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THE MOVING CURVE IDEAL AND THE REES
ALGEBRA
DAVID A. COX
Abstract. These are notes for a lecture given at Ohio University
on June 3, 2006. An important topic in commutative algebra is
the Rees algebra of an ideal in a commutative ring. The Rees algebra encodes a lot of information about the ideal and corresponds
geometrically to a blow-up. One can represent the Rees algebra
as the quotient of a polynomial ring by an ideal. This ideal is
generated by the defining relations of the Rees algebra. The surprise is that these defining relations were discovered by computer
scientists working in computer-aided geometric design, where the
ideal is called the moving curve ideal. A second surprise will be
the appearance of adjoint curves.
1. The Rees Algebra
Let R be a Noetherian commutative ring. Given an ideal I ⊂ R and
an R-module M, we can build three graded rings:
• The Rees algebra R(I) = R ⊕ I ⊕ I 2 ⊕ · · ·
• The associated graded grI (R) = R/I ⊕ I/I 2 ⊕ I 2 /I 3 ⊕ · · ·
• The symmetric algebra SymR (M) = R ⊕ M ⊕ Sym2 (M) ⊕ · · ·
These are related as follows:
- grI (R) = R(I) ⊗R R/I.
- There are natural surjections
Φ
SymR (I) −→ R(I)
χ
SymR/I (I/I 2 ) −→ grI (R)
- Proj(R(I)) is the blow-up of V(I) ⊂ Spec(R).
Here are two basic results:
Theorem Φ is an isomorphism ⇐⇒ χ is an isomorphism.
Theorem If R is local and grI (R) is Cohen-Macaulay, then R(I) is
Cohen-Macaulay.
Example 1.1 R is a regular local ring and I ⊂ R is generated by
a regular sequence f1 , . . . , fr . Then one can prove that we have a
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DAVID A. COX
commutative diagram
(R/I)[x1 , . . . , xr ]
−−−−−−−−→
&
grI (R)
%
SymR/I (I/I 2 )
where all of the maps are isomorphisms. Since χ is the arrow % in the
diagram, it follows that
(1) Φ : SymR (I) → R(I) is an isomorphism.
(2) R(I) is Cohen-Macaulay.
The Defining Equations of R(I)
Let I = hf1 , . . . , fr i be an arbitrary ideal of R. Then xi 7→ fi gives
a surjective R-algebra homomorphism
R[x1 , . . . , xr ] −→ R(I).
This map is graded, so its kernel K is homogeneous. The generators
of K are the defining equations of the Rees algebra.
Example 1.2 Let K1 be the graded piece of K in degree 1. Then
K1 = {a1 x1 + · · · + ar xr | ai ∈ R, a1 f1 + · · · + ar fr = 0}
' Syz(f1 , . . . , fr ).
Using hK1 i ⊂ K, one can prove that there is a commutative diagram
R[x1 , . . . , xr ]/hK1 i −→ R[x1 , . . . , xr ]/K
↓
↓
Φ
SymR (I)
−→
R(I)
where the vertical maps are isomorphisms.
Example 1.3 Let R = k[s, t], k a field, and I = hs2 , t2 i. We claim
that
R(I) ' R[x, y]ht2 x − s2 yi.
To see why, note that s2 , t2 form a regular sequence. This implies:
(1) Syz(s2 , t2 ) is generated by the syzygy (t2 )s2 +(−s2 )t2 = 0. Thus
hK1 i = ht2 x − s2 yi.
(2) SymR (I) ' R(I) by the graded version of Example 1. By
Example 2, we conclude that K = hK1 i.
Example 1.4 Let R = k[s, t] and I = hs2 , st, t2 i. We claim that
R(I) ' R[x, y, z]htx − sy, ty − sz, xz − y 2i.
We will prove this two ways.
THE MOVING CURVE IDEAL AND THE REES ALGEBRA
3
First proof of the claim. Let e be a new variable and consider
J = hx − s2 e, y − ste, z − t2 ei ⊂ R[x, y, z, e].
The map (x, y, z, e) 7→ (s2 e, ste, t2 e, e) gives exact sequences
0 → J
⊂ R[x, y, z, e] → R[e] → 0
↑
↑
0 → K ⊂ R[x, y, z] → R(I) → 0
where the first vertical arrow is inclusion and the second the is natural
injection
n
∞
n n
R(I) = ⊕∞
n=0 I ' ⊕n=0 I e ⊂ R[e].
It is an easy exercise to prove that
K = J ∩ R[x, y, z].
Since J is given explicitly, we can compute K on a computer using
standard methods in elimination theory.
Second proof of the claim. We begin with the Hilbert-Burch resolution
A
B
0 −→ R(−3)2 −→ R(−2)3 −→ I −→ 0,
where


t
0
t  , B = (s2 , st, t2 ).
