {one-dimensional subspaces of V }, the projective space of V
n+� � {�}��∼
�K
Pnn = PK n+�
=
P = PK n+� = �K n+� � {�}��∼ ×
where v ∼ w ⇐⇒ ∃λ ∈ K ×∶ v = λw.
Projective space
Let V be a K -vector space.
P(V ) = {one-dimensional subspaces of V }, the projective space of V
Pn = PK n+� = �K n+� � {�}��∼
where v ∼ w ⇐⇒ ∃λ ∈ K ×∶ v = λw.
Points of Pn are denoted in homogeneous coordinates [Z� , . . . , Z n ] where
[Z� , . . . , Z n ] = [λZ� , . . . , λZ n ] for λ ∈ K × .
Projective space
Projective space
Let
Let V be
be aa K -vector
-vector space.
space.
P(V ) = {one-dimensional
one-dimensional subspaces
subspaces of
of V },, the
the projective
projective space
space of
of V
n+� = �K n+�
n+� � {�}��∼
Pnn = PK n+�
×
×∶ v = λw.
where
v
∼
w
⇐⇒
∃λ
∈
K
where
n
Points
Points of
of Pn are
are denoted
denoted in
in homogeneous
homogeneous coordinates
coordinates [Z�� , . . . , Z nn ] where
where
×
[Z�� , . . . , Z nn ] = [λZ�� , . . . , λZ nn ] for
for λ ∈ K × .
Projective space
Projective space
Let
Let V be
be aa K -vector
-vector space.
space.
P(V ) = {one-dimensional
one-dimensional subspaces
subspaces of
of V },, the
the projective
projective space
space of
of V
n+� = �K n+�
n+� � {�}��∼
Pnn = PK n+�
×
×∶ v = λw.
where
v
∼
w
⇐⇒
∃λ
∈
K
where
n
Points
Points of
of Pn are
are denoted
denoted in
in homogeneous
homogeneous coordinates
coordinates [Z�� , . . . , Z nn ] where
where
×
[Z�� , . . . , Z nn ] = [λZ�� , . . . , λZ nn ] for
for λ ∈ K × .
Projective varieties
A polynomial F ∈ K[Z� , . . . , Z n ] is not a function on Pn , since in general
F(Z� , . . . , Z n ) ≠ F(λZ� , . . . , λZ n ).
If F is homogeneous of degree d , then
F(λZ� , . . . , λZ n ) = λd F(Z� , . . . , Z n ).
So given a set M of homogeneous polynomials in K[Z� , . . . , Z n ], it makes sense to define
V(M) = �p ∈ Pn ∶ f (p) = � for all f ∈ M�, a projective variety.
The projective (resp. affine) varieties in Pn (resp. An ) form the closed sets of a topology, the
Zariski topology. Projective space is covered by the open subsets
The Zariski topology
The projective (resp. affine) varieties in Pn (resp. An ) form the closed sets of a topology, the
Zariski topology. Projective space is covered by the open subsets
U i = �[Z� , . . . , Z n ] ∈ Pn ∶ Z i ≠ �� = �[Y� , . . . , Yi−� , �, Yi+� , . . . , Yn ] ∈ Pn �.
The map
U i → An , [Z� , . . . , Z n ] � (Z��Z i , . . . , Z i−��Z i , Z i+��Z i , . . . , Z n )
is a homeomorphism. The inverse map is
An → U i , (z� , . . . , z i−� , z i+� , z n ) � [z� , . . . , z i−� , �, z i+� , . . . , z n ].
Thus Pn is covered by n + � copies of An .
How to think about projective space
Projective line
Exact picture
[Ha, p.4]
How to think about projective space
Projective line
Real topological picture
Exact picture
[Ha, p.4]
(O. Alexandrov - Wikimedia Commons)
How to think about projective space
Projective line
Real topological picture
(O. Alexandrov - Wikimedia Commons)
Exact picture
[Ha, p.4]
Intuitive picture
How to think about projective space
Projective plane
Exact picture
[Univ. of Toronto Math Network]
We think of projective space over
an algebraically closed field just
as real affine space, together with
the idea that taking intersections
always works perfectly.
How to think about projective space
Projective plane
Real topological picture - Boy's surface
(virtualmathmuseum.org)
Exact picture
[Univ. of Toronto Math Network]
We think of projective space over
an algebraically closed field just
as real affine space, together with
the idea that taking intersections
always works perfectly.
How to think about projective space
Projective plane
Real topological picture - Boy's surface
(virtualmathmuseum.org)
Exact picture
[Univ. of Toronto Math Network]
We think of projective space over
an algebraically closed field just
as real affine space, together with
the idea that taking intersections
always works perfectly.
Intuitive picture
[freehomeworkmathhelp.com]
How to think about projective space
How to think about projective space
Projective
plane
Projective
plane
Real topological picture - Boy's surface
(virtualmathmuseum.org)
Exact picture
[Univ. of Toronto Math Network]
We think of projective space over
algebraically
closed
field
just
Weanthink
of projective
space
over
real affine space,
together
with
anasalgebraically
closed
field just
that
taking
intersections
asthe
realidea
affine
space,
together
with
always
works
perfectly.
the
idea that
taking
intersections
always works perfectly.
Intuitive picture
[freehomeworkmathhelp.com]
Linear Spaces
If W ⊂ V is a linear subspace, then PW ⊂ PV is a projective subspace, a linear space of dimension dim PW = dim W − � in PV .
dim PW
dim PW
dim PW
dim PW
=�
=�
=�
= dim PV − �
point
line
plane
hyperplane
If L = PW , L′ = PW ′, write
LL′ = P(W + W ′).
We have
dim LL′ = dim L + dim L′ − dim L ∩ L′ .
Linear Spaces
Linear Spaces
If W ⊂ V is a linear subspace, then PW ⊂ PV is a projective subspace, a linear space of dimension dim PW = dim W − � in PV .
dim PW
dim PW
dim PW
dim PW
=�
=�
=�
= dim PV − �
point
line
plane
hyperplane
If L = PW , L′ = PW ′, write
LL′ = P(W + W ′).
We have
dim LL′ = dim L + dim L′ − dim L ∩ L′ .
Dimension
Let X be a variety. X is reducible if it is the union of two proper, non-empty closed subvarieties;
otherwise it is called irreducible.
The Dimension of X is the largest integer k such that there exists a chain
� � X� � X� � � � X k−� � X
of irreducible closed subvarieties.
In particular, dim Pn = dim An = n.
If X ⊂ Pn or X ⊂ An is irreducible:
dim X
dim X
dim X
dim X
dim X
=�
=�
=�
=�
= n−�
point
curve
surface
threefold
hypersurface
Theorem. The hypersurfaces in Pn are exactly the varieties defined by a single equation.
The hypersurfaces in P� are the plane projective curves.
Dimension
Dimension
Let X be a variety. X is reducible if it is the union of two proper, non-empty closed subvarieties;
otherwise
it is called
Let X be a variety.
X isirreducible.
reducible if it is the union of two proper, non-empty closed subvarieties;
The
Dimension
of X irreducible.
is the largest integer k such that there exists a chain
otherwise
it is called
The Dimension
largest
� � X� � X�of� X�is�the
X k−�
� X integer k such that there exists a chain
� � X� � closed
X� � �subvarieties.
� X k−� � X
of irreducible
of irreducible closed
subvarieties.
In particular, dim Pn = dim An = n.
In particular,
dim Pnn = dim An = n.
n
If X ⊂ P or X ⊂ A is irreducible:
Ifdim
X ⊂XP=n �or X ⊂point
An is irreducible:
dim
curve
dim X
X=
= ��
point
dim
surface
dim X
X=
= ��
curve
dim
threefold
dim X
X=
= ��
surface
dim
dim X
X=
= �n − � hypersurface
threefold
dim X = n − � hypersurface
Theorem. The hypersurfaces in Pn are exactly the varieties defined by a single equation.
n are exactly the varieties defined by a single equation.
�
Theorem.
The
hypersurfaces
in
P
The hypersurfaces in P are the plane projective curves.
The hypersurfaces in P� are the plane projective curves.
Dimension
Dimension
Let X be a variety. X is reducible if it is the union of two proper, non-empty closed subvarieties;
otherwise
it is called
Let X be a variety.
X isirreducible.
reducible if it is the union of two proper, non-empty closed subvarieties;
The
Dimension
of X irreducible.
is the largest integer k such that there exists a chain
otherwise
it is called
The Dimension
largest
� � X� � X�of� X�is�the
X k−�
� X integer k such that there exists a chain
� � X� � closed
X� � �subvarieties.
� X k−� � X
of irreducible
of irreducible closed
subvarieties.
In particular, dim Pn = dim An = n.
n = dim An = n .
In particular,
dim
P
If X ⊂ Pn or X ⊂ An is irreducible:
Ifdim
X ⊂XP=n �or X ⊂point
An is irreducible:
dim
curve
dim X
X=
= ��
point
dim
surface
dim X
X=
= ��
curve
dim
threefold
dim X
X=
= ��
surface
dim
dim X
X=
= �n − � hypersurface
threefold
dim X = n − � hypersurface
Theorem. The hypersurfaces in Pn are exactly the varieties defined by a single equation.
n are exactly the varieties defined by a single equation.
Theorem.
The
hypersurfaces
in
P
�
The hypersurfaces in P are the plane projective curves.
The hypersurfaces in P� are the plane projective curves.
Dimension
Dimension
Dimension
Let X be a variety. X is reducible if it is the union of two proper, non-empty closed subvarieties;
Let
X be a variety.
X isirreducible.
reducible if it is the union of two proper, non-empty closed subvarieties;
otherwise
it
is
called
Let
X be a variety.
