DISCLAIMER: These unpolished lecture notes were designed for in-class use only, not as reference material.
They are only web-posted in response to student request.
1
Dimension
Def. The dimension of an irreducible qp var X is the
transcendence degree of k(X) over k. The dimension
of a general qp var is the maximum dim of each of its
components
Note: if U ⊂ X is open, then k(U ) = k(X), so dim(U ) =
dim(X)
Recall that if k ⊂ K is a field extension, the transcendence degree tr deg(K/k) of K over k is the maximum
number of elts of K which are alg indep over k
If tr deg(K/k) = n and x1, . . . , xn ∈ K are alg indep
over k, then K is an alg ext of k(x1, . . . , xn)
Example. Since k(An) = k(x1, . . . , xn), we have
tr deg(k(An/k)/k) = n and dim An = n.
Since An ⊂ Pn is open, it follows that dim Pn = n
Def. If Y ⊂ X are qp vars, the codimension of Y in X
is dim X − dim Y
2
Main Results of this section:
Thm 0. If f : X → Y is finite, then dim X = dim Y
Thm 1. If X ⊂ Y closed subset, then dim X ≤ dim Y .
If Y is irred and dim X = dim Y , then X = Y
Thm 2. (purity) Every irreducible component of a hypersurface in An or Pn has codim 1
Thm 3. (Geometric version of Krull’s Hauptidealsatz)
Suppose X ⊂ An is closed with all components of dimension n − 1. Then X is a hypersurface and IX is
principal
Similarly, there is a projective version of Theorem 3,
which generalized to products of projective spaces
Thm 4. Let X ⊂ Pn1 × · · · × Pnk be closed with all
P
components of dimension ni − 1. Then X is defined by
a single equation, separately homogeneous in each of the
k sets of homogeneous coordinates of the respective Pni
3
Examples. 1. Irred affine plane curves have dimension 1. Hence all plane curves have dim 1
2. If X is a point, then k(X) = k and so dim X = 0.
Therefore the dim of any finite set of points is zero.
Conversely, if dim X = 0, show X is a finite set
WLOG, may assume X ⊂ An
Coord fcns xi ∈ k[X] are algebraic over k
Therefore only finitely many values are possible
3. dim(X × Y ) = dim X + dim Y .
Idea: get a transcendence basis for k(X × Y ) by concatenating the pullbacks to X × Y of transcendence bases for
k(X) and k(Y )
See Shafarevich Example 4 P. 68 for details
4
4. dim G(k, n) = k(n − k)
Exhibit an open subset of G(k, n − k) isomorphic to
Ak(n−k)
Illustrate with G(2, 4) ,→ P5
Denote the Pl¨ucker coords by wij , i < j
Let Uij ⊂ P5 be the affine piece wij 6= 0
Claim: G(2, 4) ∩ Uij ' A4
Just check for U12
Consider φ : A4 → G(2, 4)
(a1, a2, a3, a4) 7→ span(e1 + a1e3 + a2e4, e2 + a3e3 + a4e4)
To compute Pl¨ucker coordinates, write
(e1 + a1e3 + a2e4) ∧ (e2 + a3e3 + a4e4) =
e1∧e2+a3e1∧e3+a4e1∧e4−a1e2∧e3−a2e2∧e4+(a1a4−a2a3)e3∧e4
i.e. (1, a3, a4, −a1, −a2, a1a4 − a2a3) ∈ U12 ⊂ P5
Clearly a regular embedding
Explicit computation in proof of closedness of Pl¨ucker
image shows precisely that U12 ∩ G(2, 4) = φ(A4)
5
Thm 0. If f : X → Y is finite, then dim X = dim Y
Pf. WLOG may assume X and Y are irred and affine
Then k[X] is a finite k[Y ]-module
k(X) is a finite extension of k(Y )
Tr deg(k(X)) = Tr deg(k(Y ))
Thm 1. If X ⊂ Y closed, then dim X ≤ dim Y . If Y
is irred and dim X = dim Y , then X = Y
Pf. WLOG may assume X and Y are irred and affine
X ⊂ Y ⊂ AN , dim Y = n
Any t1, . . . , tn+1 ∈ k[Y ] must be alg dep over k
Same conclusion for restrictions to t1, . . . , tn+1 ∈ k[X]
dim X = Tr deg(k(X)) ≤ n = dim Y
Now suppose dim X = dim Y , Y irred
To show X = Y suffices to show IX ⊂ k[Y ] is the zero
ideal
Let 0 6= u ∈ k[Y ] with u = 0 on X
Pick some coord fcns t1, . . . , tn ∈ k[X] alg indep
(therefore still alg indep viewed in k[Y ])
u is alg over k(t1, . . . , tn)
U = u solves an irred eqn F (U ) = 0, F ∈ k[Y ][U ]
Restrict to X, get contradiction; see text for details
6
Thm 2. (purity) Every irreducible component of a hypersurface in An or Pn has codim 1
Pf. Only need do An.
