Chapter 3 Elimination Theory §3. Implicitization In this section we ...

§3. Implicitization
Chapter 3 Elimination Theory
In this section we undertake to give a solution to the following problem:
(1.0.2)
The Implicitization Problem: Let Vp be a subset of k n given parametrically as
x1 = f1(t1, . . . , tm),
.
xn = fn(t1, . . . , tm).
If the fi are polynomials (or rational functions) in the variables tj, Vp will be (part of) an
affine variety V . The problem is to find a system of polynomial equations in the xi that
define the minimal variety V which contains Vp, i.e. for which Vp ⊂ V .
Polynomial Parametrizations
Example 2.8.0.5. Consider the tangent surface of the twisted cubic in R3 . The twisted cubic in R3 has parametric form R(t) = (t, t2 , t3 ). Its tangent at t is •
R(t) = (1, 2t, 3t3 ); so the tangent surface of the
twisted cubic has parametrization (t, u) ↦→ R(t) + u •
R(t). Componentwise this surface is parametrized by
x = t + u,
y = t 2 + 2tu,
z = t 3 + 3t 2 u.
To eliminate t, u compute a Gröbner basis G for the associated ideal relative to lex order with the variables
ordered by t > u > x > y > z. Using Mathematica 3.0 the result is
G = {−3 x 2 y 2 + 4 y 3 + 4 x 3 z − 6 x y z + z 2 ,
2 u y 3 + x y 3 − 4 x 2 y z + 5 y 2 z − 2 u z 2 − 2 x z 2 ,
− 2 u y 2 − x y 2 + 2 u x z + 2 x 2 z − y z,
u x y − x 2 y + 2 y 2 − u z − x z,
2 u x 2 − 2 x 3 − 2 u y + 3 x y − z,
u 2 − x 2 + y, t + u − x}.
Only one of these Gröbner basis elements, the first one listed which we refer to as g1, doesn’t involve t or u.
Setting it equal to zero gives
(2.8.0.6) −3 x 2 y 2 + 4 y 3 + 4 x 3 z − 6 x y z + z 2 = 0
which is the equation of a surface containing the tangent surface to the twisted cubic. Of course as far as we
know at present there may be points on (2.8.0.6) which aren’t on the tangent surface to the twisted cubic,
and we do not know that (2.8.0.6) describes the smallest variety which contains the twisted cubic.
The Basic Maps. Starting with the system (1.0.2) above, let t = (t1, . . . , tm) and
f(t) = (f1(t), . . . , fn(t)). let i: k m → k n+m be the map t ↦→ (t, f(t)) and let πm : k n+m → k n be the map
(a1, . . . , an+m) ↦→ (am+1, . . . , an+m). F : k m → k n is the map F = πm ◦ i or t ↦→ f(t). The coordinate
functions on k m+n will be denoted, respectively, by t1, . . . , tm, x1, . . . , xn. At the risk of some ambiguity we
will denote the function (t1, . . . , tm, x1, . . . , xn) ↦→ fi(t1, . . . , tm) by fi also, leaving it to the reader to infer
the interpretation from the context. In a slight extension of the preceding notation x = (x1, . . . , , xn) and
x − f(t) will refer either (i) to the apparent n-vector valued function on k m+n or without ambiguity to (ii)
the set of functions {x1 − f1(t1, . . . , tm), . . . , xn − fn(t1, . . . , tm)} on k m+n . When the fi’s are polynomials
I = 〈x − f(t)〉 will thus denote the ideal in k[t1, . . . , tm, x1, . . . , xn] = k[t, x] generated by the polynomials
xi − fi(t), 1 ≤ i ≤ n.
1

Elementary Observation. i(k m ) = V(I).
Proof. (⊂): Suppose t ∈ k m , then i(t) = (t, f(t)). Call this point q ∈ k m+n . Then x(q) = f(t) and
f(q) = f(t) also (when the fi’s are regarded as functions on k m+n . Thus x(q) − f(q) = 0 and q ∈ V(I).
(⊃): If (t, y) ∈ V(I), then for each 1 ≤ i ≤ n, 0 = (xi − fi)(t, y) = yi − fi(t); so y = f(t) and our point
has the form (t, f(t)) which puts it in i(k m ).
One consequence of this observation is that i(k m ) is an affine variety. We will denote it by V = i(k m ).
Our goal is to describe the smallest affine variety in k n which contains πm(V ) = F (k m ).
Theorem 3.3.1 (Polynomial Implicitization). If k is an infinite field, let F : k m → k n be defined
as above where we suppose that fi ∈ k[t1, . . . , tm]. Let I be the ideal I = 〈x1 − f1, . . . , xn − fn〉 ⊂
k[t1, . . . , tm, x1, . . . , xn] and let Im = I ∩ k[x1, . . . , xn] be the m-th elimination ideal. Then V(Im) is the
smallest variety in k n containing F (k m ).
Proof. Since V(g1, . . . , gr) = V(g1) ∩ · · · ∩ V(gr) when g1, . . . , gr ∈ k[x1, . . . , xn], our conclusion is
equivalent to the assertion: “If a polynomial g ∈ k[x1, . . . , xn] vanishes on F (k m ), then it vanishes on
V(Im).” This is just statement (i) of the Closure Theorem when k is algebraically closed and we have
deferred its proof to later. Taking it as known when k is algebraically closed, we must prove it in general
whenever k is an infinite field.
To do this we need some preparation. Let K be an algebraically closed extension field of k. Start with
{x − f(t)} and using these polynomials, whose coefficients are in k, and a monomial ordering in which the
ti’s come before the xj’s, compute a Gröbner basis G for I. G is unaffected by whether the computations are
carried out using k as the field or using K because computing a Gröbner basis will never take us out of the field
of coefficients of the starting ideal basis. We will use Ĩ and Ĩm to indicate the ideal in K[t1, . . . , tm, x1, . . . , xn]
generated by the polynomials {x−f(t)} and the corresponding m-th elimination ideal of Ĩ, respectively. Then
a basis for Im consists of G∩k[x1, . . . , xn] and a basis for Ĩm consists of G∩K[x1, . . . , xn]. Which is, of course,
the same as G ∩ k[x1, . . . , xn]. Nowhere does it make any difference whether the computations are carried
out over k or over K. This will produce polynomials b1(x), . . . , bs(x) in k[x1, . . . , xn] with the property that
( Ĩm) Ĩm = { ˜ ψ ∈ K[x]: ˜ ψ(x) = ˜ϕ1(t, x)b1(x) + · · · + ˜ϕs(t, x)bs(x), ˜ϕi(t, x) ∈ K[t, x], 1 ≤ i ≤ s}
and
(Im) Im = {ψ ∈ k[x]: ψ(x) = ϕ1(t, x)b1(x) + · · · + ϕs(t, x)bs(x), ϕi(t, x) ∈ k[t, x], 1 ≤ i ≤ s}.
The only restriction on the ˜ϕi(t, x)’s and the ϕj(t, x)’s is that the variables t1, . . . , tm cancel out in the
expressions for ˜ ψ and ψ above. This is easily seen to be equivalent to saying
( Ĩm) Ĩm = { ˜ ψ ∈ K[x]: ˜ ψ(x) = ˜ h1(x)b1(x) + · · · + ˜ hs(x)bs(x), ˜ hi(x) ∈ K[x], 1 ≤ i ≤ s}
and
(Im) Im = {ψ ∈ k[x]: ψ(x) = h1(x)b1(x) + · · · + hs(x)bs(x), hi(x) ∈ k[x], 1 ≤ i ≤ s}.
Using the fact that these hi’s and ˜ hj’s can be chosen rather arbitrarily, we have established that
and
a point q ∈ K n is in V( Ĩm) iff bi(q) = 0, 1 ≤ i ≤ s
a point p ∈ k n is in V(Im) iff bi(p) = 0, 1 ≤ i ≤ s.
2