X isirreducible.
reducible if it is the union of two proper, non-empty closed subvarieties;
otherwise
it
is
called
The
Dimension
of X irreducible.
is the largest integer k such that there exists a chain
otherwise
it
is
called
The Dimension of X is the largest integer k such that there exists a chain
The Dimension
largest
� � X� � X�of� X�is�the
X k−�
� X integer k such that there exists a chain
� � X� � X� � � � X k−� � X
� � X� � closed
X� � �subvarieties.
� X k−� � X
of irreducible
of irreducible closed subvarieties.
of irreducible closed
subvarieties.
In particular, dim Pn = dim An = n.
In particular, dim Pnn = dim Ann = n.
In particular,
dim Pn = dim A = n.
n
If X ⊂ P or X ⊂ A is irreducible:
If X ⊂ Pn or X ⊂ An is irreducible:
Ifdim
X ⊂XP=n �or X ⊂point
An is irreducible:
dim
X
=
��
point
dim
X
=
curve
dim
X
=
�
point
dim
X
=
�
curve
dim
X
=
�
surface
dim
X
=
�
curve
dim
X
=
�
surface
dim
X
=
�
threefold
dim
X
=
�
surface
dim
X
=
�
threefold
dim
X
=
n
−
�
hypersurface
dim
X
=
�
dim X = n − � threefold
hypersurface
dim X = n − � hypersurface
Theorem. The hypersurfaces in Pn are exactly the varieties defined by a single equation.
Theorem. The hypersurfaces in Pn are exactly the varieties defined by a single equation.
n are exactly the varieties defined by a single equation.
Theorem.
The
hypersurfaces
in
P
�
The hypersurfaces in P are the plane projective curves.
The hypersurfaces in P� are the plane projective curves.
The hypersurfaces in P� are the plane projective curves.
Points
Proposition �.�. Any finite set of d points in Pn is described by polynomials of degree at most d .
Proof. Let Γ = {p� , . . . , pd }. For q ∉ Γ, let L q,i be a linear form with L q,i (p i ) = � and L q,i (q) ≠ �.
Put
Fq = L q,��L q,d .
Then Γ = V(Fq ∶ q ∉ Γ).
Definition. Let p� , . . . , pd ∈ Pn . If d � n + �, the points p i are independent if
dim(p��pd ) = d − �,
�
otherwise dependent.
If d > n + �, the p i are in (linearly) general position if no n + � of them are dependent (i.e. lie in
a hyperplane).
Points
n
Proposition
�.�.
Any
finite
set
of
d
points
in
P
Proposition �.�. Any finite set of d points in Pn is
is described
described by
by polynomials
polynomials of
of degree
degree at
at most
most d
d ..
Proof.
be a linear form with L q,i (p
(q) ≠ �.
Proof. Let
Let Γ
Γ=
= {p
{p�� ,, .. .. .. ,, ppdd }}.. For
For q
q ∉∉ Γ
Γ,, let
let L
L q,i
(p ii )) =
= �� and
and L
L q,i
q,i be a linear form with L q,i
q,i (q) ≠ �.
Put
Put
F
.
Fqq =
=L
L q,�
�L q,d
q,��L
q,d .
Then
Then Γ
Γ=
= V(F
V(Fqq ∶∶ qq ∉∉ Γ)
Γ)..
n
Definition.
Definition. Let
Let p
p�� ,, .. .. .. ,, ppdd ∈∈ P
Pn .. If
If d
d�
�n
n+
+ ��,, the
the points
points p
p ii are
are independent
independent if
if
dim(p���pdd ) = d − �,
otherwise dependent.
�
�
If d > n + �, the p ii are in (linearly) general position if no n + � of them are dependent (i.e. lie in
a hyperplane).
Points
Proposition �.�. Any finite set of d points in Pnn is described by polynomials of degree at most d .
Proposition
Proposition �.�.
�.�. Any
Any finite
finite set
set of
of d
d points
points in
in P
Pn is
is described
described by
by polynomials
polynomials of
of degree
degree at
at most
most d
d ..
Proof. Let Γ = {p� , . . . , pd }. For q ∉ Γ, let L q,i be a linear form with L q,i (p i ) = � and L q,i (q) ≠ �.
Proof.
Let Γ = {p , . . . , p }. For q ∉ Γ, let L q,i be
(p ) = � and L q,i (q)
Proof.
be aa linear
linear form
form with
with L
L q,i
(q) ≠
≠ ��..
q,i (p ii ) = � and L q,i
Put Let Γ = {p�� , . . . , pdd }. For q ∉ Γ, let L q,i
Put
Put
Fq = L q,��L q,d .
F
.
Fqq =
=L
L q,�
�L q,d
q,��L
q,d .
Then Γ = V(Fq ∶ q ∉ Γ).
�
Then
�
Then Γ
Γ=
= V(F
V(Fqq ∶∶ qq ∉∉ Γ)
Γ)..
�
Definition. Let p� , . . . , pd ∈ Pnn . If d � n + �, the points p i are independent if
Definition.
Definition. Let
Let p
p�� ,, .. .. .. ,, ppdd ∈∈ P
Pn .. If
If d
d�
�n
n+
+ ��,, the
the points
points p
p ii are
are independent
independent if
if
dim(p��pd ) = d − �,
dim(p���pdd ) = d − �,
otherwise dependent.
otherwise dependent.
If d > n + �, the p i are in (linearly) general position if no n + � of them are dependent (i.e. lie in
If d > n + �, the p ii are in (linearly) general position if no n + � of them are dependent (i.e. lie in
a hyperplane).
a hyperplane).
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
Theorem �.�. Any collection of at most �n points in general position in Pn can
can be
be described
described by
by
quadratic
quadratic forms.
forms.
Proof.
= �n
Γ = {p� , . . We
. , pmay
�n }. assume that Γ contains exactly �n points.
Proof. May
Let Γassume
⊂ Pn be�Γ�
such
a ,collection.
n
Let
q
∈
P
Let q ∈ Pn be
be such
such that
that
F�
= ��
F�ΓΓ =
= �� �⇒
�⇒ F(q)
F(q) =
holds
holds for
for all
all quadratic
quadratic forms
forms F
F .. We
We must
must show
show q
q ∈∈ Γ
Γ..
(�)
�Γ � = �Γ � = n, then Γ spans a hyperplane H = V(L i ) and
H
∪H� =byV(L
L�).
�
�
(�) IfIf ΓΓ == ΓΓ��∪Γ
∪ �Γwith
)
,
defined
a
linear
� with ��Γ�� = ��Γ�� = n , theni Γi spans a hyperplanei H i = V(L
i
So
L� LL�,(q)
hypothesis.
H� ∪=H��by
. hypothesis. Hence q ∈ H ∪ H .
form
and=H� by
∪H
= V(L L Hence
). So LqL∈ (q)
i
�
�
� �
� �
�
�
(�)
the of
property
q ∈ pproperty
��p k . q ∈ p��p .
(�) Let
Let {p
{p�� ,, .. .. .. ,, pp kk }} ∈∈ Γ
Γ be
be minimal
a minimalwith
subset
Γ with the
k
Claim:
k =can
� ( find
⇐⇒such
q = ap�subset
)
By (�), we
with k � n.
Take Σ ⊂ Γ � {p� , . . . , p k }, Σ = n − k + �. By hypothesis, p� , . . . , p k , Σ span a hyperplane H that
Claim: k = � ( ⇐⇒ q = p�)
does not contain p�. So q ∉ H , since p� ∈ qp��p k .
Take
Γ �in
{pthe
. , p k } with �Σ�
= n − kby+ the
�. Byremaining
hypothesis,
the n points
{p� , .that
. . , pqklies
} ∪ Σonspan
� , . .hyperplane
By (�),Σq⊂lies
spanned
n points.
It follows
the
ahyperplane
hyperplanespanned
H that does
. Since
p� ∈ qpp��p, .k., .it, follows
that
q ∉ H . of all such
by p�not
andcontain
any n −p��of
the points
p
.
The
intersection
�n
k+�
By
(�), q lies inis the
by the remaining n points. It follows that q lies on the
hyperplanes
justhyperplane
p�, hence q spanned
= p�.
�
hyperplane spanned by p� and any n − � of the points p k+� , . . . , p�n . The intersection of all such
hyperplanes is just p�, hence q = p�.
�
n can be described by
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
be
described
by
n can
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
quadratic
forms.
quadratic
quadratic forms.
forms.
Proof.
May
assume
�Γ�
=
�n
,, Γ
=
{p
..
� ,, .. .. .. ,, p
�n }
Proof.
May
assume
�Γ�
=
�n
Γ
=
{p
p
}
n
� We may
�n assume that Γ contains exactly �n points.
Proof. Let Γ ⊂ P be such a collection.
n be such that
Let
q
∈
P
n
Let
q
∈
P
Let q ∈ Pn be
be such
such that
that
F�
�� �⇒
F(q)
=
��
Γ=
F�
=
�⇒
F(q)
=
F�ΓΓ = � �⇒ F(q) = �
holds
for
all
quadratic
forms
F
.. We
must
show
qq ∈∈ Γ
..
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds for all quadratic forms F . We must show q ∈ Γ.
(�)
If
Γ
=
Γ
�Γ
�Γ
n
,, then
Γ
aa hyperplane
H
V(L
and
H
V(L
..
� ∪Γ
� with
��� =
��� =
� ∪H
�=
�L
�)
i spans
i=
i)
(�)
If
Γ
=
Γ
∪Γ
with
�Γ
=
�Γ
=
n
then
Γ
spans
hyperplane
H
=
V(L
)
and
H
∪H
=
V(L
L
)
�
�
�
�
�
�
�
�
i
i
i
(�)
IfL ΓL =(q)
Γ� ∪= Γ�� by
with
��� = ��� Hence
= n, then
ΓHi spans
a hyperplane H i = V(L i ), defined by a linear
So
hypothesis.
q
∈
∪
H
.
