4 Projective space and projective varieties
One of the motivations for projective geometry is to understand perspective.
It is well-known phenomenon that parallel lines such as railway tracks ap- pear to meet at a point, called the vanishing point. It was not until the Renaissance that painters were able to give a geometrically correct represen- tation of objects. The painting below, The delivery of keys by Pietro Perug- ino in the Sistine Chapel (source https://en.wikipedia.org/wiki/File:
Entrega_de_las_llaves_a_San_Pedro_(Perugino).jpg) is a good exam- ple of the use of perspective, the parallel lines on the ground converge towards a point near the centre of the painting.
Projective geometry gives a nicer theory than affine geometry in many ways, for example, two lines in the projective plane always meet, or more generally two curves defined by degree m and degree n equations, resp., have mn intersection points counted with multiplicity.
Projective geometry also has important practical applications, images are
combined by applying projective transformations in photo stitching, which
is used in panoramic photos, Google Street View and virtual reality.
4.1 Basic properties of projective space and projective varieties
The definition of projective space is also motivated by the idea that if a camera is at the origin of the co-ordinate system, all points of a ray through the origin will be mapped to the same point of the image, so points of the image correspond to lines through the origin.
Definition. The n-dimensional projective space over a field K, denoted by P
n(K) or by P
nif the field is understood, is the set of equivalence classes of K
n+1\ {(0, 0, . . . , 0)} under the equivalence relation (x
0, x
1, . . . , x
n) ∼ (λx
0, λx
1, . . . , λx
n) for any λ ∈ K \ {0}. (The points of P
ncorrespond to lines through the origin in K
n+1.)
The equivalence class of a point (X
0, X
1, . . . , X
n) ∈ K
n+1\ {(0, 0, . . . , 0)} is denoted by (X
0: X
1: . . . : X
n). X
0, X
1, . . . X
nare called homogeneous co-ordinates on P
n.
Example: (1 : 2 : 3), (2 : 4 : 6) and (−1 : −2 : −3) are the same point in P
2, this is a peculiarity of projective space that the co-ordinates of a point are not unique.
Two ways of looking at projective space
1. Let U
0= {(X
0: X
1: . . . : X
n) ∈ P
n| X
06= 0}. (X
0: X
1: . . . : X
n) = (1 : X
1/X
0: . . . : X
n/X
0) in U
0and X
1/X
0, X
2/X
0, . . . , X
n/X
0can take arbitrary values in K, so the points of U
0are in bijection with the points of A
n. The set P
n\ U
0= {(X
0: X
1: . . . : X
n) ∈ P
n| X
0= 0} is clearly a copy of P
n−1. Therefore P
n= A
n∪ P
n−1as a set.
If n = 1, then
U
0= {(X
0: X
1) ∈ P
1| X
06= 0} = { 1 : X
1X
0| X
1∈ K, X
06= 0} = {(1 : x) | x ∈ K}, so we can identify U
0with A
1via (X
0: X
1) ↔ X
1/X
0∈ A
1.
P
1\ U
0= {(X
0: X
1) ∈ P
1| X
0= 0} = {(0 : X
1) | X
1∈ K \ {0}} = {(0 : 1)}, because (0 : X
1) = (0 : 1) for any K \ {0}. This is the decomposition, P
1= A
1∪ P
0, as P
0consists of just a single point.
We can identify A
1with K and denote the single point of P
0by ∞, this way
we can identify P
1with K ∪ {∞}.
If n = 2, then
U
0= {(X
0: X
1: X
2) ∈ P
2| X
06= 0} = { 1 : X
1X
0: X
2X
0| X
1, X
2∈ K, X
06= 0}
= {(1 : x : y) | x, y ∈ K}
so we can identify U
0with A
2via (X
0: X
1: X
2) ↔ (X
1/X
0, X
2/X
0) ∈ A
2. The complement of U
0is
P
2\U
0= {(X
0: X
1: X
2) ∈ P
2| X
0= 0} = {(0 : X
1: X
2) | (X
1, X
2) ∈ K
2\{(0, 0)}}, this is a copy of P
1.