Now to the proof proper. When g ∈ k[x1, . . . , xn] each line below implies the line which follows it.
g vanishes on F (k m );
g ◦ F vanishes on k m ;
g ◦ F is the zero polynomial ;
g ◦ F vanishes on K m ;
g vanishes on F (K m ); this implies the next line by Statement (i) of the Closure Theorem;
g vanishes on {q ∈ K n : Ĩm(q) = {0}};
g vanishes on {q ∈ K n : b1(q) = 0, . . . , bs(q) = 0};
g vanishes on {p ∈ k n : b1(p) = 0, . . . , bs(p) = 0};
g vanishes on {p ∈ k n : Im(p) = {0}};
V(Im) ⊂ V(g). where the varieties are in k n .
The implicitization problem is then solved by the following recipe: Form the ideal
I = 〈xi − fi(t1, . . . , tm): 1 ≤ i ≤ n〉 and choose a Gröbner basis G with respect to a monomial order in
which the t’s precede the x’s.. The polynomials in G which don’t involve t1, . . . , tm, say Gm, then form a
basis for the ideal Im, and the minimal variety containing the range F (k m ) = {f(t): t ∈ k m } is the variety
V(Im) = V(Gm).
This was precisely the procedure followed to get (2.8.0.6), and it shows that the minimal variety in R 3
containing the tangent surface to the twisted cubic is the locus of points (x, y, z) satisfying g1(x, y, z) = 0 or
(2.8.0.6) −3 x 2 y 2 + 4 y 3 + 4 x 3 z − 6 x y z + z 2 = 0.
We still don’t know, however, if the twisted cubic actually fills up this locus. To answer this question we must
see whether each partial solution (x, y, z) ∈ V(g), where g is the polynomial on the lefthand side of (2.8.0.6),
extends to (t, u, x, y, z) ∈ V(I) which is the graph of the tangent space to the twisted cubic. We work first
over C, so that we can use the Extension Theorem. As usual we will extend by adding one component at a
time.
Suppose (x, y, z) ∈ V(I2) which is V(g1). If the variables are ordered t > u > x > y > z, I2 = I ∩k[x, y, z]
is the first elimination ideal of I1 = I ∩ k[u, x, y, z]. We also know from the Elimination Theorem that I1 is
the ideal in C[u, x, y, z] generated by the basis
G = {−3 x 2 y 2 + 4 y 3 + 4 x 3 z − 6 x y z + z 2 ,
2 u y 3 + x y 3 − 4 x 2 y z + 5 y 2 z − 2 u z 2 − 2 x z 2 ,
− 2 u y 2 − x y 2 + 2 u x z + 2 x 2 z − y z,
u x y − x 2 y + 2 y 2 − u z − x z,
2 u x 2 − 2 x 3 − 2 u y + 3 x y − z,
u 2 − x 2 + y}.
The coefficients of the leading terms in u are
− 3 x 2 y 2 + 4 y 3 + 4 x 3 z − 6 x y z + z 2 ,
2y 3 − 2z 2 ,
2xz − 2y 2 ,
xy − z,
2x 2 − 2y,
1.
These are the gi’s in the Extension Theorem. Since the variety they generate is ∅, every partial solution in
I2 extends to a partial solution (u, x, y, z) ∈ V(I1). Further, if (x, y, z) ∈ R 3 , then (u, x, y, z) ∈ R 4 as is easily
seen by noticing that the second polynomial in G above only involves u to the first power.
3

Let’s see if we can extend another step from V(I1) to V(I). Here
G = {−3 x 2 y 2 + 4 y 3 + 4 x 3 z − 6 x y z + z 2 ,
2 u y 3 + x y 3 − 4 x 2 y z + 5 y 2 z − 2 u z 2 − 2 x z 2 ,
− 2 u y 2 − x y 2 + 2 u x z + 2 x 2 z − y z,
u x y − x 2 y + 2 y 2 − u z − x z,
2 u x 2 − 2 x 3 − 2 u y + 3 x y − z,
u 2 − x 2 + y,
t + u − x}.
forms a basis and the last of these basis elements has a coefficient of the highest power of t which is 1 and
thus never vanishes. Again the Extension Theorem (or Corollary 3.0.4) guarantees that any partial solution
(u, x, y, z) ∈ V(I1) extends to (t, u, x, y, z) ∈ V(I). Furthermore, (t, u, x, y, z) ∈ R 5 if (u, x, y, z) ∈ R 4 , seen
here by looking at the polynomial t + u − x. Thus F (k 5 ) = V(I2) and the parametrization fills up the variety.
Rational Parametrizations
Example 3.3.2 Consider the following rational parametrization:
(3.3.3) (u, v) F
2 u v2
↦→ , , u .
v u
Treating this as in the polynomial case we would introduce the coordinate variables (x, y, z) and write this
in the form
Then we would consider the two maps
and
i
x = u2
v ,
y = v2
u ,
z = u.
k 2
−→ k 5
(u, v) ↦→ u, v, u2
v2
, , u
v u
k 5
π2
−→ k 3
(u, v, x, y, z) ↦→ (x, y, z),
whose composition is the map F of (3.3.3). The variety V ⊂ k5 should be something like
V ′
x − u2
v2
v , y − u , z − u , but since the relations defining V are not then polynomial relations we would
modify them by clearing denominators to V vx − u2 , uy − v2 , z − u . If we set I = 〈vx−u 2 , uy −v 2 , z −u〉,
a Gröbner basis for I computed relative to lex order with u > v > x > y > z is
G = {x 2 y z − z 4 , −x y z + v z 2 , v x − z 2 , v 2 − y z, u − z}.
To eliminate u, v we look at the second elimination ideal I2 = I ∩ k[x, y, z]. According to our results a
Gröbner basis for I2 is given by G ∩ k[x, y, z] which in this case yields I2 = 〈x2 y z − z4
〉. The associated
2
u
variety is V(I2) = V(x2y − z3 v2
) ∪ V(z). Now the set v , u , u : uv = 0 is actually a subset of V(x2y − z3 )
alone, and there are in fact no points in it which are in V(z); so to get the minimal variety containing the
image set of F we certainly don’t want to include the V(z). What this says is that we have to be a bit more
sophisticated.
4

and
We do this by including an extra variable as follows: Consider the two maps
j
k 2
−→ k 6
(u, v) ↦→ uv, u, v, u2
v2
, , u
v u
k 6
π3
−→ k 3
(t, u, v, x, y, z) ↦→ (x, y, z),
J is now defined as the ideal J = 〈t−uv, vx−u 2 , uy −v 2 , z −u〉 (REMARK!!! It makes no difference in H
and seems to make better sense for what follows if J is defined by J = 〈1−tuv, vx−u 2 , uy −v 2 , z −u〉). For
either choice of J the Gröbner basis H for this ideal computed relative to lex order with t > u > v > x > y > z
is
H = {x 2 y − z 3 , −x y + v z, v x − z 2 , v 2 − y z, u − z, −x + t z 3 , −v + t y z 2 , −1 + t x y}.
The third elimination ideal J3 = J ∩ k[x, y, z] now has Gröbner basis {x2 y − z3 } and the variety V(x2y − z3
)
2
u v2
is in fact the minimal variety containing the set of points v , u , u : uv = 0 . We give a slightly varied
description: Let F = π3 ◦ j and let W = {(u, v): uv = 0} be the set where at least one of the denominators
is zero. Then V(J3) is the smallest variety in k3 which contains the set F (k2 − W ). This is the content of
the following theorem.
Theorem 3.3.4 (Theorem 2) (Rational Implicitization). If k is an infinite field, consider the
rational parametrization F : (k m − W ) → k n given by
(t1, . . . , tm) F
f1(t1, . . . , tm)
↦→
g1(t1, . . . , tm) , . . . , fn(t1,
. . . , tm)
,
gn(t1, . . . , tm)
where fi, gi ∈ k[t1, . . . , tm], and W = V(g1(t) · · · gn(t)). Let J be the ideal
J = 〈g1(t)x1 − f1(t), . . . , gn(t)xn − fn(t), 1 − yg1(t) · · · gn(t)〉 ⊂ k[y, t1, . . . , tm, x1, . . . , xn],
and let Jm+1 = J ∩ k[x1, . . . , xn] be the (m + 1)-st elimination ideal. Then V(Jm+1) is the smallest variety
in k n containing F (k m − W ).
Proof. Either taken as similar to the proof of the Polynomial Implicitization Theorem or deferred.
Chapter 3 §3 Exercises
§3.3.1.
Prove the “Elementary Observation” that i(k m ) = V(I).
Solution. This is done in the text section §3.3.0.
§3.3.2.
When k = C, the conclusion of Theorem 1 can be strengthened. Namely, one can show that there is a
variety W ⊂
= V(Im) such that V(Im) − W ⊂ F (Cm ). Prove this using the Closure Theorem.
Solution. The Closure Theorem guarantees the existence of an affine variety W ⊂
= V(Im) such V(Im)−W ⊂
πm(V ). We know that πm(V ) = F (C m ). Putting this together with the statement in the previous sentence
completes the argument.
5