� L�(q) = � by hypothesis. Hence q ∈ H� ∪ H�.
So
L
� = ��by hypothesis. Hence q ∈ H� ∪ H�.
form� L�i , and H� ∪ H� = V(L� L�). So L� L�(q)
(�)
Let
{p
∈∈ Γ
be
minimal
with
the
property
qq ∈∈ pp��p
� ,, .. .. .. ,, p
k}
k ..
(�)
Let
{p
p
}
Γ
be
minimal
with
the
property
�p
�
�
k
(�)
Let {p
, p kk } q∈ Γ= be
a minimal subset of Γ with the property
q ∈ p��p k .
�=, .�.(. ⇐⇒
Claim:
k
p
)
�)
Claim:
k
=
�
(
⇐⇒
q
=
p
�subset with k � n .
By
(�),
we
can
find
such
a
Take
Take Σ
Σ⊂
⊂Γ
Γ�
� {p
{p�� ,, .. .. .. ,, pp kk }},, Σ
Σ=
=n
n−
− kk +
+ ��.. By
By hypothesis,
hypothesis, p
p�� ,, .. .. .. ,, pp kk ,, Σ
Σ span
span aa hyperplane
hyperplane H
H that
that
Claim:
k=
� ( ⇐⇒p�q. So
= pq�)∉ H , since p� ∈ qp��p k .
does
not
contain
does not contain p�. So q ∉ H , since p� ∈ qp��p k .
Take
Γ �in
{pthe
. , p k } with �Σ�
= n − kby
�. Byremaining
hypothesis,
the n points
{p� , .that
. . , pqqklies
} ∪ Σon
By
spanned
n
It
the
� , . .hyperplane
By (�),
(�),Σq
q⊂lies
lies
in
the
hyperplane
spanned
by+ the
the
remaining
n points.
points.
It follows
follows
that
lies
onspan
the
ahyperplane
hyperplanespanned
H that does
. Since
p� ∈ qppp�k+�
�p,, ..k.., ..it,, follows
that
q ∉ H . of
by
and
any
the
pp�n .. The
intersection
all
such
hyperplane
spanned
by p
p��not
andcontain
any n
n−
−p���of
of
the points
points
The
intersection
of
all
such
�n
k+�
hyperplanes
just
pp�,, hence
q spanned
= p�.
�
By
(�), q lies inis
by the remaining n points. It follows that q lies on the
hyperplanes
is the
justhyperplane
�
� hence q = p�.
hyperplane spanned by p� and any n − � of the points p k+� , . . . , p�n . The intersection of all such
hyperplanes is just p�, hence q = p�.
�
n
n can
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
be
described
by
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
quadratic
forms.
quadratic
forms.
quadratic
quadratic forms.
forms.
n be such a collection. We may assume that Γ contains exactly �n points.
Proof.
Let
Γ
⊂
P
Proof.
May
assume
�Γ�
=
�n
,, Γ
=
{p
..
� ,, .. .. .. ,, p
�n }
Proof.
May
assume
�Γ�
=
�n
Γ
=
{p
p
}
n
� We may
�n assume that Γ contains exactly �n points.
Proof. Let Γ ⊂ P be such a collection.
n
n be
Let
q
∈
P
such
that
Let
q
∈
P
be
such
that
n
Let
q
∈
P
be
such
n
Let q ∈ P be such that
that
F�
=
�� �⇒
F(q)
=
��
Γ
F�
=
�⇒
F(q)
=
Γ
F�
= ��
F�ΓΓ =
= �� �⇒
�⇒ F(q)
F(q) =
holds
for
all
quadratic
forms
F
.. We
must
show
qq ∈∈ Γ
..
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds
holds for
for all
all quadratic
quadratic forms
forms F
F .. We
We must
must show
show q
q ∈∈ Γ
Γ..
(�)
If ΓΓ == ΓΓ��∪Γ
∪ �Γwith
�Γ
= ��Γ
=, n
, then
Γi spans
a hyperplane
HV(L
defined
byV(L
a linear
� with
�=� �Γ
�=� n
i = V(L
i ), H
(�)
If
�Γ
�
�
then
Γ
spans
a
hyperplane
H
=
)
and
∪H
=
..
�
�
�
�L
�)
i
i
i
(�)
If
Γ
=
Γ
∪Γ
with
�Γ
�
=
�Γ
�
=
n
,
then
Γ
spans
a
hyperplane
H
=
V(L
)
and
H
∪H
=
V(L
L
)
�
�
�
�
�
�
�
�
i
i
i
(�)
IfL ΓL
=
Γ� ∪=H
���by
with
�Γ=��V(L
= ���L��Hence
=). nSo
, then
Γ(q)
a hyperplane
H
, defined
by a linear
form
L
,
and
∪
H
L
L
=
�
by
hypothesis.
Hence
q
∈
∪
H
.
i spans
i = V(L
i )H
�
�
�
�
�
i
So
(q)
hypothesis.
q
∈
H
∪
H
.
� L�(q) = � by hypothesis. Hence q ∈ H� ∪ H�.
So
L
� = ��by hypothesis. Hence q ∈ H� ∪ H�.
form� L�i , and H� ∪ H� = V(L� L�). So L� L�(q)
(�)
Let
{p
∈∈ Γ
be
aminimal
minimalwith
subset
of
Γ with the
property
q ∈ p��p k .
�� ,, .. .. .. ,, p
kk }
(�)
Let
{p
p
}
Γ
be
the
property
q
∈
p
�p
.
�
k
(�)
Let
{p
,, .. .. .. ,, pp k }} ∈∈ Γ
be
minimal
with
the of
property
q ∈ pproperty
��can
��p k . q ∈ p��p .
(�)
Let
{p
Γ
be
a
minimal
subset
Γ
with
the
By
(�),
we
find
such
a
subset
with
k
�
n
.
k q = p�)
k
Claim:
k
=
�
(
⇐⇒
Claim:
k =can
� ( find
⇐⇒such
q = ap�subset
)
By
(�),
we
with
k��
n. hypothesis, p , . . . , p , Σ span a hyperplane H that
Claim:
k
=
�
({p
⇐⇒
q
=
p
)
Take
Σ
⊂
Γ
�
,
.
.
.
,
p
}
,
Σ
=
n
−
k
+
.
By
�
Take Σ ⊂ Γ � {p�� , . . . , p kk }, Σ = n − k + �. By hypothesis, p�� , . . . , p kk , Σ span a hyperplane H that
Claim:
k Γ=
� ({p
⇐⇒
q. ,So
=p pq}�)∉with
does
not
contain
p
..
Take
Σ
⊂
�
,
.
.
.
�Σ� = npp��−∈∈kqp
+ ���.�p
By khypothesis,
the n points {p� , . . . , p k } ∪ Σ span
�
�
does not contain p�. Sokq ∉ H
H ,, since
since
qp
�p
k
Take
Γ �in
{p
, . .hyperplane
. ,does
p k } with
�Σ�
= n −p�kby
�. Byremaining
the
n points
{p�q, ∉.that
.H
. ,. pqqklies
} ∪ Σon
By
(�),
spanned
the
It
the
aByhyperplane
Hthe
not contain
.+
Since
phypothesis,
, it follows
that
�that
� ∈ qp��pn
(�),Σq
q⊂lies
lies
in
the
hyperplane
spanned
by
the
remaining
nkpoints.
points.
It follows
follows
that
lies
onspan
the
ahyperplane
hyperplane
that
does
. Since
p� ∈ qppp�k+�
�p,,n..kpoints.
that
q ∉that
H . q lies
by
and
any
of
the
.., ..it,, follows
pp�n .. It
The
intersection
of
all
such
By
(�), q lies spanned
inHthe
hyperplane
spanned
the points
remaining
follows
on
the
hyperplane
spanned
by p
p��not
andcontain
any n
n−
−p���by
of
the
points
The
intersection
of
all
such
�n
k+�
hyperplanes
just
pp�by
, hence
qq spanned
=
pp�..n − � by
�
By
(�), q lies spanned
inis
the points
remaining
follows
that q lies
onsuch
the
hyperplane
p� and
of the
p k+� ,n. points.
. . , p�n . It
The
intersection
of all
hyperplanes
is the
justhyperplane
=any
�
�, hence
�
hyperplane
p� and
hyperplanesspanned
is just p�by
, hence
q =any
p�.n − � of the points p k+� , . . . , p�n . The intersection of all such
�
hyperplanes is just p�, hence q = p�.
�
n can be described by
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
n
n can
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
be
described
by
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
quadratic
forms.
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
quadratic
forms.
quadratic
forms.
quadratic
forms.
quadratic
forms.
n be such a collection. May assume �Γ� = �n , Γ = {p , . . . , p }.
Proof.
Let
Γ
⊂
P
n be such a collection. We may assume that Γ contains
�
�n �n points.
Proof.
Let
Γ
⊂
P
exactly
Proof.
May
assume
�Γ�
=
�n
,
Γ
=
{p
,
.
.
.
,
p
}
.
� , . . . , p�n }.
Proof.
May
assume
�Γ�
=
�n
,
Γ
=
{p
n
� We may
�n assume that Γ contains exactly �n points.
Proof. LetnΓ ⊂ P be such a collection.
Let
qq ∈∈ P
be
such
that
n
n
Let
P
be
such
that
Let
q
∈
P
be
such
that
n
Let
q
∈
P
be
such
that
n
Let q F�
∈ P = be
such
thatF(q) = �
�
�⇒
Γ=�
F�
�⇒
F(q)
=
��
Γ
F�
=
�
�⇒
F(q)
=
Γ
F�
=
�� �⇒
F(q)
= ��
Γ
F�
=
�⇒
F(q)
=
Γ all quadratic forms F . We must show q ∈ Γ.
holds
for
holds
for
all
quadratic
forms
F
.. We
must
show
qq ∈∈ Γ
..