To understand these points, consider the parallel lines x+y = 0 and x+y−2 = 0 in A
2= U
0. As x = X
1/X
0and y = X
2/X
0, we can rewrite the equations in terms of homogeneous co-ordinates as X
1X
0+ X
2X
0= 0 and X
1X
0+ X
2X
0− 2 = 0.
After multiplying by X
0we get X
1+ X
2= 0 and X
1+ X
2− 2X
0= 0. The solutions of this system of linear equations are X
0= 0, X
1= −X
2, which correspond to the point (0 : 1 : −1) ∈ P
1. Therefore these parallel lines in A
2= U
0intersect in P
2\ U
0. Any other line parallel to them also contains (0 : 1 : −1).
x+y=0 x+y-2=0 H0:1:-1L
-4 -2 2 4x
-4 -2 2 4 y
Similarly, all lines with direction vector (X
1, X
2) intersect at (0 : X
1: X
2) ∈
P
2\ U
0.
This is true in general for arbitrary n, the points of P
n\U
0= P
n−1correspond to directions of lines in A
nor equivalence classes of parallel lines. (0 : X
1: . . . : X
n) is the point where all the lines with direction vector (X
1, X
2, . . . , X
n) in A
nmeet. (The single point of P
0corresponds to the only line in A
1.) For this reason, these points are often called points at infinity, but this does not mean that they are intrinsically different, the disctinction depends on the choice of co-ordinates.
The youtube video https://www.youtube.com/watch?v=q3turHmOWq4 (also linked to from the Animations and videos section of the course website) contains a good explanation of the projective plane.
2. The other approach is is to consider the sets
U
i= {(X
0: X
1: . . . : X
n) ∈ P
n| X
i6= 0},
0 ≤ i ≤ n. Each one is a copy of A
nand their union is P
n, since there is no point (0 : 0 : . . . : 0). These n + 1 copies are glued together by identifying their points via rational maps ϕ
ij: U
i99K U
jso that P ∈ U
iis the same point of P
nas ϕ
ij(P ) ∈ U
j.
For example, if n = 1, let t = X
1/X
0be the co-ordinate on U
0, and u = X
0/X
1the co-ordinate on U
1. We have the rational maps ϕ
01: U
099K U
1, t 7→ 1/t, and ϕ
10: U
199K U
0, u 7→ 1/u. A point t ∈ U
0is the same point in P
1as ϕ
01(t) ∈ U
1, whenever ϕ
01(t) is defined and similarly, a point u ∈ U
1is the same point in P
1as ϕ
10(u) ∈ U
0, whenever ϕ
10(u) is defined.
If n = 2, then let x = X
1/X
0, y = X
2/X
0be the co-ordinates on U
0, u = X
0/X
1, v = X
2/X
1the co-ordinates on U
1and s = X
0/X
2, t = X
1/X
2the co-ordinates on U
2. Now we have u = 1/x and v = y/x, which give the rational map ϕ
01: U
099K U
1, (x, y) 7→ (1/x, y/x), so that (x, y) ∈ U
0and ϕ
01(x, y) ∈ U
1are the same point in P
2whenever the latter is defined.
The point (2 : 3 : 4) ∈ P
2is the same as (1 : 3/2 : 2), (2/3 : 1 : 4/3) or (1/2 : 3/4 : 1), therefore (3/2, 2) ∈ U
0, (2/3, 4/3) ∈ U
1and (1/2, 3/4) ∈ U
2correspond to the same point of P
2and they get identified in the gluing process.
In general, if F ∈ K[X
0, X
1, . . . , X
n] and P = (X
0: X
1: . . . : X
n) ∈ P
n, F (P ) cannot be defined, since F (λX
0, λX
1, . . . , λX
n) will give different values for different values of λ ∈ K \ {0}. However, if F is homogeneous of degree d, then F (λX
0, λX
1, . . . , λX
n) = λ
dF (X
0, X
1, . . . , X
n), so we can tell whether F (P ) = 0 or not, and this is enough to define projective algebraic varieties.
Definition. A polynomial is called homegeneous if and only if all of its terms
have the same degree.
E. g., x
3+ 2y
2z − 3xz
2is homogeneous of degree 3, while x
3− y
2is not homogeneous.