WLOG X = Z(F ), F irreducible
Claim: X is irred
If X = X1 ∪ X2, take Fi ∈ IXi − IX
Then F1F2 vanishes on X
By Nullstellensatz, (F1F2)k ∈ IX = (F )
F divides F1 or F2, contradiction
WLOG may assume xn occurs in F
Let ti ∈ k[X] restriction of xi to X
Claim: t1, . . . , tn−1 alg indep in k[X]
This implies dim X = n − 1
Suppose G(t1, . . . , tn−1) = 0
Then G(x1, . . . , xn−1) vanishes on X
F divides G
Impossible because F contains xn and G does not
7
Thm 3. (Geometric version of Krull’s Hauptidealsatz)
Suppose X ⊂ An is closed with all components of dimension n − 1. Then X is a hypersurface and IX is
principal
Pf. WLOG may assume X irred
X 6= An since dim X = n − 1
So IX 6= 0. Let 0 6= F ∈ IX
Since X is irred, some factor of F vanishes on X
So may assume F irred
Then X ⊂ Z(F ), Z(F ) proven to be irred
Since dim X = dim Z(F ), conclude X = Z(F ), a hypersurface
To show IX is principal,
recall
r
I(X) = I(Z(F )) = (F )
So if G ∈ IX we have F divides Gk for some k
Therefore F divides G
IX = (F )
8
Hypersurface sections (intersections with a hypersurface)
Let X ⊂ Pn be a proj var
F ∈ k[x0, . . . , xn] homog, not identically zero on X
XF := X ∩ Z(F ) ⊂ X hypersurface section of X
We might think dim XF = dim X − 1, but this is false
Ex. X = Z(x1x2), F = x1
dim XF = dim X = 1
What went “wrong” is that F vanishes identically on a
component of X
This is essentially the only thing that can go wrong
Thm. Let X ⊂ Pn irred, proj, F a homog poly not
identically zero on X. Then dim XF = dim X − 1
Furthermore, if X = ∪Xi is a union of irreducible components, either Xi ⊂ Z(F ) or not
Will see that if Xi ⊂ Z(F ), then dim(Xi ∩ Z(F )) =
dim Xi since they are equal
Otherwise we are in the case of the theorem, and dim((Xi)F ) =
dim Xi − 1
9
Example. In P3 consider
F1 = x0x3 − x1x2, F2 = x0, F3 = x1
Then X1 := Z(F1) = (P3)F1 is a quadric hypersurface
(isomorphic to P1 × P1 by Segre, hence irred)
dim X1 = 2 = dim P3 − 1
Next, consider X2 := (X1)F2 = Z(F1, F2)
Note (x0x3 − x1x2, x0) = (x0, x1x2)
So X2 = L1 ∪ L2 is reducible, the union of the two lines
Li = Z(x0, xi) for i = 1, 2
Consistent with the theorem: dim X2 = 1 = dim X1 − 1
Now, F3 = x1 is not identically zero on X2 so we can
define
X3 = (X2)F3 = Z(F1, F2, F3)
But (x0x3 − x1x2, x0, x1) = (x0, x1), so X3 = L1
dim X3 = dim X2
No contradiction to theorem, since X2 is reducible
Note (L2)F3 = {(0, 0, 0, 1)} and so dim(L2)F3 = 0 =
dim L2 − 1
10
More interesting example: twisted cubic
C = {(s3, s2t, st2, t3)} ⊂ P3
Can check that IC = (x0x3 − x1x2, x0x2 − x21, x1x3 − x22)
First hypersurface section: X1 := Z(x0x3−x1x2), smooth
irred quadric isomorphic to P1 × P1
Second hypersurface section:
X2 = Z(x0x3 − x1x2, x0x2 − x21)
not only contains C, but it also contains the line L =
Z(x0, x1).