§3.3.3.
Give an example to show that Exercise 3.3.2 is false over R. Hint: t 2 is always positive.
Solution. Consider the parametrization t ↦→ t 2 mapping R 1 → R 1 . The map i: R → R 2 is the map
t ↦→ (t, t 2 ) and its image i(R 1 ) is the variety V (V is just the parabola with equation x = t 2 if t, x denote
the coordinate functions in R 2 .) Now F (R) = {x ∈ R: x ≥ 0}. I = 〈x − t 2 〉 and the only polynomial in this
ideal which does not involve t is 0; so I1 = {0} and V(I1) = R.
If Exercise 3.3.2 were true for this example there would be a variety W ⊂
= R such that R−W ⊂ {x ∈ R: x ≥
0}. This means that W ⊃ {x ∈ R: x < 0}. The polynomials which vanish on W would have to vanish on
the negative reals. Only the zero polynomial does this; so W must contain R contradicting the assumption
that W ⊂
= R. No such W can exist.
§3.3.4.
In the text we proved that over C the tangent surface to the twisted cubic is defined by the equation
(2.8.0.6) −3 x 2 y 2 + 4 y 3 + 4 x 3 z − 6 x y z + z 2 = 0.
We want to show that the same is true over R. If (x, y, z) is a real solution of the above equation, then we
proved (using the Extension Theorem) that there are t, u ∈ C such that
(3.3.4.1)
x = t + u,
y = t 2 + 2tu,
z = t 3 + 3t 2 u.
Use the Gröbner basis given in the text to show that t and u are real. This will prove that (x, y, z) is on the
tangent surface in R 3 . Hint: First show that u is real.
Solution. The ideal I = 〈x − t − u, y − t 2 − 2tu, z − t 3 − 3t 2 u〉 has Gröbner basis
G = {−3 x 2 y 2 + 4 y 3 + 4 x 3 z − 6 x y z + z 2 ,
2 u y 3 + x y 3 − 4 x 2 y z + 5 y 2 z − 2 u z 2 − 2 x z 2 ,
− 2 u y 2 − x y 2 + 2 u x z + 2 x 2 z − y z,
u x y − x 2 y + 2 y 2 − u z − x z,
2 u x 2 − 2 x 3 − 2 u y + 3 x y − z,
u 2 − x 2 + y, t + u − x}.
If (x, y, z) ∈ R 3 the vanishing of the second polynomial in G shows that u is real and then the vanishing of
the last polynomial in G shows that t is real.
In summary. Every (x, y, z) which is complex and satisfies (2.8.0.6) can be augmented to (t, u, x, y, z),
a point in V and hence a zero of the polynomials in G, by adjoining a pair of complex numbers, t, u. Now
if (x, y, z) are real to start with then t, u are necessarily real also and it follows that a real triple (x, y, z)
satisfying (2.8.0.6) has the form (3.3.4.1) for suitable choice of (real) t, u. What we have shown is that the
locus (2.8.0.6) and (3.3.4.1) describe exactly the same point set, and this happens whether the coordinates
have values in C or in R.
§3.3.5.
In the parametrization of the tangent surface of the twisted cubic, show that the parameters t and u
are uniquely determined by x, y, and z. Hint: The argument is similar to what you did in Exercise 4.
Solution. Suppose (t, u, x, y, z) satisfy (3.3.4.1). Then they constitute a zero of the polynomials in G. u is
determined by (x, y, z) from the second equation and t is then determined by u, x from the last equation.
6

§3.3.6.
Let S be the parametric surface defined by
(3.3.6.1)
x = uv,
y = u 2 ,
z = v 2 .
(a) Find the equation of the smallest variety V that contains S.
Solution. By Theorem 3.0.1, the Polynomial Implicitization Theorem, this is the variety V(I2), where
I = 〈x − uv, y − u 2 , z − v 2 〉 and the variables are ordered u > v > x > y > z.
Using Mathematica:
H={x-u*v, y-u^2, z-v^2}
followed by
GroebnerBasis[H,{u,v,x,y,z}, MonomialOrder->{{1,1,1,1,1},
{0,0,0,0,-1},{0,0,0,-1,0},{0,0,-1,0,0},{0,-1,0,0,0}}]
which is the DegreeReverseLexicographic order or just using the Lexicographic order in the command
GroebnerBasis[H,{u,v,x,y,z}]
Leads to the output
(3.3.6.2) {x 2 − y z, v x − u z, u x − v y, v 2 − z, u v − x, u 2 − y}
and from this it follows that V has equation x 2 − yz = 0.
(b) Over C, use the Extension Theorem to prove that S = V . Hint: The argument is similar to what
we did for the tangent surface of the the twisted cubic.
Solution. What we have to do is to show that every (x, y, z) which satisfies x 2 − yz = 0 can be written in
the form (3.3.6.1) for some choice of u, v. Since we know (3.3.6.1) and the vanishing of the expressions in
(3.3.6.2) say exactly the same things about (u, v, x, y, z) we will work with (3.3.6.2).
A Direct Argument. Given (x, y, z) with x 2 − yz = 0, choose u ∈ {± √ y} and v ∈ {± √ z}. Then
(uv) 2 = yz = x 2 ; so uv = ±x. Choose the signs so that uv = x. It follows that uv = x. All the
expressions of (3.3.6.2) are thus zero except possibly for the second and third. Now vx = v · uv = uv 2 = uz
and ux = u 2 v = yv; so the second and third expressions are zero too. Thus this choice of u, v leads to
(u, v, x, y, z) satisfying (3.3.6.2) and hence (3.3.6.1).
Using the Extension Theorem. If we eliminate u first to get to I1 and then v to get to I2 we have
the Gröbner bases
(3.3.6.3)
I = 〈x 2 − yz, u 2 − y, −vx + uz, ux − vy, uv − x, v 2 − z〉,
I1 = 〈x 2 − yz, v 2 − z〉,
I2 = 〈x 2 − yz〉.
Now the Extension Theorem guarantees that given (x, y, z) ∈ V(I2), or equivalently satisfying x 2 − yz = 0,
there is a v such that (v, x, y, z) ∈ V(I1) because the set of leading coefficients of v in the polynomials
of the given basis for I1 is {x 2 − yz, 1} and they have no common (x, y, z)-zero. Similarly, making another
application of the Extension Theorem, this time from V(I1) to V(I), we see that the set of leading coefficients
of u in the polynomials in the given basis for I contains a 1, so that they don’t have a common zero. This
means that we can find a u such that (u, v, x, y, z) ∈ V(I). This (u, v, x, y, z) satisfies (3.3.6.1) and shows
that S = V .
(c) Over R, show that S only covers the “half” of V . What parametrization would cover the other half?
Solution. Looking at the “Direct Argument” in the solution to part (b) of this Exercise 3.3.6, we see that
we can find real u and v only when y and z are both nonnegative. The equation x 2 = yz shows that y and
z both have the same sign which leaves the “half” of V where y and z are both negative that is not covered
by the parametrization (3.3.6.1).
7

Consider then the parametrization
(3.3.6.4)
x = uv,
y = −u 2 ,
z = −v 2 .
A Gröbner basis for J = 〈x − uv, y + u 2 , z + v 2 〉 is given by
K = {x 2 − y z, v 2 + z, v x + u z, u x + v y, u v − x, u 2 + y}.
So the smallest variety which contains the points (x, y, z) having the form (3.3.6.4) is still V , the locus of
x 2 − yz = 0. This time we choose u ∈ {± √ −y} and v ∈ {± √ −z} with signs chosen so that uv = x as
before. We get an (u, v, x, y, z) for which the expressions of K are zero and which, accordingly, satisfies
(3.3.6.4). Thus (3.3.6.4) is the parametrization which fills out the other “half” of V .
Remark. Make the coordinate change s = z+y
√ , t =
2 z−y
√ , or y =
t s−t √ , z =
2 s+t √ . The s, t, x coordinate
2
axes are obtained by rotating the y, z, x axes π
4 radians in the “counterclockwise” (when viewed from the
positive side of the x axis) sense about the x-axis. The variety x2−yz = 0 has s, t, x-equation x2
2 2
s −t − 2 = 0
or 2x2 + t2 = s2 . This is an “eliptical cone” whose central axis is the s-axis, i.e. the line (t = 0, x = 0) or,
equivalently, (y = z, x = 0). Its intersection with the yz-coordinate plane is the union of the lines y = 0,
x = 0 and z = 0, x = 0. The portions where (s ≥ 0 or z ≥ 0, y ≥ 0) and (s ≤ 0 or z ≤ 0, y ≤ 0) are
the two “nappes” of this cone. In the complex case (3.3.6.1) covers both these nappes. In the real case
they are parametrized separately with (3.3.6.1) parametrizing the portion where y ≥ 0, z ≥ 0 and (3.3.6.4)
parametrizing the nappe where y ≤ 0, z ≤ 0.
§3.3.7.
Let S be the parametric surface
(3.3.7.1)
x = uv,
y = uv 2 ,
z = u 2 .
(a) Find the equation of the smallest variety V that contains S.
Solution. We know that if I = 〈x − uv, y − uv 2 , z = u 2 〉 and the variables are ordered u > v > x > y > z,
then V = V(I2). Mathematica shows a Gröbner basis for I to be
G = {x 4 − y 2 z, −x 3 + v y z, v x − y, −x 2 + v 2 z, −x 2 + u y, u x − v z, u v − x, u 2 − z}.
We read directly from G that V = V(x 4 − y 2 z). So the equation of V is x 4 − y 2 z = 0 . From this we can
easily extract the elimination ideals
I = 〈x 4 − y 2 z, −x 3 + v y z, v x − y, −x 2 + v 2 z, −x 2 + u y, u x − v z, u v − x, u 2 − z〉
I1 = 〈x 4 − y 2 z, −x 3 + v y z, v x − y, −x 2 + v 2 z〉
I2 = 〈x 4 − y 2 z〉
(b) Over C, show that V contains points which are not on S. Determine exactly which points of V are not
on S. Hint: Use lexicographic order with u > v > x > y > z.
Solution. What we are asked to do is find those points (x, y, z) with x 4 − y 2 z = 0 for which there are no
auxiliary variables u, v such that (u, v, x, y, z) is a zero of each of the polynomials in G.
8