Definition. An ideal I / K[X
0, X
1, . . . , X
n] is called homogeneous if and only if it can be generated by homogeneous elements. (Warning: This does not mean that all elements of the ideal are homogeneous polynomials.)
Definition. Let I /K[X
0, X
1, . . . , X
n] be a homogeneous ideal. The projective algebraic variety defined by I is the set
V(I) = {(X
0: X
1: . . . : X
n) ∈ P
n| F (X
0, X
1, . . . , X
n) = 0, ∀F ∈ I, F homogeneous}.
Two ways of looking at projective varieties
1. Let’s consider the variety V = V(hX
1X
2− X
02i) ⊂ P
2. If X
06= 0, we can divide X
1X
2− X
02= 0 by X
02to get (X
1/X
0)(X
2/X
0) − 1 = 0, which can be written as xy − 1 = 0 in terms of the affine co-ordinates x = X
1/X
0and y = X
2/X
0on U
0= A
2, so V
0= V ∩ U
0is a hyperbola. The points V \ V
0are the points of V with X
0= 0. By substituting X
0= 0 into X
1X
2− X
02= 0 we obtain X
1X
2= 0, therefore the points at infinity are (0 : 1 : 0), which corresponds to lines parallel to the x-axis, and (0 : 0 : 1), which corresponds to lines parallel to the y-axis. The asymptotes of the hyperbola pass through these points at infinity. V consists of the hyperbola with two points at infinity corresponding to the asymptotes.
xy-1=0 H0:0:1L
H0:1:0L
-4 -2 2 4x
-4 -2 2 4 y
Starting with the equation xy − 1 = 0 we can recover X
1X
2− X
02= 0 by
substituting x = X
1/X
0and y = X
2/X
0into xy − 1 = 0 and multiplying it
by X
02.
Let’s now consider the variety V = V(hX
12− X
0X
2i) ⊂ P
2. If X
06= 0, we can divide X
12− X
0X
2= 0 by X
02to get (X
1/X
0)
2− (X
2/X
0) = 0, which can be written as x
2− y = 0 in terms of the affine co-ordinates x = X
1/X
0and y = X
2/X
0on U
0= A
2, so V
0= V ∩ U
0is a parabola. The points V \ V
0are the points of V with X
0= 0. By substituting X
0= 0 into X
12− X
0X
2= 0 we obtain X
12= 0, therefore the only points at infinity is (0 : 0 : 1), which corresponds to lines parallel to the y-axis. Unlike the hyperbola, the parabola has no asympotes, but for large x and y, its tangent direction approaches the direction of the y-axis.
x2-y=0 (0:0:1)
-3 -2 -1 1 2 3 x
2 4 6 8 y
Starting with the equation x
2− y = 0 we can recover X
12− X
0X
2= 0 by substituting x = X
1/X
0and y = X
2/X
0into x
2− y = 0 and multiplying it by X
02.
Given a homogeneous polynomial F ∈ K[X
0, X
1, . . . , X
n], F [X
0, X
1, . . . , X
n]
X
0deg F= F [1, x
1, x
2, . . . , x
n]
where x
i= X
i/X
0, 1 ≤ i ≤ n, is called the dehomogenisation of F with respect to X
0. If we dehomogenise all homogeneous elements of an homo- geneous ideal J / K[X
0, X
1, . . . , X
n], we obtain an ideal defining the affine algebraic variety V
0= V ∩ U
0⊆ A
n, called the affine piece X
06= 0 of V = V(J ) ⊆ P
n.
If we start with a polynomial f (x
1, x
2, . . . , x
n),
X
0deg ff (X
1/X
0, X
2/X
0, . . . , X
n/X
0) ∈ K[X
0, X
1, . . . , X
n]
is a homogeneous polynomial, the homogenisation of f . Homogenisation can also be done at the level of ideals, the projective algebraic variety in P
ndefined by the ideal generated by the homogenisation of the elements of an ideal J / K[x
1, x
2, . . . , x
n] is called the projective closure of V(J ) ⊆ A
n. The affine piece X
06= 0 of the projective closure is exactly V(J), while the points at infinity correspond to asymptotic directions of V(J ) as we have seen in the example.
Homogenisation and dehomogenisation are one-sided inverses of each other.