A = −s
0 −s
The Hilbert-Burch Theorem tells us that
I2 (A) = ht2 , −st, s2 i = I,
where I2 means that we take the ideal generated by 2 × 2 minors. In
this case, the 1 × 1 minors give the ideal
I1 (A) = hs, ti.
The above resolution implies that Syz(s2 , st, t2 ) is generated by the
columns of A. This shows that K1 is generated by
p = tx − sy
q = ty − sz.
If we express p and q in terms of I1 (A) = hs, ti, we get
p = (−y)s + (x)t
q = (−z)s + (y)t.
Taking the determinant of the coefficients gives
−y x
det
= −y 2 + xz.
−z y
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DAVID A. COX
This is easily seen to lie in K2 . Since I1 (A) = hs, ti is generated by
a regular sequence, a 1996 PAMS paper of Morey and Ulrich implies
that K is generated by K1 and the above element of K2 .
2. The Moving Curve Ideal
Let a, b, c ∈ R = R[s, t] be homogeneous polynomials of degree n
with gcd(a, b, c) = 1. The map
(s, t) 7−→ a(s, t), b(s, t), c(s, t)
gives a curve parametrization P1 → P2 . To view this map in concrete
terms, set t = 1 and note that
b(s,1)
a(s, 1), b(s, 1), c(s, 1) ∼ a(s,1)
,
,
1
.
c(s,1) c(s,1)
Thus we have R 99K R2 given by s 7→ a(s,1)
, b(s,1) (the dash in the
c(s,1) c(s,1)
arrow indicates that this map may fail to be defined on all of R because
of the denominator).
A moving curve of degree m is given by a polynomial
X
Aijk (s, t)xi y j z k , Aijk (s, t) ∈ R.
i+j+k=m
This follows the parametrization given by a, b, c if
X
Aijk (s, t)a(s, t)ib(s, t)j c(s, t)k ≡ 0.
i+j+k=m
This says that the point on the parametrized curve always lies on the
corresponding moving curve.
Example 2.1 A moving line Ax+By +Cz follows the parametrization
⇐⇒ Aa + Bb + Cc = 0 ⇐⇒ (A, B, C) ∈ Syz(a, b, c). Note that a, b, c
generate I.
Example 2.2 A moving conic Ax2 + Bxy + Cxz + Dy 2 + Eyz + F z 2
follows the parametrization ⇐⇒ Aa2 + Bab + Cac + Db2 + Ebc +
F c2 = 0 ⇐⇒ (A, B, C, D, E, F ) ∈ Syz(a2 , ab, ac, b2 , bc, c2 ). Note that
a2 , ab, ac, b2 , bc, c2 generate I 2 .
The set of all moving curves that follow the parametrization is an
ideal in R[x, y, z] = C[s, t, x, y, z]. This is the moving curve ideal, denoted MC. I want to emphasize that this ideal was discovered by the
geometric modeling community.
Amazing Fact The moving curve ideal is precisely the ideal of relations defining the Rees algebra R(I).
THE MOVING CURVE IDEAL AND THE REES ALGEBRA
5
Example 2.3 x = s2 , y = st, z = t2 gives a parametrization of the
conic xz −y 2 = 0. Affinely, this is the parabola x = y 2 parametrized by
x = s2 , y = s. Then K1 = MC1 is generated by the moving lines tx−sy
and ty − sz, which give the affine lines x − sy = 0 and y − s = 0. This
makes it easy to draw the picture of how these moving lines determine
the parametrization.
By Example 1.4, we know that K = MC is generated by the moving
lines tx − sy, ty − sz plus xz − y 2 . Note that xz − y 2 = 0 is the implicit
equation of the conic.
To analyze the general case, recall that the ideal I = ha, b, ci ⊂ R
has a Hilbert-Burch resolution
0 → R(−n − µ) ⊕ R(−2n + µ) → R3 (−n) → I → 0.
In other words, the syzygy module Syz(a, b, c) is free with generators
p, q of degree µ and n − µ. We assume µ ≤ n − µ, and we can regard
p, q as moving lines
p = Ax + By + Cz
q = Ex + F y + Gz.
On easily sees that p, q are the degree 1 generators of the moving curve
ideal MC.
Question What can we say about the other generators of K = MC?
Here are two examples with n = 4 to illustrate what can happen.
Example 2.4 (“Example 1” on the transparencies) Consider
a = 6s2 t2 − 4t4 , b = 4s3 t − 4st3 , z = s4 .
This has µ = 2, which is the generic value of µ when n = 4. The
moving curve ideal has five generators:
• Two moving lines of degree 2 in s, t:
p = st x + ( 21 s2 − t2 )y − 2st z
q = s2 x − st y − 2t2 z.