If xy = 0 then the point
in order,
x 2
y
y
, x , x, y, z ∈ V(I) as we now verify. Taking the equations after the first
−x 3 + vyz = −x 3 + y 1
· yz =
x x · x 4 − y 2 z = 0,
vx − y = y
· x − y = 0,
x
−x 2 + v 2 z = −x 2 + y2 1
z =
x2 x2 (−x4 + yz) = 0,
−x 2 + uy = −x 2 + x2
y
ux − vz = x2 y
· x −
y x
u 2 2 2 x
− z =
y
· y = 0,
· z = 1
x 2 y (x4 − y 2 z) = 0,
− z = 1
y 2 · (x4 − y 2 z) = 0.
This leaves just the points on the xz and yz coordinate planes for consideration. If x = 0, the relation
x 4 − y 2 z = 0 becomes y 2 z = 0 and we need only consider points on the y or z axes. The relations for u, v
become
(3.3.7.2a)
(3.3.7.2b)
(3.3.7.2c)
(3.3.7.2d)
y 2 z = 0,
vyz = 0,
y = 0,
v 2 z = 0,
uy = 0,
vz = 0,
uv = 0,
u 2 − z = 0.
Case I (y = 0): If it turns out that y = 0, we can satisfy this system by choosing v = 0 and u = √ z.
That is, the point ( √ z, 0, 0, 0, z) ∈ V(I) and projects (under π2) onto (x, y, z) = (0, 0, z).
Case II (y = 0, z = 0): Here there is trouble. (0, y, 0) ∈ V(I2) and yet there is no point (u, v, 0, y, 0)
which is in V(I). So the portion of V which is not on S consists of the non zero points on the y-axis.
Remark. The surface x 4 − y 2 z = 0 is the union of the two surfaces x 2 − y √ z = 0 and x 2 + y √ z = 0.
Slicing it with the plane z = const will give two butting parabolas when const > 0, the y-axis when const = 0
and the empty set when const < 0. The parabolas flatten out and approach the line y = 0 as const ↑ +∞.
The trace on the plane x = x0 is the curve z = which splits into two hyperbola-like components.
(x0) 4
y 2
§3.3.8.
The Enneper surface is defined parametrically by
(3.3.8.1)
x = 3u + 3uv 2 − u 3 ,
y = 3v + 3u 2 v − v 3 ,
z = 3u 2 − 3v 2 .
(a) Find the equation of the smallest variety V containing the Enneper surface. It will be a very
complicated equation!
9

Solution. Letting I = 〈x − 3u − 3uv 2 − u 3 , y − 3v + 3u 2 v − v 3 , z − 3u 2 − 3v 2 〉 and calculating a Gröbner
basis for I using lex order with u > v > x > y > z gives
(g1)
(g2)
(g3)
(g4)
(g5)
(g6)
G = {57395628 x 2 + 2125764 x 4 + 19683 x 6 + 57395628 y 2 − 12754584 x 2 y 2 +
59049 x 4 y 2 + 2125764 y 4 + 59049 x 2 y 4 + 19683 y 6 − 172186884 z+
57395628 x 2 z + 1121931 x 4 z + 57395628 y 2 z − 6259194 x 2 y 2 z+
1121931 y 4 z − 229582512 z 2 + 24091992 x 2 z 2 + 196830 x 4 z 2 +
24091992 y 2 z 2 − 1023516 x 2 y 2 z 2 + 196830 y 4 z 2 − 131797368 z 3 +
5432508 x 2 z 3 + 10935 x 4 z 3 + 5432508 y 2 z 3 − 56862 x 2 y 2 z 3 +
10935 y 4 z 3 − 42515280 z 4 + 691092 x 2 z 4 + 691092 y 2 z 4 −
8424324 z 5 + 46656 x 2 z 5 + 46656 y 2 z 5 − 1049760 z 6 +
1296 x 2 z 6 + 1296 y 2 z 6 − 80352 z 7 − 3456 z 8 − 64 z 9 ,
− 354294 x 2 − 6561 x 4 − 8503056 v y+
2480058 y 2 − 39366 x 2 y 2 + 314928 v y 3 − 32805 y 4 + 1062882 z−
216513 x 2 z − 729 x 4 z − 6613488 v y z + 1436859 y 2 z − 5832 x 2 y 2 z+
104976 v y 3 z − 5103 y 4 z + 944784 z 2 − 50301 x 2 z 2 − 2020788 v y z 2 +
303993 y 2 z 2 + 8748 v y 3 z 2 + 341172 z 3 − 5103 x 2 z 3 −
303264 v y z 3 + 27945 y 2 z 3 + 64152 z 4 − 189 x 2 z 4 − 22356 v y z 4 +
945 y 2 z 4 + 6642 z 5 − 648 v y z 5 + 360 z 6 + 8 z 7 ,
13122 v + 243 v x 2 − 4374 y + 81 x 2 y−
729 v y 2 + 81 y 3 + 8748 v z + 27 v x 2 z − 1944 y z − 135 v y 2 z+
2106 v z 2 − 270 y z 2 + 216 v z 3 − 12 y z 3 + 8 v z 4 ,
39366 x 2 + 729 x 4 + 1180980 v y+
4374 v x 2 y − 354294 y 2 + 5832 x 2 y 2 − 48114 v y 3 + 5103 y 4 − 118098 z+
19683 x 2 z + 761076 v y z − 155277 y 2 z − 8748 v y 3 z − 91854 z 2 + 3402 x 2 z 2 +
177876 v y z 2 − 21384 y 2 z 2 − 27702 z 3 + 189 x 2 z 3 + 17820 v y z 3 −
945 y 2 z 3 − 4050 z 4 + 648 v y z 4 − 288 z 5 − 8 z 6 ,
− 6377292 v + 2187 v x 4 + 2125764 y − 157464 x 2 y − 729 x 4 y − 1180980 v y 2 +
393660 y 3 − 5832 x 2 y 3 + 50301 v y 4 − 5103 y 5 − 7085880 v z + 2125764 y z−
65610 x 2 y z − 708588 v y 2 z + 161838 y 3 z + 8748 v y 4 z−
3228012 v z 2 + 826686 y z 2 − 8019 x 2 y z 2 − 157464 v y 2 z 2 +
21141 y 3 z 2 − 769824 v z 3 + 157464 y z 3 − 189 x 2 y z 3 −
15876 v y 2 z 3 + 945 y 3 z 3 − 101088 v z 4 + 15066 y z 4 −
648 v y 2 z 4 − 6912 v z 5 + 648 y z 5 − 192 v z 6 + 8 y z 6 ,
972 v 2 + 27 x 2 − 324 v y + 27 y 2 − 81 z + 270 v 2 z − 54 v y z − 18 z 2 +
18 v 2 z 2 − z 3 , −1458 v − 27 v x 2 + 486 y + 324 v 2 y−
27 v y 2 − 810 v z + 135 y z + 54 v 2 y z − 144 v z 2 + 9 y z 2 − 8 v z 3 ,
1458 v 2 − 324 x 2 + 243 v 2 x 2 − 4860 v y + 1134 y 2 + 243 v 2 y 2 + 972 z+
162 v 2 z − 108 x 2 z − 1458 v y z + 135 y 2 z + 540 z 2 − 108 v y z 2 +
84 z 3 + 4 z 4 ,
10