In fact, not hard to check X2 = C ∪ L
Third hypersurface section: X3 = C
In fact, is clearly not possible to write IC = (G1, G2) for
any G1, G2
Even though C has codim 2, we need at least 3 equations
Projective varieties X ⊂ Pn of codimension r with IX =
(F1, . . . , Fr ) are called complete intersections
The twisted cubic is not a complete intersection
Complete intersections are relatively easy to analyze, as
they are iteratively understood as hypersurface sections
11
There is a weaker notion of a set-theoretic complete intersection:
X = Z(F1, . . . , Fr ), codim(X) = r
Example: Z(x0x2 − x21, x0) ⊂ P3 is the line L given by
x0 = x1 = 0
However, (x0x2 − x21, x0) = (x0, x21) 6= (x0, x1)
12
The theorem has many easy useful corollaries
Cor. 1. A projective variety X has closed subvarieties
of any dimension s < dim X
Pf. Iterate the process of taking hypersurface sections
and irred comps thereof
Cor. 2. dim X is the maximum number n of irred closed
subsets Xi of X,
X ⊃ X1 ⊃ X2 ⊃ · · · ⊃ Xn ,
all inclusions proper
Cor. 3. Let X ⊂ PN be projective and let s be the
maximal dimension of a linear subspace E ⊂ X which is
disjoint from X. Then dim X = N − s − 1
Pf. Let n = dim X and suppose E is a linear subspace
of dim s ≥ N − n
E can be defined by ≤ n lin eqns so dim(X ∩ E) ≥
n−n=0
i.e. X ∩ E is nonempty
To finish proof, need only exhibit an E of dim N − n − 1
disjoint from X
Get this by applying theorem repeatedly to linear forms
13
Example. Twisted cubic curve C = {(s3, s2t, st2, t3) has
dimension 1:
Any hyperplane ax0 + bx1 + cx2 + dx3 = 0 intersects C
when (s, t) ∈ P1 satisfies as3 + bs2t + cst2 + dt3 = 0
C is disjoint from line x0 = x3 = 0
Cor. 4. Given X ⊂ Pn and F1, . . . , Fr homog, then
dim (X ∩ Z(F1) ∩ . . . ∩ Z(Fr )) ≥ dim X − r
This corollary has an immediate corollary, so powerful
that it is elevated to a Proposition
Prop. Any r ≤ n homogeneous forms on Pn have a
common solution
Example. Any two plane curves have at least one intersection point
In fact, B´ezout’s thm says that the number of intersection
points, counted by multiplicities, is the product of the
degrees of the curves
14
Now let’s prove the theorem:
Thm. Let X ⊂ Pn irred, proj, F a homog poly not
identically zero on X. Then dim XF = dim X − 1
Pf. Choose forms F0 = F, F1, . . . , Fn inductively as follows:
Letting Xi = X ∩ Z(F0, . . . , Fi), choose Fi+1 so as to
not vanish identically on any component of Xi
This implies that each component of Xi+1 is a proper
closed subset of a component of Xi
In particular, dim Xi+1 < dim Xi
Therefore Xn is empty
Then have reg finite map f = (F0, . . . , Fn) : X → Pn
Therefore surjective
If thm is not true, then dim XF ≤ n−2 and consequently