Homogenisation followed by dehomogenisation always yields the same poly- nomial, and similarly for varieties, taking the projective closure of an affine variety and then the affine piece X
06= 0 gives the original variety.
Dehomogenising and then homogenising a homogeneous polynomial will give the polynomial divided by the highest power of X
0dividing it. (For example, the dehomogenisation of X
0X
1is x
1, whose homogenisation is just X
1.) For varieties this means that taking projective closure of the affine piece X
06= 0 of a projective algebraic variety will give the union of irreducible components of the original variety not contained in the hyperplane X
0= 0.
(Irreducibility for projective varieties will be defined later, but it is completely analogous to the affine case.)
If a projective variety V has no irreducible components contained in the hyperplane X
0= 0, then V is the projective closure the affine piece V
0= V ∩ U
0and the points of V correspond to the points of V
0and the asympotic directions of V
0.
2. The other approach is to consider all the affine pieces V
i= V ∩ U
i, 0 ≤ i ≤ n, of a projective variety V . These are affine varieties and they are glued together by identifying the points via the rational maps ϕ
ij: U
i99K U
jdefined previously. For example, by homogenising the equation y
2−x
3−x
2= 0 of the nodal cubic in A
2, we obtain X
0X
22−X
13−X
0X
12= 0. The projective curve defined by this equation is the projective closure V , whose affine piece V
0= V ∩ U
0is original affine nodal cubic curve. Substituting X
0= 0 into the equation gives X
13= 0, therefore the only point with X
0= 0 is (0 : 0 : 1), the asymptotic direction of the y-axis.
x and y can be expressed as x = X
1/X
0, y = X
2/X
0in terms of the homo-
geneous co-ordinates. The affine co-ordinates on the other affine pieces are
u = X
0/X
1, v = X
2/X
1on U
1and s = X
0/X
2, t = X
1/X
2on U
2. Deho-
mogenising X
0X
22− X
13− X
0X
12= 0 with respect to X
1gives uv
2− 1 − u = 0,
dehomogenising it with respect to X
2gives s − t
3− st
2= 0. Therefore the
projective curve has the affine pieces shown below.
y
2− x
3− x
2= 0 uv
2− 1 − u = 0 s − t
3− st
2= 0
-10 -5 5 10 x
-10 -5 5 10 y
-10 -5 5 10 u
-10 -5 5 10 v
-10 -5 5 10 s
-10 -5 5 10 t
They are glued together by the maps ϕ
01: U
099K U
1, (x, y) 7→ (1/x, y/x) and ϕ
02: U
099K U
2, (x, y) 7→ (1/y, x/y).
Definition. Let F ∈ K[X
0, X
1, . . . , X
n] be a polynomial. The degree i ho- mogeneous part of F , denoted by F
[i], is the sum of all the terms of degree i in F . If i < 0 or i > deg F , F
[i]is defined to be 0.
It is clear from this definition that F = F
[0]+ F
[1]+ . . . + F
[deg F ].
Example: Let F = X
0+X
0X
2−X
12+X
0X
1X
2∈ K[X
0, X
1, X
2], then F
[0]= 0, F
[1]= X
0, F
[2]= X
0X
2− X
12and F
[3]= X
0X
1X
2.
Lemma 4.1 The ideal I / K[X
0, X
1, . . . , X
n] is homogeneous if and only if for any F ∈ I, all the homogeneous parts of F are also elements of I.
Proof. Assume that I is homogeneous, let F
1, F
2, . . . , F
rbe a set of ho- mogeneous generators for I. Let F ∈ I, then F =
r
P
i=1
F
iG
ifor some G
i∈ K[X
0, X
1, . . . , X
n], 1 ≤ i ≤ r. Then F
[d]=
r
P
i=1
F
iG
[d−deg Fi]∈ I for every d, 0 ≤ d ≤ deg F , so all the homogeneous parts of F are also elements of I.