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds
holds for
for all
all quadratic
quadratic forms
forms F
F .. We
We must
must show
show q
q ∈∈ Γ
Γ..
(�)
IfIf ΓΓ == ΓΓ� ∪Γ
�Γ��Γ
� =� �Γ
� =� n=, n
then
Γi spans
a hyperplane
H i =HV(L
H
V(L
�Γwith
��Γ
� ∪H� =by
� L�).
i ) and
(�)
∪
with
=
,
then
Γ
spans
a
hyperplane
=
V(L
)
,
defined
a
linear
�∪Γ� with
�
�= �Γ�� �= n , then Γi spans
i
i
i H� ∪H� = V(L� L�).
(�)
If
Γ
=
Γ
�Γ
�
aa hyperplane
H
=
V(L
)
and
�
�
i
i
(�)
If
Γ
=
Γ
∪Γ
with
�Γ
�
=
�Γ
�
=
n
,
then
Γ
spans
hyperplane
H
=
V(L
)
and
H
∪H� =byV(L
L�).
So
L�ΓLL�=,(q)
�� by
hypothesis.
qLi∈ Γ(q)
Hi spans
.a hyperplane
�� ∪=�H
��Γ
���L� Hence
�
�
i H
i
� ∪=H
�by
(�)
If
Γ
Γ
with
�
=
=
n
,
then
=
V(L
)
,
defined
a
linear
form
and
∪
H
=
V(L
)
.
So
L
�
hypothesis.
Hence
q
∈
H
∪
H
.
�
i
i �
�
� �Hence q
� ∈� H� ∪ H�.
�
i (q) = ��by hypothesis.
So
L
L
�
�
So
L
L
(q)
=
�
by
hypothesis.
Hence
q
∈
H
∪
H
.
� = ��by hypothesis. Hence q ∈ H� ∪ H�.
form� L�i , and H� ∪ H� = V(L� L�). So L� L�(q)
(�)
Let
{p
}} ∈∈ Γ
be
aa minimal
subset
of
Γ
with
the
property
qq ∈∈ pp��p
..
� ,, .. .. .. ,, p
k
k
(�)
Let
{p
p
Γ
be
minimal
subset
of
Γ
with
the
property
�p
�� , . . . , p kk } ∈ Γ be minimal with the property q ∈ p��p k .
�
k
(�)
Let
{p
(�)
Let
{p
,
.
.
.
,
p
}
∈
Γ
be
minimal
with
the
property
q
∈
p
�p
.
By
(�),
we
find
aa subset
with
kk �
n
.. of Γ with the property
��can
�
k }such
k
(�)
Let
{p
,
.
.
.
,
p
∈
Γ
be
a
minimal
subset
q ∈ p��p k .
By
(�),
we
can
find
such
subset
with
�
n
k
Claim:
k
=
�
(
⇐⇒
q
=
p
)
�)
Claim:
k
=
�
(
⇐⇒
q
=
p
��subset
Claim:
k
=
�
(
⇐⇒
q
=
p
))
By
(�),
we
can
find
such
a
with
k��
n. hypothesis, p , . . . , p , Σ span a hyperplane H that
Claim:
k
=
�
(
⇐⇒
q
=
p
Take
Σ
⊂
Γ
�
{p
,
.
.
.
,
p
}
,
Σ
=
n
−
k
+
.
By
�
� , . . . , p k }, Σ = n − k + �. By hypothesis, p� , . . . , p k , Σ span a hyperplane H that
Take
Σ
⊂
Γ
�
{p
�, . . q
kp},) Σ = n − k + �. By hypothesis, p� , . . . , p k , Σ span a hyperplane H that
Take
Σ
⊂
Γ
�
{p
.
,
p
Claim:
k
=
�
(
⇐⇒
=
�
� the nk points {p , . . . , p } ∪ Σ span
kq
�∉
does
not
contain
p
.
So
H
,
since
p
∈
qp
�p
.
Take
Σ
⊂
Γ
�
{p
,
.
.
.
,
p
}
with
�Σ�
=
n
−
k
+
�
.
By
hypothesis,
�
�
�
k
�
�
k
k
does
not
contain
p
.
So
q
∉
H
,
since
p
∈
qp
�p
.
�
�
�
k
does
not
contain
p.hyperplane
q} ∉with
H , contain
since
p�−p∈kby
qp
�p
. ∈ qp �pn points.
�.. ,So
k
Take
Σ
⊂
Γ
�
{p
,
.
p
�Σ�
=
n
+
�
.
By
hypothesis,
the
n points
{p�q, ∉.that
.H
. ,. pqqklies
} ∪ Σon
By
(�),
q
lies
in
the
spanned
the
remaining
It
the
aByhyperplane
H
that
does
not
.
Since
p
, it follows
that
�
k
�
� nkpoints.
(�), q lies in the hyperplane spanned �by the remaining
It follows
follows
that
lies
onspan
the
(�),
in
hyperplane
spanned
by
the
remaining
follows
on
the
aBy
hyperplane
that
does
. Since
p� ∈ qppp�k+�
�p,,n
that
q ∉that
H . qq lies
hyperplane
by
and
any
of
the
..kpoints.
.., ..it,, follows
pp�n .. It
The
intersection
of
all
such
By
(�), q
q lies
lies spanned
inHthe
the
hyperplane
spanned
the points
remaining
n
points.
It
follows
that
lies
on
the
hyperplane
spanned
by p
p��not
andcontain
any n
n−
−p���by
of
the
points
The
intersection
of
all
such
�n
k+�
hyperplane
pp� and
any
n
−
�� by
of
the
points
pp k+� ,,n.. points.
.. .. ,, pp�n .. It
The
intersection
of
all
such
hyperplanes
just
pp�by
,, hence
q
=
p
.
�
By
(�), q lies spanned
inis
spanned
the
remaining
follows
that
q
lies
on
the
�
hyperplane
spanned
by
and
any
n
−
of
the
points
The
intersection
of
all
such
hyperplanes
is the
justhyperplane
hence
q
=
p
.
�
�
�n
k+�
�
�
hyperplanes
is
,, hence
qq =
pp�..n − � of the points p , . . . , p�n . The intersection of all such
�
hyperplane
p� and
hyperplanesspanned
is just
just p
p��by
hence
=any
�
k+�
�
hyperplanes is just p�, hence q = p�.
�
n
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
quadratic
forms.
quadratic
forms.
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
quadratic
forms.
quadratic
forms.
quadratic
forms.
quadratic
forms.
n
n
Proof.
Let
Γ
⊂
P
such
aaa collection.
May
assume
�Γ� =that
�n, Γ
Γ contains
= {p� , . . .exactly
, p�n }. �n
n be
Proof.
Let
Γ
⊂
P
be
such
collection.
We
may
assume
points.
Proof.
Let
Γ
⊂
P
be
such
collection.
We
may
assume
that
Γ
contains
exactly
�n
points.
Proof.
May
assume
�Γ�
=
�n
,
Γ
=
{p
,
.
.
.
,
p
}
.
�
�n
Proof.
= �n
Γ = {p� , . . We
. , pmay
�n }. assume that Γ contains exactly �n points.
Proof. May
LetnΓassume
⊂ Pn be�Γ�
such
a ,collection.
n
Let
qqq ∈∈∈ P
be
such
that
n
Let
P
be
such
that
n
Let
P
be
such
that
Let
q
∈
P
be
such
that
n
Let
q
∈
P
be
such
that
n
Let q F�
∈ P = be
such
thatF(q) = �
�
�⇒
Γ
F�
=
��� �⇒
F(q)
=
���
F�
=
�⇒
F(q)
=
Γ
Γ
F�
=
�⇒
F(q)
=
Γ=�
F�
�⇒
F(q)
= ��
Γ
F�
=
�
�⇒
F(q)
=
Γ all
holds
for
quadratic
forms
F
... We
must
show
qqq ∈∈∈ Γ
...
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds
for
all
quadratic
forms
F
.. We
must
show
qq ∈∈ Γ
..
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds for all quadratic forms F . We must show q ∈ Γ.
(�)
IfIf
ΓΓ
==
ΓΓ
with
�Γ��Γ
� �=�� �Γ
� �=�� n=
,n
then
Γi spans
a hyperplane
H i =H
V(L
H
V(L
��∪Γ
�Γ
��Γ
� ∪H� =by
� L�).
i ) and
(�)
∪
with
=
,
then
Γ
spans
a
hyperplane
=
V(L
)
,
defined
a
linear
�
i
i
i
(�)
If
Γ
=
Γ
∪
Γ
with
�Γ
=
�Γ
=
n
,
then
Γ
spans
a
hyperplane
H
=
V(L
)
,
defined
by
a
linear
�∪Γ� with
�
�= �Γ�� �= n , then Γi spans
i
i
i H� ∪H� = V(L� L�).
(�)
If
Γ
=
Γ
�Γ
�
aa hyperplane
H
=
V(L
)
and
�
�
i
i
(�)
If
Γ
=
Γ
∪Γ
with
�Γ
�
=
�Γ
�
=
n
,
then
Γ
spans
hyperplane
H
=
V(L
)
and
H
∪H
=byV(L
L�).
So
L�ΓLL
���by
hypothesis.
Hence
q� L
Hi spans
H���by
.a hyperplane
�� ∪=�H
��Γ
����L
�
�
�
i∈�Γ
i H
i
�=
� ∪=
form
,i ,(q)
and
∪
H
=
V(L
)
.
So
L
(q)
hypothesis.
Hence
q
∈
H
∪
H
.