Xn−1 is empty
Therefore f −1(0, 0, . . . , 0, 1) is empty
Contradicts surjectivity of f
15
A stronger result is in fact true
Thm. Every component of XF has dimension dim X −1
Pf. Shafarevich Pp. 74–75
Cor. If X ⊂ Pn is an irred qp var, F a form not
vanishing identically on X, then every irred comp of XF
has codimension 1
Pf. Apply the theorem to X ⊂ Pn
Each component of X F has codim 1
Since Z(F ) is closed and nonempty open subsets of X
are dense in X, we see that
XF = X ∩ Z(F ) = X ∩ Z(F )
Let Z ⊂ XF be an irred comp
Then Z is an irred comp of XF
dim Z = dim Z = dim X − 1
Example. X = A20, F = x0, XF empty, no contradiction
Cor. Let X ⊂ Pn be an irred qp var of dim n and let
F1, . . . , Fr be forms. The every comp of X∩Z(F1, . . . , Fr )
has dim at least n − r
Pf. Induction on r
16
Prop. Let X, Y ⊂ PN be irred qp vars of dims m, n.
Then each comp of X ∩ Y has dim at least m + n − N .
If X and Y are proj, and m + n ≥ N , then X ∩ Y is
nonempty.
Pf. Via the diagonal embedding X ,→ ∆ ⊂ X × X,
identify X ∩ Y with (X × Y ) ∩ ∆.
dim(X × Y ) = n + m
Since dimension is local, may assume X, Y ⊂ AN
∆ ⊂ AN × AN is a (complete) intersection of N hypersurfaces xi = yi, i = 1, . . . , N
If X, Y are proj, construct the affine cones C(X) and
C(Y ) in AN +1
The affine cone C(X) is defined as Z(I(X)), where I(X)
is the homogeneous ideal of X
Z(I(X)) ⊂ AN +1 is understood as the zero set of the
polynomials in (x0, . . . , xN ), now identified as the ordinary (non-homogeneous) coords of AN +1
C(X) and C(Y ) are qp vars in PN +1 of dims m+1, n+1
dim(C(X) ∩ C(Y )) ≥ (m + 1) + (n + 1) − (N + 1) =
m+n−N +1
If m + n ≥ N , then dim C(X) ∩ C(Y ) ≥ 1
C(X) ∩ C(Y ) contains a nonzero point of AN +1
17
The coordinates of this point are then the homog coords
of a point of X ∩ Y
18
We now turn to the dimension of the fibers of a reg map
and apply it to study lines on surfaces
Let X, Y be qp vars and f : X → Y regular
Def. The fiber of f over y ∈ Y is f −1(y)
Example. Consider X ⊂ P2 × P2 with equation
x0y0 + x1y1 + x2y2 = 0.
with projection π : X → P2, π(x, y) = y
The fiber π 1(y), y = (y0, y1, y2), is identified with line
Ly = Z(x0y0 + x1y1 + x2y2 = 0) ⊂ P2
(y0, y1, y2) in the above equation are fixed elts of k
Thus, y ∈ P2 (the dual P2) parametrizes lines in P2
X describes the family of all lines in P2
Since X ⊂ P2 × P2 is a hyperplane section in the Segre
embedding P2 ×P2 ,→ P8, we have dim X = 2+2−1 =
3.
19
Theorem. Let f : X → Y be a reg surjective map,
dim X = n, dim Y = m. Then n ≥ m and
• dim F ≥ n − m for any component F of any fiber of
f
• There is a nonempty open subset U ⊂ Y such that
the fiber over any y ∈ U has dimension exactly n −
m.