(g7)
(g8)
(g9)
(g10)
(g11)
(g12)
(g13)
− 3 v + 2 v 3 + y − v z,
1458 u − 486 x + 162 v 2 x + 27 v x y + 27 u y 2 + 810 u z − 216 x z + 36 v 2 x z+
144 u z 2 − 24 x z 2 + 8 u z 3 ,
54 v 2 + 9 u x − 9 v y − 9 z + 6 v 2 z − z 2 ,
18 u v − 3 v x − 3 u y + 4 u v z,
81 u − 27 x + 9 v 2 x + 9 u v y + 27 u z − 6 x z + 2 u z 2 ,
9 u + 6 u v 2 − 3 x + u z,
3 u 2 + 3 v 2 − z}
The equation of the smallest variety V containing the Enneper surface is g1(x, y, z) = 0.
(b) Over C, use the Extension Theorem to prove that the above equations (3.3.8.1) parametrize the
entire surface V . Hint: There are lots of polynomials in the Gröbner basis. Keep looking – you will find
what you need.
Solution. Apparently with the order u > v > x > y > z
I = 〈g1, . . . , g13〉
I1 = 〈g1, g2, g3, g4, g5, g6, g7〉
I2 = 〈g1〉,
Now given a point (x, y, z) ∈ V(I2) on the variety V the Extension Theorem states that over C we can find a
v such that (v, x, y, z) ∈ V(I1) because the coefficient of the hignest power of v in g7 is the constant 2 which
never vanishes. Starting with this (v, x, y, z) ∈ V(I1) the Extension theorem guarantees that we can find a
complex number u such that (u, v, x, y, z) ∈ V(I) because the coefficient of the highest power of u in g13 is
3 which never vanishes. Thus every (x, y, z) ∈ V gives rise to a (u, v, x, y, z) ∈ V(I) which hence satisfies
(3.3.8.1) and shows that this point (x, y, z) is “covered” by the parametrization (3.3.8.1).
§3.3.9.
The Whitney umbrella surface is given parametrically by
(3.3.9.1)
x = uv,
y = v,
z = u 2 .
(A picture of this surface is given in CLO but we leave it out.)
(a) Find the equation of the smallest variety containing the Whitney umbrella.
Solution. Again using Mathematica, a Gröbner basis for the ideal I = 〈x − uv, y − v, z − u 2 〉 computed
with lex order and u > v > x > y > z is
{x 2 − y 2 z, v − y, −x + u y, u x − y z, u 2 − z}.
So the equation of the smallest variety containing the Whitney umbrella is x 2 − y 2 z = 0 .
11

(b) Show that the parametrization fills up the variety over C but not over R. Over R, exactly what
points are omitted?
Solution. Here we have
I = 〈x 2 − y 2 z, v − y, −x + u y, u x − y z, u 2 − z〉,
I1 = 〈x 2 − y 2 z, v − y〉,
I2 = 〈x 2 − y 2 z〉.
A point (x, y, z) ∈ V = V(I2) can be augmented to a point (v, x, y, z) ∈ V(I1) because the coefficient of the
highest power of v in the polynomial v − y is 1 which never vanishes. Note here that if x, y, z are real, then
so is v.
Continuing, a point (v, x, y, z) ∈ V(I1) can be augmented to a point (u, v, x, y, z) ∈ V(I) because the
coefficient of the highest power of u in u 2 − z is 1 which (again!) never vanishes. Here if z < 0 the
augmentation requires that u be non real. So it is the points (x, y, z) with x 2 − y 2 z = 0 and z < 0 which are
not reached by the real form of the parametrization (3.3.9.1).
(c) Show that the parameters u and v are not always uniquely determined by x, y, z. Find the points
where uniqueness fails and explain how your answer relates to the picture.
Solution. Given (x, y, z) ∈ V(I2) = V , v is determined by the equation v − y = 0. u is determined by the
equations −x + u y = 0, u x − y z = 0, and u 2 − z = 0. u too is certainly unique unless both y = 0 and
x = 0, that is u will be unique except possibly for points on the z-axis. There, in fact, ( √ z, 0, 0, 0, z) and
(− √ z, 0, 0, 0, z) are points extending (0, 0, z) with two different values for the coordinate u. I guess in the
picture shown that the z-axis is the set of singular points where the surface self-intersects.
§3.3.10.
Consider the curve in C n parametrized by xi = fi(t), where f1, . . . , fn are polynomials in C[t]. This
gives the ideal
I = 〈x1 − f1(t), . . . , xn − fn(t)〉 which is an ideal in C[t, x1, . . . , xn].
(a) Prove that the parametric equations fill up all of the variety V(I1) ⊂ C n .
Restatement. What we are to show is that if (a1, . . . , an) is a zero of each h(x1, . . . , xn) of the form
h(x1, . . . , xn) = w1(t, x1, . . . , xn)(x1 − f1(t)) + · · · + wn(t, x1, . . . , xx)(xn − fn(t)),
where wi ∈ C[t, x1, . . . , xn], 1 ≤ i ≤ n, the main requirement being that h doesn’t involve the variable t,
then there is a u ∈ C, with a1 = f1(u), . . . , an = fn(u).
As an Illustration. Consider the curve x = t, y = t 2 . The point (2, 4) is a zero of each
One of these h’s is, for example,
h(x, y) = b(t, x, y)(x − t) + c(t, x, y)(y − t 2 ), b, c ∈ C[t, x, y].
y − x 2 = −(x + t)(x − t) + 1 · (y − t 2 ).
The result we are to establish says that there is a u with (2, 4) = (u, u 2 ). True here with u = 2 of course.
Solution. We are going to cheat and use the Extension Theorem. Here with the variables ordered by
t > x1 > x2 > · · · > xn we are asked to extend the partial solution (a1, . . . , an) from I1 to I, from (a1, . . . , an)
to (u, a1, . . . , an), and the Extension Theorem says that this is possible (over C) if the coefficients of the
highest powers of t in the polynomials xi − fi(t) do not have a common zero. In this case, where t is just a
single variable, these coefficients are constants (they do not involve x1, . . . , xn); so either (i) one of them is
nonzero in which case the Extension Theorem guarantees the existence of a u or (ii) all of them are zero in
which case the statement to be shown is trivial.
12

(b) Show that the conclusion of part (a) may fail if we let f1, . . . , fn be rational functions. Hint: See §3
of Chapter 1.
Solution. Consider the parametric equations
(3.3.10.b1)
x = t
1 + t ,
y = 1 − 1
.
t2 A simple calculation verifies that t = x
1−x and substituting this last expression for t in the second equation
yields y = 1 −
1−x 2 2
x or x y − 2x + 1 = 0.
Remark. The Mathematica command sequence
H={(1+t)*x-t,t^2y-t^2+1}
GroebnerBasis[H,{t,x,y}]
gives the result
{1 − 2 x + x 2 y, 1 − t + t y − x y, −t + x + t x},
which is in keeping with the preceding calculations to eliminate t from (3.3.10.b1).
Anyway, the point (1, 1) is on the curve x 2 y−2x+1 = 0 but cannot be represented in the form (3.3.10.b1)
by any t ∈ C.
(c) Even if all of the fj’s are polynomials, show that it may not be true that the parametric equations
fill up all of the variety V(I1) if we work over R.
Solution. Let the parametric equations be: x = t 2 , y = t 2 . Then I = 〈x − t 2 , y − t 2 〉 and I1 = 〈x − y〉. The
point (−1, −1) ∈ V(I1), but can not be represented in the form (t 2 , t 2 ) for any real t.
§3.3.11.
This problem is concerned with the proof of
Theorem 3.3.4 (Theorem 2) (Rational Implicitization). If k is an infinite field, consider
the rational parametrization F : (k m − W ) → k n given by
(t1, . . . , tm) F
f1(t1, . . . , tm)
↦→
g1(t1, . . . , tm) , . . . , fn(t1,
. . . , tm)
,
gn(t1, . . . , tm)
where fi, gi ∈ k[t1, . . . , tm], and W = V(g1(t) · · · gn(t)). Let J be the ideal
J = 〈g1(t)x1 − f1(t), . . . , gn(t)xn − fn(t), 1 − yg1(t) · · · gn(t)〉 ⊂ k[y, t1, . . . , tm, x1, . . . , xn],
and let Jm+1 = J ∩ k[x1, . . . , xn] be the (m + 1)-st elimination ideal. Then V(Jm+1) is the smallest
variety in k n containing F (k m − W ).
Restatement of Conclusion. If f ∈ k[x1, . . . , xn] vanishes on F (k m − W ) then f ∈ Jm+1.
(a) Let k be an infinite field and let f, g ∈ k[t1, . . . , tm]. Assume that g = 0 and that f vanishes on
k m − V(g). Prove that f is the zero polynomial. Hint: Consider the product fg.
Solution. The product fg is zero, both on V(g) and k m − V(g); so it is zero on k m . Since k is an infinite
field it follows that fg = 0 is the zero polynomial. But k[t1, . . . , tm] is an integral domain; so either (i) g = 0
or (ii) f = 0. Since (i) has been ruled out it must be that f = 0 is the zero polynomial.
(b) Prove Theorem 2 using the hints given in the text.
13