(�)
If
Γ
Γ
with
�
=
�
=
n
,
then
=
V(L
)
,
defined
a
linear
�
�
�
�
i
form
L
and
H
∪
H
=
V(L
L
)
.
So
L
L
(q)
=
by
hypothesis.
Hence
q
∈
H
∪
H
.
�
i
i �
�by hypothesis.
�
� �Hence q
� ∈� H� ∪ H�.
�
So
L
L
(q)
=
�
�
�
So
L
L
(q)
=
�
by
hypothesis.
Hence
q
∈
H
∪
H
.
� = ��by hypothesis. Hence q ∈ H� ∪ H�.
form� L�i , and H� ∪ H� = V(L� L�). So L� L�(q)
(�)
Let
{p
}}} ∈∈∈ Γ
be
aaa minimal
subset
of
Γ
with
the
property
qqq ∈∈∈ ppp���p
...
�� ,,, ... ... ... ,,, p
k
k
(�)
Let
{p
p
Γ
be
minimal
subset
of
Γ
with
the
property
�p
(�)
Let
{p
p
Γ
be
minimal
subset
of
Γ
with
the
property
�p
k
k
�� , . . . , p kk } ∈ Γ be minimal with the property q ∈ p��p k .
�
k
(�)
Let
{p
(�)
Let
{p
,
.
.
.
,
p
}
∈
Γ
be
minimal
with
the
property
q
∈
p
�p
.
By
(�),
we
find
aaa subset
with
kkk �
n
... of Γ with the property
��can
�
k }such
k
By
(�),
we
can
find
such
subset
with
�
n
(�)
Let
{p
,
.
.
.
,
p
∈
Γ
be
a
minimal
subset
q ∈ p��p k .
By
(�),
we
can
find
such
subset
with
�
n
k
Claim:
k
=
�
(
⇐⇒
q
=
p
)
�)
Claim:
k
=
�
(
⇐⇒
q
=
p
��subset
Claim:
k
=
�
(
⇐⇒
q
=
p
)))
By
(�),
we
can
find
such
a
with
k��
n. hypothesis, p , . . . , p , Σ span a hyperplane H that
Claim:
k
=
�
(
⇐⇒
q
=
p
�
Claim:
k
=
�
(
⇐⇒
q
=
p
Take
Σ
⊂
Γ
�
{p
,
.
.
.
,
p
}
,
Σ
=
n
−
k
+
.
By
�
� , . . . , p k }, Σ = n − k + �. By hypothesis, p� , . . . , p k , Σ span a hyperplane H that
Take
Σ
⊂
Γ
�
{p
�, . . q
kp},) Σ = n − k + �. By hypothesis, p� , . . . , p k , Σ span a hyperplane H that
Take
Σ
⊂
Γ
�
{p
.
,
p
Claim:
k
=
�
(
⇐⇒
=
�
� the
kq
k points
Take
Σ
⊂
Γ
�
{p
,
.
.
.
,
p
}
with
�Σ�
=
n
−
k
+
�
..�p
By
hypothesis,
n
{p
∪
Σ
span
�∉
does
not
contain
p
So
H
,
since
p
∈
qp
.
�
�� ,, .. .. .. ,, p
Take
Σ
⊂
Γ
�
{p
,
.
.
.
,
p
}
with
�Σ�
=
n
−
k
+
�
By
hypothesis,
the
n
points
{p
p
}
∪
Σ
span
k
kk }
�
�
�
k
�
k
does
not
contain
p
.
So
q
∉
H
,
since
p
∈
qp
�p
.
�
�
�
k
does
not
contain
p.hyperplane
q} ∉with
H , contain
since
p�−pp∈�kby
qp
�p
.� ∈∈ qp
�.. ,So
k
aTake
H
that
does
not
.
Since
p
,, it
that
Σ
⊂
Γ
�
{p
,
.
p
�Σ�
=
n
+
�
.
By
hypothesis,
the
n points
{p�qq, ∉∉.that
.H
. ,.. pqqklies
} ∪ Σon
By
(�),
q
lies
in
the
spanned
the
remaining
It
the
aByhyperplane
hyperplane
H
that
does
not
contain
.
Since
p
�pn
it follows
follows
that
H
�
k
� qp���p
(�), q lies in the hyperplane spanned �by the remaining
nkkpoints.
points.
It follows
follows
that
lies
onspan
the
(�),
in
hyperplane
spanned
by
the
remaining
follows
on
the
aBy
hyperplane
that
does
. Since
p� ∈ qppp�k+�
�p,,n
that
q ∉that
H . qqq lies
hyperplane
by
and
any
of
the
..kpoints.
.., ..it,, follows
pp�n .. It
The
intersection
of
all
such
By
(�),
qq lies
lies
in
the
hyperplane
spanned
the
remaining
n
points.
It
follows
that
lies
on
the
By
(�), q
lies spanned
inHthe
the
hyperplane
spanned
by
the points
remaining
n
points.
It
follows
that
lies
on
the
hyperplane
spanned
by p
p��not
andcontain
any n
n−
−p���by
of
the
points
The
intersection
of
all
such
�n
k+�
hyperplane
ppp�� and
any
n
−
��� by
of
the
points
ppp k+�
,,,n... points.
... ... ,,, ppp�n
... It
The
intersection
of
all
such
hyperplanes
just
pp�by
,, hence
q
=
p
.
�
hyperplane
spanned
by
and
any
n
−
of
the
points
The
intersection
of
all
such
By
(�), q lies spanned
inis
spanned
the
remaining
follows
that
q
lies
on
the
�
�n
hyperplane
spanned
by
and
any
n
−
of
the
points
The
intersection
of
all
such
k+�
hyperplanes
is the
justhyperplane
hence
q
=
p
.
�
�
�n
k+�
�
�
hyperplanes
is
,,, hence
qqq =
ppp��...n − � of the points p , . . . , p�n . The intersection of all such
�
hyperplanes
is
just
pp���by
hence
=
�
hyperplane
p� and
hyperplanesspanned
is just
just p
hence
=any
�
k+�
�
hyperplanes is just p�, hence q = p�.
�
n
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
n
quadratic
forms.
quadratic
forms.
Theorem
�.�.
Any
collection
of
at
most
�n
points
in
general
position
in
P
can
be
described
by
quadratic
forms.
quadratic
forms.
quadratic
forms.
quadratic
forms.
n
n
Proof.
Let
Γ
⊂
P
such
aaa collection.
May
assume
�Γ�
=
�n
,, Γ
Γ
=
{p
,, pp�n
}}.. �n
n be
�� ,, .. .. ..exactly
Proof.
Let
Γ
⊂
P
be
such
collection.
May
We
may
assume
assume
�Γ�
=
that
�n
Γ
contains
=
{p
points.
�n
Proof.
Let
Γ
⊂
P
be
such
collection.
We
may
assume
that
Γ
contains
exactly
�n
points.
Proof.
May
assume
�Γ�
=
�n
,
Γ
=
{p
,
.
.
.
,
p
}
.
�
�n
Proof.
= �n
Γ = {p� , . . We
. , pmay
�n }. assume that Γ contains exactly �n points.
Proof. May
LetnΓassume
⊂ Pn be�Γ�
such
a ,collection.
n
Let
qqq ∈∈∈ P
be
such
that
n
Let
P
be
such
that
n
Let
P
be
such
that
Let
q
∈
P
be
such
that
n
Let
q
∈
P
be
such
that
n
Let q F�
∈ P = be
such
thatF(q) = �
�
�⇒
Γ
F�
=
��� �⇒
F(q)
=
���
F�
=
�⇒
F(q)
=
Γ
Γ
F�
=
�⇒
F(q)
=
Γ=�
F�
�⇒
F(q)
= ��
Γ
F�
=
�
�⇒
F(q)
=
Γ all
holds
for
quadratic
forms
F
... We
must
show
qqq ∈∈∈ Γ
...
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds
for
all
quadratic
forms
F
.. We
must
show
qq ∈∈ Γ
..
holds
for
all
quadratic
forms
F
We
must
show
Γ
holds for all quadratic forms F . We must show q ∈ Γ.
(�)
If
=
Γ
with
�Γ
�� �=
�Γ
�� �=
n
,, n
then
Γ
aa hyperplane
H
V(L
and
H
V(L
..
���∪Γ
��Γ
���Γ
���Γ
�� ∪H
�� =
�� L
��)
ii spans
ii =
ii )
(�)
IfIf Γ
Γ
Γ
=
=
Γ
Γ
∪Γ
∪
with
with
�Γ
=
�
�Γ
=
=
�
n
=
then
,
then
Γ
spans
Γ
spans
hyperplane
a
hyperplane
H
=
H
V(L
=
V(L
)
and
)
,
H
defined
∪H
=
by
V(L
a
linear
L
)
�
i
i
i
(�)
Γ
=
Γ
∪
Γ
with
�Γ
�
=
�Γ
�
=
n
,
then
Γ
spans
a
hyperplane
H
=
V(L
)
,
defined
by
a
linear
�∪Γ� with
�
�= �Γ�� �= n , then Γi spans
i
i
i H� ∪H� = V(L� L�).
(�)
If
Γ
=
Γ
�Γ
�
aa hyperplane
H
=
V(L
)
and
�
�
i
i
(�)
If
Γ
=
Γ
∪Γ
with
�Γ
�
=
�Γ
�
=
n
,
then
Γ
spans
hyperplane
H
=
V(L
)
and
H
∪H
=byV(L
L�).
So
L
����by
hypothesis.
Hence
qq� L
H
∪
H
.
�� ∪=
�H
��Γ
����L
�
�
�
i∈
i
i
��ΓL
��=
�
�
So
form
L
L
,i ,(q)
(q)
and
=
by
∪
H
hypothesis.
=
V(L
Hence
)
.