• Two moving conics of degree 1 in s, t:
s xy − t y 2 − 2t xz − s yz + 4t z 2
s x2 − t xy + 21 s y 2 − 2s xz + t yz.
• The implicit equation:
F = y 4 + 4x3 z + 2xy 2 z − 16x2 z 2 − 6y 2z 2 + 16xz 3 .
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DAVID A. COX
This quartic curve has three nodes:
2
1
0
0
0.5
1
2
1.5
-1
-2
Example 2.5 (“Example 2” on the transparencies) Consider
a = 3s3 t − 3s2 t2 , b = 3s2 t2 − 3st3 , z = (s2 + t2 )2 .
This has µ = 1, which is rather special when n = 4. The moving curve
ideal has five generators:
• Two moving lines of degree 1,3 in s, t:
p = tx− sy
q = s3 x + (2s2 t + t3 )y + (3st2 − 3st2 )z.
• One moving conic of degree 2 in s, t:
s2 x2 + (2s2 + t2 )y 2 + (3st − 3s2 )yz.
• One moving cubic of degree 1 in s, t:
s x3 + 2s xy 2 + t y 3 − 3s xyz + 3s y 2z.
• The implicit equation:
F = x4 + 2x2 y 2 + y 4 − 3x2 yz + 3xyz 2 .
This quartic curve has a triple point:
2
1.5
1
0.5
0
-2
-1.5
-1
-0.5
0
0.5
-0.5
-1
1
THE MOVING CURVE IDEAL AND THE REES ALGEBRA
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Theorem A curve parametrized by a, b, c of degree n = 4 has a triple
point ⇐⇒ µ = 1.
The above computations of K = MC were done by computer. To
find a theoretical explanation for what’s happening, we will adapt the
determinantal method used in Example 1.4.
Let us illustrate the basic idea using Example 2.4. Recall that we
have the moving lines
p = st x + ( 12 s2 − t2 ) y − 2st z
q = s2 x − st y − 2t2 z,
which we write as
p = 12 y s2 + (x − 2z) st − y t2
q = x s2 − y st − 2z t2 .
Now proceed as follows:
• Express p, q in terms of s, t2 to obtain
p = (tx + 21 sy − 2tz)s + (−y)t2
q = (sx − ty)s + (−2z)t2 .
Taking the determinant gives
tx + 21 sy − 2tz −y
.
det
sx − ty
−2z
This is the first moving conic generator of MC!
• Express p, q in terms of s2 , t to obtain the second moving conic
generator!
• Express the two moving conics in terms of s, t gives
(xy − yz)s + (4z 2 − y 2 − 2xz)t
(x2 + 21 y 2 − 2xz)s + (yz − xy)t.
Taking the determinant gives
xy − yz
4z 2 − y 2 − 2xz
det
.
x2 + 12 y 2 − 2xz
yz − xy
This is the implicit equation!
Conjecture For any a, b, c ∈ k[s, t] with n = 4, µ = 2 that gives
a proper parametrization, the above procedure computes the minimal
generators of the moving curve ideal K = MC.
By a “proper parametrization”, we mean a parametrization that is
generically one-to-one.
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DAVID A. COX
Here is some evidence for the conjecture:
• (Geometric Modeling) In 1997, Sederberg, Goldman and Du
proved that the determinant constructed from the moving conic
generators gives the implicit equation when n = 4 and µ = 2.
• (Commutative Algebra) In 1997, Jouanolou proved that if p and
q are forms of degree 2 in s, t whose coefficients are variables,
then the above procedure computes the minimal generators of
the saturation of hp, qi with respect to hs, ti (which coincides
with K = MC in our situation).
3. Adjoint Curves
Let C ⊂ P2 is defined by an irreducible equation of degree n, say
F (x, y, z) = 0. Here are two facts from the classical theory of plane
curves:
• The genus of C is
1X
1
νp (νp − 1),
g = (n − 1)(n − 2) −
2
2 p
where the sum is over all singular points p of C (including infinitely near points) and νp is the multiplicity of C at p.
• g = 0 ⇐⇒ C is birationally equivalent to P1 ⇐⇒ C can be
parametrized by some a, b, c.
Example 3.1 Consider the quartic curve of Example 2.4:
2
1
0
0
0.5
1
1.5
2
-1
-2
The three nodes have νp = 2, so that the genus is
1
1
g = 3 · 2 − (2 · 1 + 2 · 1 + 2 · 1) = 0.
2
2
THE MOVING CURVE IDEAL AND THE REES ALGEBRA
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Example 3.1 Consider the quartic curve of Example 2.5:
2
1.5
1
0.5
0
-2
-1.5
-1
-0.5
0
0.5
1
-0.5
-1
The triple point has νp = 3, so that the genus is
1
1
g = 3 · 2 − 3 · 2 = 0.