In the previous example, the fibers of π : X → P2 are
lines, of dimension 1 = 3 − 2
Continuing with the example above, we have U = Y =
P2 and all fibers are irreducible
Cor. The sets Yk = {y ∈ Y | dim f −1(y) ≥ k} are
closed in Y
Pf of Cor. The theorem says Yn−m = Y and there is a
proper closed set Y 0 = Y − U ⊂ Y such that Yk ⊂ Y 0 if
k >n−m
Let Zi be the irred comps of Y 0, so closed in Y
Consider the restriction
fi : f −1(Zi) → Zi
The result follows by induction on dim Y
20
Thm. Let f : X → Y reg surjective map, X, Y projective. Suppose Y is irred and the fibers of f are all of the
same dimension and irreducible. Then X is irreducible.
The proofs of the last two theorems will be deferred until
after powerful applications are given
Now we show how these results easily imply existence and
non-existence of lines on surfaces!
Thm. For any m > 3, there exist surfaces in P3 of
degree m which do not contain any lines. Moreover, such
surfaces correspond to an open subset of PN .
Thm. Every cubic surface contains at least one line.
There is a nonempty open subset U ⊂ P19 parametrizing
cubic surfaces with only finitely many lines.
Remark. In fact, U contains the open set Usm parametrizing smooth cubic surfaces, and any smooth cubic surface
contains exactly 27 lines.
21
Let S ⊂ P3 be a surface of degree m.
We have seen how to identify S with a point of PN ,
N = m+3
−1
3
Let L ⊂ P3 be a line
Can identify L with a point of G = G(2, 4):
P1s contained in P3 are in 1-1 correspondence with 2dimensional subspaces of k 4
22
Let I be the incidence correspondence
N
I = (L, S) ∈ G × P | L ⊂ S ⊂ G × PN
Lemma. I is a closed subset of G × PN , hence is a
projective variety
Pf. Identify L = span(v, w) with the projectivization of
P
v ∧ w = i<j pij ei ∧ ej in terms of Pl¨ucker coords pij
Put pij = −pji for i > j and put pii = 0
Equivalently, pij = xiyj − xj yi, where v = (x0, . . . , x3)
and w = (y0, . . . , y3)
Claim. L consists of all points (z0, . . . , z3) with zj =
P
i ai pij as the ai range over k
For example, consider z = yk v − xk w ∈ L
Then zj = yk xj − xk yj = −pkj
which is of claimed form, ak = −1 and ai = 0 for i 6= k
The general case is obtained from these by linear combinations
23
Now let F be a form of deg m defining S
P
Then L ⊂ S iff F ( i aipij ) = 0 as in identity in ai
This is a homog poly eqn in the ai whose coeffs are homog
polys in the pij and linear in the coeffs of F (the homog
coords on PN )
I is obtained by setting all of these coeffs to zero
Resulting equations are bihomogeneous in the Pl¨ucker
coords and the homog coords of PN , so I closed
Lemma. I is irreducible of dimension N + 3 − m
Pf. Let π : I → G be the projection.
π is surjective. The fiber over L ∈ G is the set of hypersurfaces of deg m which containing L
Letting L = Z(`1, `2) with `1, `2 linear forms, we let
G, H be any homog forms of deg m − 1
Then F = `1G + `2H defines a hyp of deg m containing
L
24
Claim. The fibers of π are all projective spaces of dim
N −m−1
The lemma follows immediately from this claim
Since the F as above form a vector space, the fibers are
projective space and we just have to compute the dimension
G, H are chosen from a vector space of dim m+2
3
But we have overcounted, since forms `1`2K with deg K =
m − 2 can be written in either the form `1G or `2H
So the set of all F forms a vector space of dimension
m + 2 m + 1 m + 3
−m−1 = N −m
=
−
2
3
3
3
and the claim follows
Thm. For any m > 3, there exist surfaces in P3 of
degree m which do not contain any lines. Moreover, such
surfaces correspond to an open subset of PN .