So
L
∈
(q)
H
∪
=
H
�
by
.
hypothesis.
Hence
q
∈
H
∪
H
.
(�)
If
Γ
Γ
with
�
=
�
=
n
,
then
Γ
spans
a
hyperplane
H
=
V(L
)
,
defined
a
linear
�
�
�
�
�
�
�
i
form
L
and
H
∪
H
=
V(L
L
)
.
So
L
L
(q)
=
�
by
hypothesis.
Hence
q
∈
H
∪
H
.
�
i ∪H .
i
i �
�by hypothesis.
�
� �Hence q
� ∈� H
�
So
L
L
(q)
=
�
�
�
�
�
So
L
L
(q)
=
�
by
hypothesis.
Hence
q
∈
H
∪
H
.
� = ��by hypothesis. Hence q ∈ H� ∪ H�.
form� L�i , and H� ∪ H� = V(L� L�). So L� L�(q)
(�)
Let
{p
}}} ∈∈∈ Γ
be
aaa minimal
subset
of
Γ
with
the
property
qqq ∈∈∈ ppp���p
...
�� ,,, ... ... ... ,,, p
k
k
(�)
Let
{p
p
Γ
be
minimal
subset
of
Γ
with
the
property
�p
(�)
Let
{p
p
Γ
be
minimal
subset
of
Γ
with
the
property
�p
k
k
�� , . . . , p kk } ∈ Γ be minimal with the property q ∈ p��p k .
�
k
(�)
Let
{p
(�)
Let
{p
,
.
.
.
,
p
}
∈
Γ
be
minimal
with
the
property
q
∈
p
�p
.
By
(�),
we
find
aaa subset
with
kkk �
n
... of Γ with the property
��can
�
k }such
k
By
(�),
we
can
find
such
subset
with
�
n
(�)
Let
{p
,
.
.
.
,
p
∈
Γ
be
a
minimal
subset
q ∈ p��p k .
By
(�),
we
can
find
such
subset
with
�
n
k
Claim:
k
=
�
(
⇐⇒
q
=
p
)
�)
Claim:
k
=
�
(
⇐⇒
q
=
p
��subset
Claim:
k
=
�
(
⇐⇒
q
=
p
)))
By
(�),
we
can
find
such
a
with
k��
n. hypothesis, p , . . . , p , Σ span a hyperplane H that
Claim:
k
=
�
(
⇐⇒
q
=
p
�
Claim:
k
=
�
(
⇐⇒
q
=
p
Take
Σ
⊂
Γ
�
{p
,
.
.
.
,
p
}
,
Σ
=
n
−
k
+
.
By
�
� , . . . , p k }, Σ = n − k + �. By hypothesis, p� , . . . , p k , Σ span a hyperplane H that
Take
Σ
⊂
Γ
�
{p
�, . . q
kp},) Σ = n − k + �. By hypothesis, p� , . . . , p k , Σ span a hyperplane H that
Take
Σ
⊂
Γ
�
{p
.
,
p
Claim:
k
=
�
(
⇐⇒
=
�
�� , . .the
kkq
kk ,points
Take
Σ
⊂
Γ
�
{p
{p
,
,
.
.
.
.
.
.
,
,
p
p
}
}
with
Σ
=
�Σ�
n
−
=
k
n
+
−
�
.kkBy
+
�
hypothesis,
..�p
By
hypothesis,
p
. , pn
Σ span{p
a��hyperplane
,, .. .. .. ,, pp kk }} ∪
Σ
Hspan
that
�,∉
does
not
contain
p
So
H
,
since
p
∈
qp
.
�
�
Take
Σ
⊂
Γ
�
{p
,
.
.
.
,
p
}
with
�Σ�
=
n
−
+
�
By
hypothesis,
the
n
points
{p
∪
Σ
span
k
�
�
�
k
�
k
does
not
contain
p
.
So
q
∉
H
,
since
p
∈
qp
�p
.
�. So q ∉ H , since p� ∈ qp��p k .
does
not
contain
p
��. ,So
does
aTake
hyperplane
not
contain
H
that
p
does
not
H , contain
since
..qp
Since
ppkkhypothesis,
.�� ∈∈ qp
,, it
that
Σ
⊂
Γ
�
{p
,
.
.
p kq} ∉with
�Σ�
= np��−pp∈��kby
+
��.�p
Byremaining
the
n points
{p�qq, ∉∉.that
.H
. ,.. pqqklies
} ∪ Σon
By
(�),
hyperplane
spanned
the
It
the
aByhyperplane
does
not
contain
Since
qp���p
�pn
it follows
follows
that
H
�that
(�), q
q lies
lies in
inHthe
the
hyperplane
spanned
by
the
remaining
nkkpoints.
points.
It follows
follows
that
lies
onspan
the
(�),
in
hyperplane
spanned
by
the
remaining
follows
on
the
aBy
hyperplane
that
does
. Since
p� ∈ qppp�k+�
�p,,n
that
q ∉that
H . qqq lies
hyperplane
by
and
any
of
the
..kpoints.
.., ..it,, follows
pp�n .. It
The
intersection
of
all
such
By
(�),
qq lies
lies
in
the
hyperplane
spanned
the
remaining
n
points.
It
follows
that
lies
on
the
By
(�), q
lies spanned
inHthe
the
hyperplane
spanned
by
the points
remaining
n
points.
It
follows
that
lies
on
the
hyperplane
spanned
by p
p��not
andcontain
any n
n−
−p���by
of
the
points
The
intersection
of
all
such
�n
k+�
hyperplane
ppp�� and
any
n
−
��� by
of
the
points
ppp k+�
,,,n... points.
... ... ,,, ppp�n
... It
The
intersection
of
all
such
hyperplanes
just
pp�by
,, hence
q
=
p
.
�
hyperplane
spanned
by
and
any
n
−
of
the
points
The
intersection
of
all
such
By
(�), q lies spanned
inis
spanned
the
remaining
follows
that
q
lies
on
the
�
�n
hyperplane
spanned
by
and
any
n
−
of
the
points
The
intersection
of
all
such
k+�
hyperplanes
is the
justhyperplane
hence
q
=
p
.
�
�
�n
k+�
�
�
hyperplanes
is
,,, hence
qqq =
ppp��...n − � of the points p , . . . , p�n . The intersection of all such
�
hyperplanes
is
just
pp���by
hence
=
�
hyperplane
p� and
hyperplanesspanned
is just
just p
hence
=any
�
k+�
�
hyperplanes is just p�, hence q = p�.
�
Projective equivalence
The group PGLn+� K = (GLn+� K)�K × I acts on Pn . Two varieties X, Y ⊂ Pn are projectively
equivalent if there exists A ∈ PGLn+� K such that A ⋅ X = Y .
Any two ordered sets of n + � points in general position in Pn are projectively equivalent.
The group PGL� K acts on P� = A� ∪ {∞} through Möbius transformations:
az + b
a b
�
� ∈ PGL� K induces z �
.
c d
cz + d
Two sets of four points in P� are projectively equivalent if and only if they have the same crossratio, defined by
(z� − z�)(z� − z�)
λ(z� , z� , z� , z�) =
.
(z� − z�)(z� − z�)
Projective equivalence
Projective equivalence
The group PGLn+� K = (GLn+� K)�K × I acts on Pn . Two varieties X, Y ⊂ Pn are projectively
equivalent if there exists A ∈ PGLn+� K such that A ⋅ X = Y .
Any two ordered sets of n + � points in general position in Pn are projectively equivalent.
The group PGL� K acts on P� = A� ∪ {∞} through Möbius transformations:
az + b
a b
�
� ∈ PGL� K induces z �
.
c d
cz + d
Two sets of four points in P� are projectively equivalent if and only if they have the same crossratio, defined by
(z� − z�)(z� − z�)
λ(z� , z� , z� , z�) =
.
(z� − z�)(z� − z�)
Projective equivalence
Projective equivalence
Projective equivalence
The group PGLn+� K = (GLn+� K)�K ×× I acts on Pnn . Two varieties X, Y ⊂ Pnn are projectively
The group PGLn+� K = (GLn+� K)�K I acts on P . Two varieties X, Y ⊂ P are projectively
equivalent if there exists A ∈ PGLn+� K such that A ⋅ X = Y .
equivalent if there exists A ∈ PGLn+� K such that A ⋅ X = Y . n
Any two ordered sets of n + � points in general position in Pn are projectively equivalent.
Any two ordered sets of n + � points in general position in P are projectively equivalent.
The group PGL� K acts on P�� = A�� ∪ {∞} through Möbius transformations:
The group PGL� K acts on P = A ∪ {∞} through Möbius transformations:
a b
az + b
�a b � ∈ PGL� K induces z � az + b .
� c d � ∈ PGL� K induces z � cz + d .
c d
cz + d
Two sets of four points in P�� are projectively equivalent if and only if they have the same crossTwo sets of four points in P are projectively equivalent if and only if they have the same crossratio, defined by
ratio, defined by
(z� − z�)(z� − z�)
λ(z� , z� , z� , z�) = (z� − z�)(z� − z�) .
λ(z� , z� , z� , z�) = (z� − z�)(z� − z�) .
(z� − z�)(z� − z�)
Cross-ratio
[Krishnavedala - Wikimedia Commons]
The twisted cubic
Let v∶ P� → P�, [X� , X�] � [X�� , X�� X� , X� X�� , X��].
The image C = v(P�) is the twisted cubic in P�. It is defined by
C = V(F� , F� , F�)
where
F� = Z� Z� − Z��
F� = Z� Z� − Z� Z�
F� = Z� Z� − Z�� .
It is not defined by any two of these.
For example, F� and F� define the union of C
and the line {Z� = Z� = �}.