Assume now that F ∈ I implies that F
[0], F
[1], . . . , F
[deg F ]are also in I. Let F
1, F
2, . . . , F
rbe an arbitrary set of generators for I. We claim that the set {F
i[j]| 1 ≤ i ≤ r, 0 ≤ j ≤ deg F
i} generates I. On one hand F
i=
deg Fi
P
j=0
F
i[j]is contained in the ideal generated by this set for each i, 1 ≤ i ≤ r, so this ideal contains I. On the other hand, each of the generators is in I, so the ideal generated by them is a subset of I. Therefore the ideal generated by this set is exactly I and since these generators are homogeneous, I is a homogeneous ideal.
Proposition 4.2 (Cf. Proposition 1.3)
(i) Let V
1= V(I
1), V
2= V(I
2), . . . , V
k= V(I
k) be projective algebraic varieties in P
n. Then
V
1∪ V
2∪ . . . ∪ V
k= V(I
1∩ I
2∩ . . . ∩ I
k) = V(I
1I
2. . . I
k) is also a projective algebraic variety.
(ii) Let V
α= V(I
α), α ∈ A be projective algebraic varieties in P
n. Then
\
α∈A
V
α= V X
α∈A
I
αis also a projective algebraic variety.
Proof. Just imitate the proof of Proposition 1.3 (i) and (ii).
Warning: There is no direct analogue of Proposition 1.3 (iii) for projective varieties because P
m× P
nis very different from P
m+n.
Definition. The homogeneous ideal of a set Z ⊆ P
nis the ideal I(Z) / K[X
0, X
1, . . . , X
n] generated by the set
{F ∈ K[X
0, X
1, . . . , X
n] | F homogeneous, F (X
0, . . . , X
n) = 0 ∀(X
0: . . . : X
n) ∈ Z}.
Theorem 4.3 (Projective Nullstellensatz, cf. Theorem 1.7)
Let K be an algebraically closed field and let J / K[X
0, X
1, . . . , X
n] be a homogeneous ideal.
(i) V(J ) = ∅ if and only if J = K[X
0, X
1, . . . , X
n] or √
J = hX
0, X
1, . . . , X
ni.
(ii) I(V(J )) = √
J unless √
J = hX
0, X
1, . . . , X
ni.
Idea of proof. For any homogeneous ideal J / K[X
0, X
1, . . . , X
n] we can consider the projective algebraic variety V(J ) defined by in P
nand also the affine algebraic variety defined by J in A
n+1, called affine cone on V(J ).
This gives an almost bijective correspondence between projective algebraic varieties in P
nand certain affine algebraic varieties in A
n+1, the only failure of bijectivity is that K[X
0, X
1, . . . , X
n] and hX
0, X
1, . . . , X
ni both define the empty set as a projective variety.
Definition. A projective algebraic variety V is reducible if and only if it can be written as V = V
1∪V
2, where V
1, V
2are also projective algebraic varieties, V
16= V 6= V
2. If V is not reducible, it is called irreducible.
Proposition 4.4 (Cf. Theorem 1.8) Every projective algebraic variety V can be decomposed into a union V = V
1∪ V
2∪ . . . ∪ V
ksuch that every V
i, 1 ≤ i ≤ k, is an irreducible projective algebraic variety and V
i6⊆ V
jfor i 6= j.
The decomposition is unique up to the ordering of the components. The V
i,
1 ≤ i ≤ k, are called the irreducible components of V .
Proof. Completely analogous to the proof of Theorem 1.8.
Lemma 4.5 A homogeneous ideal I / K[X
0, X
1, . . . , X
n] is prime if and only if for any homogeneous polynomials F , G ∈ K[X
0, X
1, . . . , X
n], F G ∈ I implies F ∈ I or G ∈ I.
Proof. If I is prime then F G ∈ I implies F ∈ I or G ∈ I for all F, G ∈ K[X
0, X
1, . . . , X
n] by the definition of a prime ideal.
Let’s assume I is a homogeneous ideal which is not prime. Then there exist polynomials P / ∈ I, Q / ∈ I such that P Q ∈ I. Let d ≥ 0 be the minimal integer such that P
[d]∈ I. There exists such a d since if P /
[j]∈ I for every j ≥ 0, then P =
deg P
P
j=0
P
[j]∈ I, too. Similarly, let e ≥ 0 be the minimal integer such that Q
[e]∈ I. Then (P Q) /
[d+e]=
d+e
P
j=0