Pf. Consider the projection φ : I → PN
Since I is projective, the image of φ is closed and has dim
≤ dim I = N + 3 − m < N
φ is not surjective
25
Thm. Every cubic surface contains at least one line.
There is a nonempty open subset U ⊂ P19 parametrizing
cubic surfaces with only finitely many lines.
Pf. Repeating the above argument, we first have to show
that φ is surj
For m = 3 have dim I = N
If φ not surj, the image has dim < N , so all fibers have
positive dim
Get a contradication by exhibiting just one zero-dimensional
fiber
Ex: x1x2x3 = x30 contains exactly 3 lines
No lines intersecting affine piece A30:
x1x2x3 = 1 has no nontrivial linear parametric solns xi =
ai t + bi
Hyperplane at infinity x0 = 0 has 3 lines x0 = xi = 0,
i = 1, 2, 3
So φ is surjective
There is a U ⊂ PN = P19 over which all fibers are
zero-dimensional, hence finite
26
We now return to prove the results that the analysis of
lines on surfaces was based on
Theorem. Let f : X → Y be a reg surjective map,
dim X = n, dim Y = m. Then n ≥ m and
1. dim F ≥ n − m for any component F of any fiber of
f
2. There is a nonempty open subset U ⊂ Y such that
the fiber over any y ∈ U has dimension exactly n −
m.
Pf. (i) This statement is local
WLOG may assume Y affine, Y ⊂ AN
We showed earlier in affine space that we can take a succession of hyperplane sections, each decreasing the dimension by 1
Starting with Y , have g1 . . . , gm ∈ k[Y ] s.t. dim Z(g1, . . . , gm) =
m−m=0
Can further choose each gi with gi(y) = 0
Z(g1, . . . , gm) is a finite set containing y
By removing some points from Y , can assume Z(g1, . . . , gm) =
{y}
Then f −1(y) is defined by m eqns f ∗(g1), . . . , f ∗(gm) ∈
k[X]
So each component of f −1(y) has dim ≥ n − m
27
(ii) (Sketch) First, reduce to the affine case:
Take W ⊂ Y affine, then choose V ⊂ f −1(W ) ⊂ X
affine
Consider f : V → W
By surjectivity, k[W ] → k[V ] is injective
View k[W ] ⊂ k[V ], k(W ) ⊂ k(V )
Tr degk(W )(k(V )) = n − m
Choose generators vi for k[V ] with v1, . . . , vn−m alg indep
over k(W )
Remaining generators vi = vn−m+1, . . . are all alg dependent
Satisfy polynomial eqns
Fi(v1, . . . , vn−m, vi) = 0
with coefficients in k[W ].
In particular, have leading coefficients hi ∈ k[W ]
Show U = W − Z((hn−m+1, . . .)) satisfies the conclusion
28
Thm. Let f : X → Y reg surjective map, X, Y projective. Suppose Y is irred and the fibers of f are all of the
same dimension and irreducible. Then X is irreducible.
Pf. Express X = ∪Xi as the union of its irred comps
Xi is projective, so f (Xi) is closed and Y = ∪f (Xi)
Y is irred, so Y = f (Xi) for some i
For each i, let fi : Xi → Y be the restriction of f
If f (Xi) = Y then fi still surj
By theorem just proven have nonempty Ui ⊂ Y and
integer ni s.t. dim fi−1(y) = ni all y ∈ Ui
If f (Xi) 6= Y put Ui = Y − f (Xi)
Pick y ∈ ∩Ui
f −1(y) is irred by assumption
So f −1(y) contained in some Xi
Then f −1(y) = fi−1(y) for that value of i
n = ni
Now fi−1(y) ⊂ f −1(y) for any y ∈ Y
dim fi−1(y) ≥ ni = n by previous thm
dim f −1(y) = n
Therefore fi−1(y) = f −1(y)
So X = Xi, and X is irreducible
29
© Copyright 2024 ExpyDoc