Claudio Rocchini - Wikimedia Commons
The twisted cubic
Let v∶ P� → P�, [X� , X�] � [X�� , X�� X� , X� X�� , X��].
The image C = v(P�) is the twisted cubic in P�. It is defined by
C = V(F� , F� , F�)
where
F� = Z� Z� − Z��
F� = Z� Z� − Z� Z�
F� = Z� Z� − Z�� .
It is not defined by any two of these.
For example, F� and F� define the union of C
and the line {Z� = Z� = �}.
Claudio Rocchini - Wikimedia Commons
It is not defined by any two of these.
For example, F� and F� define the union of C
and the line {Z� = Z� = �}.
The twisted cubic
Let v∶ P� → P�, [X� , X�] � [X�� , X�� X� , X� X�� , X��].
The image C = v(P�) is the twisted cubic in P�. It is defined by
C = V(F� , F� , F�)
where
F� = Z� Z� − Z��
F� = Z� Z� − Z� Z�
F� = Z� Z� − Z�� .
It is not defined by any two of these.
For example, F� and F� define the union of C
and the line {Z� = Z� = �}.
It is not defined by any two of these.
Exercise (1.11. in [Ha])
For example, F� and F� define the union of C
and the line {Z� = Z� = �}.
Claudio Rocchini - Wikimedia Commons
Corollary �.�. If L is any secant line of C (i.e. L = pq with p, q ∈ C ), there exist µ, ν with
Q µ ∩ Qν = C ∪ L.
Proof. For any r ∈ pq, r ≠ p, r ≠ q, the space
�Fλ ∶ Fλ (r) = ��
is �-dimensional.
Let F µ , Fν be a basis. Since F µ , Fν vanish at the three points p, q, r on L, we have L ⊂ V(F µ , Fν ),
hence
Q µ ∩ Qν = V(Fµ , Fν ) = C ∪ L
by the exercise above.
�
Corollary
Corollary �.�.
�.�. If
If L
L is
is any
any secant
secant line
line of
of C
C (i.e.
(i.e. L
L=
= pq
pq with
with p,
p, qq ∈∈ C
C ),), there
there exist
exist µ,
µ, νν with
with
Q
Qµ ∩
∩Q
Qν =
=C
C∪
∪ L.
L.
µ
ν
Proof.
Proof. For
For any
any rr ∈∈ pq
pq,, rr ≠
≠ pp,, rr ≠
≠ qq,, the
the space
space
�F
�Fλ ∶∶ F
Fλ (r)
(r) =
= ��
��
λ
λ
is
is ��-dimensional.
-dimensional.
Let
Let F
Fµµ ,, F
Fνν be
be aa basis.
basis. Since
Since F
Fµµ ,, F
Fνν vanish
vanish at
at the
the three
three points
points p,
p, q,
q, rr on
on L
L,, we
we have
have L
L⊂
⊂ V(F
V(Fµµ ,, F
Fνν )),,
hence
hence
Q
Q µµ ∩
∩Q
Qνν =
= V(F
V(Fµµ ,, F
Fνν )) =
=C
C∪
∪L
L
by
by the
the exercise
exercise above.
above.
�
�
The rational normal curve
The rational normal curve
The rational normal curve in Pd is the Veronese embedding of P� of degree d . It is the image
of the map
vd ∶ P� → Pd , [X� , X�] � [X�d , X�d−� X� , . . . , X� X�d−� , X�d ].
Any d + � points on a rational normal curve are linearly independent. For given distinct points
p� , . . . , pd ∈ P�, we may assume p i ≠ [�, �] for all i , say p i = [Yi , �]. The matrix
� d d−� d−�
�
� Y� Y� Y�
⋅ Y� � �
� d d−�
�
� Y Y
�
⋅
⋅
⋅
�
� � �
�
�� ⋅
⋅
⋅
⋅
⋅
⋅ ��
�
�
� ⋅
�
⋅
⋅
⋅
⋅
�
�
�
� Yd ⋅
⋅
⋅ Yd � ��
� d
�
�
is a Vandermonde matrix with determinant ∏ i< j(Yi − Y j) ≠ �, showing that vd (p�), . . . , vd (pd )
are independent.
Any curve projectively equivalent to the rational normal curve is also a rational normal curve.
In particular, if H� , . . . , Hd is any basis of K[X� , X�]d , then
vd ∶ [X� , X�] � [H�(X� , X�), . . . , Hd (X� , X�)]
is a rational normal curve.
The
The rational
rational normal
normal curve
curve
The rational normal curve
The rational normal curve in Pdd is the Veronese embedding of P�� of degree d . It is the image
The rational normal curve in P is the Veronese embedding of P of degree d . It is the image
of the map
of the map
d−� , X d ].
vd ∶ P�� → Pdd , [X� , X�] � [X�dd , X�d−�
X
,
.
.
.
,
X
X
�
�
vd ∶ P → P , [X� , X�] � [X� , X�d−� X� , . . . , X� X��d−� , X��d ].
Any d + � points on a rational normal curve are linearly independent. For given distinct points
Any d + � points
on a rational normal curve are linearly independent. For given distinct points
p� , . . . , pd ∈ P��, we may assume p i ≠ [�, �] for all i , say p i = [Yi , �]. The matrix
p� , . . . , pd ∈ P , we may assume p i ≠ [�, �] for all i , say p i = [Yi , �]. The matrix
� d d−� d−�
�
� Y�d Y�d−� Y�d−� ⋅ Y� � �
� Z d Z d−� Z
�
⋅
Z
�
�
� Y� Y�
�
�⋅
⋅
⋅
�
� Z�d Z�d−� ⋅
⋅
⋅
� ��
�� ⋅� �⋅
⋅
⋅
⋅
⋅ �
� ⋅
⋅
⋅
⋅
⋅
⋅ ��
� ⋅
⋅
⋅
⋅
⋅
� �
� ⋅
⋅
⋅
⋅
� �
�� Y d ⋅⋅
⋅
⋅ Yd � ��
� Zdd ⋅
⋅
⋅ Zd �
� d
�
is a Vandermonde matrix with determinant ∏ i< j(Yi − Y j) ≠ �, showing that vd (p�), . . . , vd (pd )
is a Vandermonde matrix with determinant ∏ i< j(Z i − Z j) ≠ �, showing that vd (p�), . . . , vd (pd )
are independent.
are independent.
Any curve projectively equivalent to the rational normal curve is also a rational normal curve.
Any curve projectively equivalent to the rational normal curve is also a rational normal curve.
In particular, if H� , . . . , Hd is any basis of K[X� , X�]d , then
In particular, if H� , . . . , Hd is any basis of K[X� , X�]d , then
vd ∶ [X� , X�] � [H�(X� , X�), . . . , Hd (X� , X�)]
vd ∶ [X� , X�] � [H�(X� , X�), . . . , Hd (X� , X�)]
is a rational normal curve.
is a rational normal curve.
Theorem �.�. Through any d + � points in general position in Pd , there passes a unique rational
normal curve.
Construction. Let µ� , . . . , µd , ν� , . . . , νd ∈ K × and consider
d
G = �(µ i X� − ν i X�) ∈ K[X� , X�]d+�
i=�
G
Hi =
,
µ i X� − ν i X�
i = �, . . . , d.
Then H� , . . . , Hd are a basis of K[X� , X�]d . For if ∑di=� a i H i = � is any linear relation, evaluation
at [µ i , ν i ] gives a i = �.
Thus
vd ∶ [X� , X�] � [H�(X� , X�), . . . , Hd (X� , X�)]
is a rational normal curve. We find
vd ([µ� , ν�]) = [�, �, . . . , �], . . . , vd ([µd , νd ]) = [�, . . . , �, �]
µ��µd
µ��µd
vd ([�, �]) = �
,...,
� = [µ�−� , . . . , µd−�] and vd ([�, �]) = [ν�−� , . . . , νd−�].
µ�
µd
So let p� , . . . , pd+� be any d + � points in general position Pd .
We can assume p i = [e i ] for i = �, . . . , d by projective equivalence. Then pd+� and pd+� have
non-zero coordinates. We can choose [µ� , ν�], . . . , [µd , νd ] ∈ P� such that vd [�, �] = pd+� and
vd [�, �] = pd+�. Uniqueness is left as an exercise.
�
d , there passes a unique rational
Theorem
�.�.
Through
any
d
+
�
points
in
general
position
in
P
Theorem �.�. Through any d + � points in general position in Pd , there passes a unique rational
normal
normal curve.
curve.
× with [µ , ν ] ≠ [µ , ν ] for all i ≠ j and consider
Construction.
Let
µ
,
.
.
.
,
µ
,
ν
,
.
.
.
,
ν
∈
K
�
�
i i
j j
Construction. Let µ� , . . . , µdd , ν� , . . . , νdd ∈ K × and consider
d
d
G
G =
= �
(µ ii X
X�� −
− νν ii X
X��)) ∈∈ K[X
K[X�� ,, X
X��]]d+�
�(µ
d+�
i=�
i=�
G
G
H
=
i
H i = µ X − ν X ,,
µ i X� − ν i X�
i �
i �
ii =
= �,
�, .. .. .. ,, d.
d.
d
Then
H
,
.
.
.
,
H
are
a
basis
of
K[X
,
X
]
.
For
if
aa i H
�� is
any
linear
relation,
evaluation
∑
ii =
Then H�� , . . . , Hdd are a basis of K[X�� , X��]dd . For if ∑di=�
H
=
is
any
linear
relation,
evaluation
i
i=�
at
at [µ
[µ ii ,, νν ii ]] gives
gives a
a ii =
= ��..