Solution. We begin by restating the Closure Theorem
Theorem 3.2.0.3. (Theorem 3) (The Closure Theorem). Let V = V(f1, . . . , fs) ⊂ C n and
let Iℓ be the ℓ-th elimination ideal of 〈f1, . . . , fs〉. That is, Iℓ = 〈f1, . . . , fs〉 ∩ C[xℓ+1, . . . , xn]. Then:
(i) V(Iℓ) is the smallest affine variety containing πℓ(V ) ⊂ Cn−ℓ .
(ii) When V = ∅, there is an affine variety W ⊂
= V(Iℓ) such that V(Iℓ) − W ⊂ πℓ(V ).
(iii) There are affine varieties Zi ⊂ Wi ⊂ Cn−ℓ with 1 ≤ i ≤ m such that
m
πℓ(V ) = (Wi − Zi).
i=1
Note. ALL OF THE CLOSURE THEOREM THAT WE USE IN THIS PROOF is Part (i) which
states: “If a polynomial f ∈ C[xℓ+1, . . . , xn] vanishes on πℓ(V ) ⊂ C n−ℓ , then f ∈ I∩C[xℓ+1, . . . , xn].”
Proof of Rational Implicitization Theorem. First (by way of review) we let
g(t) = g1(t)g2(t) · · · gn(t) be the product of the “denominator polynomials” g1, . . . , gn and set W = V(g).
Then the maps
j : k m − W → k 1+m+n and F : k m − W → k n are defined by
(3.3.11.1)
(3.3.11.2)
(t1, t2, . . . , tm) j
1 f1(t) fn(t)
↦→ , t, , . . . , = (y, s, x), t ∈ k
g(t) g1(t) gn(t)
m − W.
(t1, t2, . . . , tm) F
f1(t) fn(t)
↦→ , . . . , , t ∈ k
g1(t) gn(t)
m − W.
We will use y, s, x for the coordinate variables on k 1+m+n , where s = (s1, . . . , sm) and x = (x1, . . . , xn). Let
J = 〈1 − yg(s), g1(s)x1 − f1(s), . . . , gn(s)xn − fn(s)〉. J is an ideal in the ring k[y, s, x]. Our first contention
is:
Contention 1. j(k m − W ) = V(J).
Proof of Contention 1: (⊂): If q ∈ j(km − W ), then q has the form on the right of (3.3.11.1) for some
t. We evaluate the polynomials in the basis for J at q, using the facts that s(q) = t, gi(s)(q) = gi(t), and
fi(s)(q) = fi(t). This gives
1
(1 − yg(s) (q) = 1 − · g(t) = 0,
g(t)
(gi(s)xi − fi(s))(q) = gi(t) · fi(t)
gi(t) − fi(t) = 0.
So it follows that j(k m − W ) ⊂ V(J).
Thus
(⊃): Suppose now that p = (b, r, z) ∈ V(J). Then
0 = (1 − yg(s))(p) = 1 − bg(r) =⇒ g(r) = 0 so r ∈ k m − W and b = 1
g(r)
and for each 1 ≤ i ≤ n, gi(r) = 0 and
0 = (gi(s)xi − fi(s))(p) = gi(r)zi − fi(r) =⇒ zi = fi(r)
gi(r) .
(b, r, z) =
This completes the proof of Contention 1.
1 f1(r) fn(r)
, r, , . . . , ∈ j(k
g(r) g1(r) gn(r)
m − W ).
14

The case where k is algebraically closed: A direct application of (i) of the Closure Theorem with
ℓ = m + 1 then shows that V(Jm+1) is the smallest affine variety which contains πm+1 (V(J)). Now notice
that
f1(r) fn(r)
πm+1 (V(J)) = , . . . , : r ∈ k
g1(r) gn(r)
m
− W = F (k m − W ).
With this identification the proof of the Rational Implicitization Theorem when k is algebraically closed is
complete.
Next suppose that k ⊂ K where K is algebraically closed. Since k may be strictly smaller than K we
cannot use the Closure Theorem directly; so our strategy will be to switch back and forth between k and K
and we will use the subscript k or K (in parentheses) to keep track of which field we are working with. Thus
V (k)(Jm+1) is the variety in k n where the polynomials of Jm+1 vanish and V (K) (Jm+1) is the larger set of
solutions in K n . Let
H = {1 − yg(s), g1(s)x1 − f1(s), . . . , gn(s)xn − fn(s)},
then
V (k)(Jm+1) = x ∈ k n : for some (y, s) ∈ k × k m , the polynomials H vanish on (y, s, x) ;
V (K) (Jm+1) = x ∈ K n : for some (y, s) ∈ K × K m , the polynomials H vanish on (y, s, x) .
Since the polynomials H which generate J don’t themselves change V (k)(Jm+1) ⊂ V (K) (Jm+1). (Note. A
way to get a basis for Jm+1 or Jm+1 is to start with the set of polynomials H and using these polynomials,
all of whose coefficients are in k (by the way), to compute a Gröbner basis G for the ideal J that they
generate in k[y, s, x]. G is unaffected by whether the computaions are carried out using k as the field or
using K because computing a Gröbner basis will never take us out of the field of coefficients of the starting
ideal basis. Then a basis for Jm+1 consists of G ∩ k[x1, . . . , xn] and nowhere does it make any difference
whether the computations are carried out over k or over K. So G ∩ k[x] = G ∩ K[x].)
This discussion shows that the same set of polynomials in k[x] serve as a Gröbner basis for the ideal
Jm+1 in k[x] and for the ideal Jm+1 in K[x]. Let these polynomials be
Collecting this information,
{b1(x), . . . , bs(x)} = G ∩ k[x] = G ∩ K[x].
(Jm+1) Jm+1 = { ˜ ψ ∈ K[x]: ˜ ψ(x) = ˜ h1(x)b1(x) + · · · + ˜ hs(x)bs(x), ˜ hi(x) ∈ K[x], 1 ≤ i ≤ s}
and
(Jm+1) Jm+1 = {ψ ∈ k[x]: ψ(x) = h1(x)b1(x) + · · · + hs(x)bs(x), hi(x) ∈ k[x], 1 ≤ i ≤ s}.
Using the fact that these hi’s and ˜ hj’s can be chosen rather arbitrarily, we have established that
and
a point q ∈ K n is in V (K) (Jm+1) iff bi(q) = 0, 1 ≤ i ≤ s
a point p ∈ k n is in V (k)(Jm+1) iff bi(p) = 0, 1 ≤ i ≤ s.
We need to prove that V (k)(Jm+1) is the smallest variety in k n containing F (k m − W (k)). Since K is
algebraically closed we can pick up here applying the closure theorem to state that: “V (K) (Jm+1) has been
shown to be the smallest variety in K n containing F (K m − W K )”.
I don’t think we actually need to do this, but to make sure our argument is complete we next show that
the variety V (k)(Jm+1) actually contains F (k m −W (k)) and that V (K) (Jm+1) actually contains F (K m −W (K) ).
15

Contention 2. (i) F (k m − W (k)) = πm+1(V (k)) ⊂ V (k)(Jm+1) and
(ii) F (K m − W (K) ) = πm+1(V (K) ) ⊂ V (K) (Jm+1).
Proof of Contention 2. Suppose f ∈ Jm+1 and p ′ = πm+1(p) with p ∈ V (k). The polynomial f only
involves the variables x1, . . . , xn; so we can regard it as being a polynomial f in k[x1, . . . , xn] or as being
a polynomial ˜ f ∈ k[y, s1, . . . , sm, x1, . . . , xn] which doesn’t involve y or s. In the latter case we have the
relations ˜ f = f ◦ πm+1 and ˜ f ∈ Jm+1. Together with p ∈ V (k)(J) and ˜ f ∈ J we have 0 = ˜ f(p) =
(f ◦ πm+1)(p) = f (πm+1(p)) = f(p ′ ), which proves the contention.
Let f ∈ k[x1, . . . , xn] be any polynomial vanishing on F (k m − W (k)). To complete the proof we must
show that f ∈ Jm+1 or, equivalently, that f vanishes on V (k)(Jm+1). First, note that f ◦ F vanishes on all
of km − W (k). Since f is a polynomial in the variables x1, . . . , xn and
f1
F =
, . . . ,
g1
fn
gn
are rational functions in t1, . . . , tm. It follows that f ◦ F is a rational function in the variables t1, . . . , tm.
Now
Contention 3.The only rational function in t1, . . . , tm which vanishes on all of k m −W (k) is the zero rational
function.
Proof of contention 3. Suppose this rational function has the form p(t)
q(t) . The polynomial t ↦→ g(t)p(t)
vanishes on km
− W (k) ∪ W(k) = km . Since k is an infinite field this implies that g(t)p(t) = 0 ∈ k[t]. Since
g(t) is not the zero polynomial and k[t] is an integral domain, this implies that p(t) is the zero polynomial
is the zero rational function.
in k[t] and p(t)
q(t)
We need one more fact to carry the argument on smoothly.
Contention 4. k n − W (k) ⊂ K n − W (K) .
Proof of contention 4.
k n − W (k) = k n ∩ {ζ ∈ k n : g(ζ) = 0}
⊂ K n ∩ {ζ ∈ K n : g(ζ) = 0}
= K n − W (K) .
Now to the proof proper. When f ∈ k[x1, . . . , xn] each line below implies the line which follows it.
f vanishes on F (k m − W (k));
f ◦ F vanishes on k m ;
f ◦ F is the zero rational function ;
f ◦ F vanishes on K m ;
f vanishes on F (K m − W (K) ); this implies the next line by Statement
f vanishes on {q ∈ K n − W (K) : Jm+1(q) = {0}};
(i) of the Closure Theorem;
f vanishes on {q ∈ K n − W (K) : b1(q) = 0, . . . , bs(q) = 0}; this implies the next line
f vanishes on {p ∈ k n − W (k) : b1(p) = 0, . . . , bs(p) = 0};
f vanishes on {p ∈ k n − W (k) : Jm+1(p) = {0}};
V (k)(Jm+1) ⊂ V (k)(f). where the varieties are in k n .
because of contention 4
which proves that V (k)(Jm+1) is the smallest variety of k n containing F (k m −W (k)), and completes the proof
of the theorem.
16