2
2
Question Given a curve with g = 0, how do we find a parametrization?
The classical answer uses adjoint curves. For simplicity, consider an
irreducible curve C of degree n and genus zero that has only nodes.
Then the number of nodes must equal 12 (n − 1)(n − 2).
Lemma Fix n − 3 smooth points on C. Then there is a one-parameter
family of curves that go through both the singular points of C and the
n − 3 chosen points.
Sketch of Proof A curve of degree n−2 has n2 coefficients. We want
such a curve to vanish at
1
n
(n − 1)(n − 2) + n − 3 =
−2
2
2
points. By linear algebra, it follows that there are at least two linearly independent curves G1 , G2 of degree n − 2 that satisfy the desired
condition. Then
G = sG1 + tG2
gives the desired one-parameter family. QED
We call G = sG1 + tG2 a one-parameter family because multiplying
G by a constant factor gives the same curve. Hence we can think of
G = sG1 + tG2 as giving a family of curves parametrized by (s, t) ∈ P1 .
We call G = sG1 + tG2 an adjoint linear system.
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DAVID A. COX
Now suppose that we have an adjoint linear system G = sG1 + tG2
as in the lemma. Where does G = 0 meet the curve C? By Bezout’s
Theorem, there must be
n(n − 2) = n2 − 2n
points of intersection. However, G = 0 meets C at n − 3 smooth
points and 21 (n − 1)(n − 2) nodes. Each of the latter has an intersection
multiplicity of at least 2, so that we get
1
n − 3 + 2 · (n − 1)(n − 2) = n2 − 2n − 1
2
points of intersection. This means there is one further point of intersection. This point moves as we vary (s, t) ∈ P1 and gives the desired
parametrization!
Let us see how this works in an example.
Example 3.3 Consider the affine curve C defined by
F (x, y) = y 4 + 4x3 + 2xy 2 − 16x2 − 6y 2 + 16x = 0.
This is the curve of Example 2.4, which has three nodes. If we fix
n − 3 = 1 further point, say the origin, then the lemma says that we
can find an adjoint linear system of affine conics (n − 2 = 2). Such a
linear system is given by
Gs,t (x, y) = sx2 − txy + 21 sy 2 − 2sx + ty = 0.
3
Here is a picture for (s, t) = (1, 10
):
2
1
0
0
0.5
1
1.5
2
-1
-2
These conics go through the singular points of the quartic curve and
the origin, plus one more point that moves.
THE MOVING CURVE IDEAL AND THE REES ALGEBRA
11
This enables us to parametrize F = 0 as follows. Compute the
resultants
• Res(F, G, y) = x(x − 1)4 (x − 2)2 (s4 x − 6t2 s2 + 4t4 )
• Res(F, G, x) = y 3(y 2 − 2)2 (s3 y − 4ts2 + 4t3 )
The constant factors show that G = 0 goes through the origin and the
singular points of F = 0. The other factors give
6s2 t2 − 4t4
4s3 t − 4st3
x=
,
y
=
.
s4
s4
This is our original parametrization of F = 0! Furthermore, Gs,t is
one of the moving conics given in Example 2.4. So this adjoint linear
system is one of the defining equations of the Rees algebra!
The general definition of adjoint curve is as follows.
Definition A curve D, possibly reducible, of degree m is adjoint to C
if at all infinitely near singular points p of C with multiplicity νp , the
curve D has multiplicity at least νp − 1.
The curve defined by G = sG1 + tG2 as in the above lemma satisfies
this definition since the curve in lemma had only nodes. This means
νp = 2, so that νp − 1 = 1. Since G = 0 goes through each node, it has
multiplicty at least 1 there, as required by the definition.
Example 3.4 For an example with νp > 2, consider the quartic curve
of Example 2.5. It has a triple point at the origin. We also saw in
Example 2.5 that the moving curve ideal contains a moving cubic of
degree 1 in s, t, which in affine coordinates is
Gs,t = s x3 + 2s xy 2 + t y 3 − 3s xy + 3s y 2z.
This has double point at the origin, so that Gs,t gives a linear system
of adjoint curves.
Conjecture Any moving curve in the moving curve ideal of degree one
in s, t and degree m ∈ {n − 1, n − 2} in x, y, z gives an adjoint linear
system on the rational curve defined by the implicit equation.
This conjecture is based on a observation of Rafael Sendra. If true,
this conjecture would imply that the defining equations of the Rees
algebra of a parametrized curve are intimately related to the geometry
of the curve.
Department of Mathematics and Computer Science, Amherst College, Amherst, MA 01002