Thus
Thus
vvd ∶∶ [X
[H
.. .. .. ,, H
(X
�� ,, X
��]] �
��(X
�� ,, X
��),
�� ,, X
��)]
d
[X
X
�
[H
(X
X
),
H
(X
X
)]
d
d
is
is aa rational
rational normal
normal curve.
curve. We
We find
find
vvd ([µ
=
[�,
�,
.. .. .. ,, �],
.. .. .. ,, vvd ([µ
,, ννd ])
=
[�,
.. .. .. ,, �,
�]
�� ,, ν
��])
d
([µ
ν
])
=
[�,
�,
�],
([µ
])
=
[�,
�,
�]
d
d
d d
µ
µ
��µ
��µ
d
d
−�] and v ([�, �]) = [ν −� , . . . , ν −�].
µ
�µ
µ
�
d , . . . , ��µ d � = [µ�−�
vvd ([�,
�])
=
�
,
.
.
.
,
µ
−�
−�] and v d ([�, �]) = [ν�−� , . . . , ν d
−�].
d
([�,
�])
=
�
,
.
.
.
,
�
=
[µ
,
.
.
.
,
µ
d
d
µ
µ
�
�
d
d
µ��
µdd
d.
So
let
p
,
.
.
.
,
p
be
any
d
+
�
points
in
general
position
in
P
d
�
So let p� , . . . , pd+�
d+� be any d + � points in general position P .
We
Then
ppd+� and
ppd+� have
We can
can assume
assume p
p ii =
= [e
[e ii ]] for
for ii =
= �,
�, .. .. .. ,, d
d by
by projective
projective equivalence.
equivalence.
Then
and
d+�
d+� have
�
non-zero
and
non-zero coordinates.
coordinates. We
We can
can choose
choose [µ
[µ�� ,, νν��],
], .. .. .. ,, [µ
[µdd ,, ννdd ]] ∈∈ P
P� such
such that
that vvdd [�,
[�, �]
�] =
= ppd+�
d+� and
vvd [�,
�
[�, �]
�] =
= ppd+�.. Uniqueness
Uniqueness is
is left
left as
as an
an exercise.
exercise.
�
d
d+�
d , there passes a unique rational
Theorem
�.�.
Through
any
d
+
�
points
in
general
position
in
P
Theorem �.�. Through any d + � points in general position in Pdd , there passes a unique rational
normal
normal curve.
curve.
× with [µ , ν ] ≠ [µ , ν ] for all i ≠ j and consider
Construction.
Let
µ
,
.
.
.
,
µ
,
ν
,
.
.
.
,
ν
∈
K
�
�
i i
j j
Construction. Let µ� , . . . , µdd , ν� , . . . , νdd ∈ K ×× and consider
d
d
G
G =
= �
(µ ii X
X�� −
− νν ii X
X��)) ∈∈ K[X
K[X�� ,, X
X��]]d+�
�(µ
d+�
i=�
i=�
G
G
H
=
i
H i = µ i X� − ν i X� ,, ii =
= �,
�, .. .. .. ,, d.
d.
µ X −ν X
i �
i �
d
d ai Hi = �
Then
H
,
.
.
.
,
H
are
a
basis
of
K[X
,
X
]
.
For
if
is
any
linear
relation,
evaluation
∑
�
Then H�� , . . . , Hdd are a basis of K[X�� , X��]dd . For if ∑ i=�
a
H
=
is
any
linear
relation,
evaluation
i
i
i=�
at
at [µ
[µ ii ,, νν ii ]] gives
gives a
a ii =
= ��..
Thus
Thus
vvd ∶∶ [X
[H
.. .. .. ,, H
�� ,, X
��]] �
��(X
�� ,, X
��),
�� ,, X
��)]
d (X
[X
X
�
[H
(X
X
),
H
(X
X
)]
d
d
is
is aa rational
rational normal
normal curve.
curve. We
We find
find
vvd ([µ
=
[�,
�,
.. .. .. ,, �],
.. .. .. ,, vvd ([µ
,, ννd ])
=
[�,
.. .. .. ,, �,
�]
� ,, ν
�])
d
([µ
ν
])
=
[�,
�,
�],
([µ
])
=
[�,
�,
�]
� �
d
d
d d
µ
µ
��µ
��µ
d
d
−�] and v ([�, �]) = [ν −� , . . . , ν −�].
µ
�µ
µ
�
d , . . . , ��µ d � = [µ�−�
vvd ([�,
�])
=
�
,
.
.
.
,
µ
−�
−�] and v d ([�, �]) = [ν�−� , . . . , ν d
−�].
d
([�,
�])
=
�
,
.
.
.
,
�
=
[µ
,
.
.
.
,
µ
d
d
µ
µ
�
�
d
d
µ��
µdd
d.
So
let
p
,
.
.
.
,
p
be
any
d
+
�
points
in
general
position
in
P
d
�
So let p� , . . . , pd+�
d+� be any d + � points in general position P .
We
Then
ppd+� and
ppd+� have
We can
can assume
assume p
p ii =
= [e
[e ii ]] for
for ii =
= �,
�, .. .. .. ,, d
d by
by projective
projective equivalence.
equivalence.
Then
and
d+�
d+� have
�
non-zero
and
non-zero coordinates.
coordinates. We
We can
can choose
choose [µ
[µ�� ,, νν��],
], .. .. .. ,, [µ
[µdd ,, ννdd ]] ∈∈ P
P� such
such that
that vvdd [�,
[�, �]
�] =
= ppd+�
d+� and
vvd [�,
�
[�, �]
�] =
= ppd+�.. Uniqueness
Uniqueness is
is left
left as
as an
an exercise.
exercise.
�
d
d+�
d , there passes a unique rational
Theorem
�.�.
Through
any
d
+
�
points
in
general
position
in
P
d
d
Theorem
�.�.
Through
any
d
+
�
points
in
general
position
in
P
passes
a
unique
rational
d ,, there
Theorem
�.�.
Through
any
d
+
�
points
in
general
position
in
P
there
passes
a
unique
rational
normal
curve.
normal
normal curve.
curve.
× with [µ , ν ] ≠ [µ , ν ] for all i ≠ j and consider
Construction.
Let
µ
,
.
.
.
,
µ
,
ν
,
.
.
.
,
ν
∈
K
�
�
i i
j j
d , ν� , . . . , ν d ∈ K ×
×
Construction.
Let
µ
,
.
.
.
,
µ
and
consider
×
�
Construction. Let µ� , . . . , µdd , ν� , . . . , νdd ∈ K and consider
(µ
νν ii X
∈∈ K[X
�
�� −
��)
�� ,, X
��]]d+�
ii X
(µ
X
−
X
)
K[X
X
�
d+�
�
i=�(µ i X� − ν i X�) ∈ K[X� , X�]d+�
i=�
i=� G
G
H
=
i
H ii = µ i X� − ν i X� ,, ii =
= �,
�, .. .. .. ,, d.
d.
µ
µ ii X
X�� −
− νν ii X
X��
d
d
Then
H
,
.
.
.
,
H
are
a
basis
of
K[X
,
X
]
.
For
if
�� is
any
linear
relation,
evaluation
∑
ii H
ii =
di=� a
Then H��� , . . . , Hddd are a basis of K[X��� , X���]ddd . For if ∑
a
H
=
is
any
linear
relation,
evaluation
∑i=�
i i
i=�
at
[µ
aa i =
��..
ii ,, ν
ii ]] gives
at
[µ
ν
gives
=
i
at [µ i , ν i ] gives a i = �.
Thus
Thus
vvd ∶∶ [X
[H
.. .. .. ,, H
(X
�� ,, X
��]] �
��(X
�� ,, X
��),
�� ,, X
��)]
d
[X
X
�
[H
(X
X
),
H
(X
X
vdd ∶ [X� , X�] � [H�(X� , X�), . . . , Hdd (X� , X�)]
)]
is
is aa rational
rational normal
normal curve.
curve. We
We find
find
vvd ([µ
=
[�,
�,
.. .. .. ,, �],
.. .. .. ,, vvd ([µ
,, ννd ])
=
[�,
.. .. .. ,, �,
�]
� ,, ν
�])
d
([µ
ν
])
=
[�,
�,
�],
([µ
])
=
[�,
�,
�]
�� ��
d
d
d
d
d
d d
d
µ
µ
��µ
��µ
d
d
−� , . . . , µ −�] and v ([�, �]) = [ν −� , . . . , ν −�].
µ
�µ
µ
�µ
��
�
d
d
vvd ([�,
�])
=
�
,
.
.
.
,
�
=
[µ
−�
−�
�
d � = [µ�−�
d, . . . ,
−� , . . . , µ d
−�] and v d ([�, �]) = [ν�−�
−� , . . . , ν d
−�].
([�,
�])
=
�
d
d
µ
µ
�
�
d
d
d
d
�
�
d
d
µ
µ
µ���
µddd
d.
So
let
p
,
.
.
.
,
p
be
any
d
+
�
points
in
general
position
in
P
d
�
d.
So let p�� , . . . , pd+�
be
any
d
+
�
points
in
general
position
P
.
in
P
d+�
d+�
We
can
assume
Then
ppd+� and
ppd+� have
We can assume p
p iii =
= [e
[e iii ]] for
for ii =
= �,
�, .. .. .. ,, d
d by
by projective
projective equivalence.
equivalence.
Then
and
have
d+�
d+�
d+�
d+�
�
non-zero
and
non-zero coordinates.
coordinates. We
We can
can choose
choose [µ
[µ��� ,, νν���],
], .. .. .. ,, [µ
[µddd ,, ννddd ]] ∈∈ P
P�� such
such that
that vvddd [�,
[�, �]
�] =
= ppd+�
and
d+�
d+�
vvd [�,
�]
=
ppd+�.. Uniqueness
is
left
as
an
exercise.
�
[�,
�]
=
Uniqueness
is
left
as
an
exercise.
�
d
d+�
d
d+�
G
G
G
=
=
=
d
d
d