§3.3.12.
Consider the following rational parametrization:
(3.3.3) (u, v) F
2 u v2
↦→ , , u .
v u
For simplicity, let k = C . Also let I = 〈vx − u 2 , uy − v 2 , z − u〉 be the ideal obtained by “clearing
denominators”.
(a) Show that I2 = 〈zx 2 y − z 4 〉
Solution. As we found before the Mathematica commands
h={v*x-u^2,u*y-v^2,z-u}
GroebnerBasis[h,{u,v,x,y,z}]
executed in sequence yield
{x 2 y z − z 4 , − (x y z) + v z 2 , v x − z 2 , v 2 − y z, u − z},
and we can read off I2 = 〈x 2 y z − z 4 〉 from this.
(b) Show that the smallest variety in C 5 containing i(C 2 − W ) is
where i: (C 2 − W ) −→ C 5 is the map
V(vx − u 2 , uy − v 2 , z − u, x 2 y − z 3 , vz − xy),
(u, v)
i
−→
u, v, u2
v
v2
, , u .
u
Hint: Show that i(C 2 − W ) = π1(V(J)), and then use the Closure Theorem.
Solution. Following the hint, let j : C 2 −→ C 6 be the map
(u, v)
j
−→
1
uv
, u, v, u2
v
v2
, , u .
u
and J = 〈1 − yuv, vx − u 2 , uy − v 2 , z − u〉. It is clear that i = π1 ◦ j and we know from previous
discussion ( Viz.,the proof of the Rational Implicitization Theorem given in the solution to Exercise 3.3.11)
that V(J) = j(C 2 − W ). Applying π1 to this last equation gives π1 (V(J)) = (π1 ◦ j)(C 2 − W ) = i(C 2 − W ).
Now the Closure Theorem guarantees that V(J1) is the smallest affine variety containing π1 (V(J)). So V(J1)
is the smallest variety containing i(C 2 − W ). This leaves us the problem of identifying V(J1). This task is
made easier by Mathematica. The Mathematica command sequence:
H={1 − t u v, v x − u 2 , u y − v 2 , z − u}
GroebnerBasis[H,{t,u,v,x,y,z}]
produces the output
So
{x 2 y − z 3 , − (x y) + v z, v x − z 2 , v 2 − y z, u − z, −x + t z 3 , −v + t y z 2 , −1 + t x y}
J = 〈x 2 y − z 3 , − (x y) + v z, v x − z 2 , v 2 − y z, u − z, −x + t z 3 , −v + t y z 2 , −1 + t x y〉
J1 = 〈x 2 y − z 3 , − (x y) + v z, v x − z 2 , v 2 − y z, u − z〉
We can rewrite J1, rearranging terms and making free use of the relation u − z = 0, as
so
J1 = 〈v x − z 2 , v 2 − y u, u − z, x 2 y − z 3 , − (x y) + v z〉;
V(J1) = V(v x − z 2 , v 2 − y u, u − z, x 2 y − z 3 , − (x y) + v z)
is the smallest variety in C 5 containing i(C 2 − W ).
17

Remark. Theorem 3.2.0.2, Theorem 2 of §3.2, tells us that
V(J1) = π1 (V(J)) ∪ (V(g1, . . . , gs) ∩ V(J1)) ,
where in this case the gi’s are the coefficients of the highest power of the variable t in a polynomial basis
for J. Disregarding those polynomials which don’t involve t, and whose “coefficients of highest power of
t” don’t add anything anyway because they form a basis for J1 and their zeros appear in π1(V(J)), these
polynomials are z 3 , yz 2 and xy. Considering this it is not clear to me at the outset that V(J1) is actually
equal to π1(V(J)). No matter, it is still the smallest variety containing π1(V(J)) and that is all part (b) of
Exercise 3.3.12 actually claims.
(c) Show that {(0, 0, x, y, 0): x, y arbitrary} ⊂ V(I) and conclude that V(I) is not the smallest variety
containing i(C 2 − W ).
Solution. Remembering that I = 〈vx − u 2 , uy − v 2 , z − u〉 it is easy to check that
{(0, 0, x, y, 0): x, y arbitrary} ⊂ V(I). On the otherhand Since
J1 = 〈v x−z 2 , v 2 −y u, u−z, x 2 y −z 3 , − (x y)+v z〉, the only points of the set {(0, 0, x, y, 0): x, y arbitrary}
which are in V(J1) are those for which xy = 0. In particular, (0, 0, 1, 1, 0) ∈ V(I) − V(J1); so V(I) = V(J1)
and by part(b) of this exercise V(I) can’t be V(J1), the smallest variety containing i(C 2 − W ).
(d) Determine exactly which portion of x 2 y − z 3 = 0 is parametrized by (3.3.3).
Suppose (x, y, z) satisfies x 2 y = z 3 .
Case z = 0. Then y = 0, yz > 0 and there are two values for x, namely, x = ±
(3.3.12.d1)
(x, y, z): x 2 y = z 3 , z = 0 =
±
z3 y
z3
, y, z
y
: yz > 0 .
. Thus the points
2
u v2
Each of these points has the form v , u , u . To see this, given a point with the form (3.3.12.d1) set u = z
and v = ± √ yz. With these choices for u and v,
u 2
v
v2
z
, , u = ±
u 3
, y, z .
y
So all the points of (3.3.12.d1) are “covered” by the parametrization.
Case z = 0. Here x2y = z3 is the graph of
the x and y coordinate axes on the coordinate plant z = 0. None
2
u
of these points have the form
§3.3.13.
v
v2 , u , u
Given a rational parametrization t ↦→
; so they are not covered by the parametrization.
f1(t)
g1(t)
fn(t)
, . . . , gn(t) , there is one case where the naive ideal I =
〈g1(t)x1 − f1(t), . . . , gn(t)xn − fn(t)〉 ⊂ k[t, x] obtained by “clearing denominators” gives the right answer.
Suppose that t = (t1) = (t), i.e. there is only one value of t. We can assume for each i that fi(t) and gi(t)
are relatively prime in k[t]; so, in particular, they have no common roots. If I ⊂ k[t, x1, . . . , xn] is as above,
then prove that V(I1) is the smallest variety containing F (k − W ), where as usual g = g1 · · · gn ∈ k[t] and
W = V(g) ⊂ k. Hint: Show that i(k − W ) = V(I), where i: (k − W ) → k n+1 is given by
t i
↦→
t, f1(t) fn(t)
, . . . ,
g1(t) gn(t)
and adapt the proof of the Polynomial Implicitization Theorem.
Solution. We prove first the following:
18
,

Contention 1. i(k − W ) = V(I).
Proof of contention 1. (⊃): Suppose p = (t, x1, . . . , xn) ∈ V(I). Then for each i, 1 ≤ i ≤ n, gi(t)xi−fi(t) =
0. Since gi and fi have no common zeros it must be that gi(t) = 0, 1 ≤ i ≤ n; so we have at the start that
t ∈ k − W . But then solving the relations gi(t)xi − fi(t) = 0 for xi shows, since t ∈ k − W , that
p = (t, x1, . . . , xn) = t, f1(t)
fn(t)
, . . . , ∈ i(k − W ).
g1(t) gn(t)
(⊂): If t ∈ k − W , then i(t) =
t, f1(t)
g1(t)
(s, x) ↦→ gi(s)xi − fi(s) vanish on i(t). This shows that i(k − W ) ⊂ V(I).
fn(t)
, . . . , gn(t) and it is an easy matter to check that the polynomials
Contention 2. If k is algebraically closed, then V(I1) is the smallest variety containing F (k − W ), where
F = π1 ◦ i: (k − W ) → k n is the map
t F
f1(t) fn(t)
↦→ , . . . , .
g1(t) gn(t)
Proof of contention 2. We know from the Closure Theorem that V(I1) is the smallest variety containing
π1(V(I)). Now π1 (V(I)) = π1 (i(k − W )) = (π1 ◦ i) (k − W ) = F (k − W ).
Next suppose that k ⊂ K where K is algebraically closed. Since k may be strictly smaller than K we
cannot use the Closure Theorem directly; so our strategy will be to switch back and forth between k and K
and we will use the subscript k or K (in parentheses) to keep track of which field we are working with. Thus
V (k)(I1) is the variety in k n where the polynomials of I1 vanish and V (K) (I1) is the larger set of solutions in
K n . (Note. A way to get a basis for I1 is to start with the set of polynomials
(3.3.13.1) {gi(s)xi − fi(s): 1 ≤ i ≤ n}
and using these polynomials, whose coefficients are in k, to compute a Gröbner basis G for the ideal I that
they generate in k[s, x]. G is unaffected by whether the computations are carried out using k as the field or
using K because computing a Gröbner basis will never take us out of the field of coefficients of the starting
ideal basis. Then a basis for I1 consists of G∩k[x1, . . . , xn] and nowhere does it make any difference whether
the computations are carried out over k or over K.)
We need to prove that V (k)(I1) is the smallest variety in k n containing F (k − W (k)). We know from
contention 2 that V (K) (I1) is the smallest variety in K n containing F (K − W (K) ).
Put in other words: Let Z (k) = V (k)(h1, . . . , hs) ⊂ kn be any variety of kn containing F (k − W (k));
so F (k − W (k)) ⊂ Z (k). We must show that V (k)(I1) ⊂ Z (k). First, note that hi = 0 on Z (k) and, hence,
hi = 0 on the smaller set F (k − W (k)). This shows that each hi ◦ F vanishes on all of k − W (k). But hi is a
polynomial in the variables x1, . . . , xn and
f1
F = , . . . ,
g1
fn
gn
are rational functions in t. It follows that hi ◦F ∈ k(t), the field of rational functions in t, and hi ◦F vanishes
on k − W (k). To continue this arguement we make use of contention 3 below in the special case where m = 1.
We can do this because k is infinite and W (k) is finite; so k − W (k) is still infinite.
Contention 3.Any rational function in t1, . . . , tm which vanishes on all of k m − W (k) is the zero rational
function.
Proof of contention 3. Suppose this rational function has the form p(t)
q(t) . The polynomial t ↦→ g(t)p(t)
vanishes on km
− W (k) ∪ W(k) = km . Since k is an infinite field this implies that g(t)p(t) = 0 ∈ k[t]. Since
g(t) is not the zero polynomial and k[t] is an integral domain, this implies that p(t) is the zero polynomial
is the zero rational function.
in k[t] and p(t)
q(t)
19

Picking up the proof where we left off, contention 3 guarantees that hi ◦ F is the zero rational function;
so hi ◦ F vanishes not only on k − W (k) but on K − W (K) , where K ⊃ k is algebraically closed. This means
that the hi’s vanish on F (K −W (K) ). Contention 2 then guarantees that V (K) (I1) ⊂ V (K) (h1, . . . , hs) because
V (K) (I1) is the smallest variety in K n which contains F (K − W (K) ). It follows that
V (k)(I1) = k n ∩ V (K) (I1) ⊂ K n ∩ V (K) (h1, . . . , hs) = V (k)(h1, . . . , hs).
This completes the proof of the fact that when there is just one parameter t, V(I1) is the smallest variety in
k n which contains F (k − W ).
§3.3.14.
The folium of Descartes can be parametrized by
x = 3t
,
1 + t3 y = 3t2
.
1 + t3 (a) Find the equation of the folium. Hint: Use Exercise 3.3.13.
Solution. Using Mathematica a Gröbner basis for I = 〈(1 + t 3 )x − 3t, (1 + t 3 )y − 3t 2 〉 is
{x 3 − 3 x y + y 3 , x 2 − 3 y + t y 2 , t x − y, −3 t + x + t 2 y}.
So the equation of the folium is x 3 − 3 x y + y 3 = 0 .
(b) Over C or R, show that the above parametrization covers the entire curve.
Solution. If (x, y) ∈ V(I1), i.e. satisfies x 3 − 3 x y + y 3 = 0, let us try to extend to a point (t, x, y) ∈ V(I).
The Extension Theorem guarantees that the extension will work over C unless (x, y)V(y 2 , x, y) = V(x, y).
That is except possibly at the point (0, 0), but if x = 0, y = 0 the extension is easy. It is (t, x, y) = (0, 0, 0).
For the real case we must reason differently and our reasoning here covers both the real and complex
cases. Let us try to extend (x, y) ∈ V(I1), and thereby satisfying x3 − 3xy + y3 = 0, to a (t, x, y) ∈ V(I)
satisfying the four equations associated with the polynomial basis of I listed above in the first offset line.
If x = 0, then y = 0 and an/the appropriate extension is to set t = 0 giving (0, 0, 0) ∈ V(I).
If x = 0, put t = y
x . y
x , x, y is easily seen to be a zero of the polynomials in the offset line.
§3.3.15.
In Exercise 16 to §3 of Chapter 1, we studied the parametric equations over R
x = (1 − t)2x1 + 2t(1 − t)wx2 + t2x3 (1 − t) 2 + 2t(1 − t)w + t2 ,
y = (1 − t)2y1 + 2t(1 − t)wy2 + t2y3 (1 − t) 2 + 2t(1 − t)w + t2 ,
where w, x1, y1, x2, y2, x3, y3 are constants and w > 0. By eliminating t show that these equations describe
a portion of a conic section. Recall that a conic section is described by an equation of the form
ax 2 + bxy + cy 2 + dx + ey + f = 0.
Hint: In most computer algebra systems, the Gröbner basis command allows polynomials to have coefficients
involving symbolic constants like w, x1, y1, x2, y2, x3, y3. Since we are over R and the denominators never
vanish, you can use Exercise 3.3.13.
20

Solution. If one runs the Mathematica command sequence
A=(1-t)^2*x1+2*t*(1-t)*w*x2+t^2*x3
B=(1-t)^2*y1+2*t*(1-t)*w*y2+t^2*y3
F=(1-t)^2+2*t*(1-t)*w+t^2
H={F*x-A,F*y-B}
GroebnerBasis[H,{t,x,y},CoefficientDomain->RationalFunctions]
21

The output is (hold your hat!)
{ x1 2 − 4 w 2 x1 x2 + 4 w 2 x2 2 − 2 x1 x3 + 4 w 2 x1 x3 − 4 w 2 x2 x3 + x3 2 y 2 +
x3 2 y1 2 − 4 w 2 x2 x3 y1 y2 + 4 w 2 x1 x3 y2 2 + 4 w 2 x2 2 y1 y3 − 2 x1 x3 y1 y3 − 4 w 2 x1 x2 y2 y3 + x1 2 y3 2 +
x y −2 x1 y1 + 4 w 2 x2 y1 + 2 x3 y1 − 4 w 2 x3 y1 + 4 w 2 x1 y2 − 8 w 2 x2 y2 + 4 w 2 x3 y2 + 2 x1 y3−
+
4 w 2 x1 y3 + 4 w 2 x2 y3 − 2 x3 y3
y −4 w 2 x2 2 y1 + 2 x1 x3 y1 + 4 w 2 x2 x3 y1 − 2 x3 2 y1 + 4 w 2 x1 x2 y2 − 8 w 2 x1 x3 y2+
+
4 w 2 x2 x3 y2 − 2 x1 2 y3 + 4 w 2 x1 x2 y3 − 4 w 2 x2 2 y3 + 2 x1 x3 y3
x 2 y1 2 − 4 w 2 y1 y2 + 4 w 2 y2 2 − 2 y1 y3 + 4 w 2 y1 y3 − 4 w 2 y2 y3 + y3 2 +
x −2 x3 y1 2 + 4 w 2 x2 y1 y2 + 4 w 2 x3 y1 y2 − 4 w 2 x1 y2 2 − 4 w 2 x3 y2 2 +
2 x1 y1 y3 − 8 w 2 x2 y1 y3 + 2 x3 y1 y3 + 4 w 2 x1 y2 y3 + 4 w 2 x2 y2 y3 − 2 x1 y3 2 ,
− 2 w x2 y1 2 − x3 y1 2 + 2 w x3 y1 2 + 2 w x1 y1 y2 + 4 w 2 x2 y1 y2 − 2 w x3 y1 y2 − 4 w 2 x1
y2 2 + x1 y1 y3 − 2 w x1 y1 y3 + 2 w x2 y1 y3 − 4 w 2 x2 y1
y3 + x3 y1 y3 + 4 w 2 x1 y2 y3 − x1 y3 2 +
t y −4 w x2 y1 + 4 w 2 x2 y1 + 4 w x3 y1 − 4 w 2 x3 y1 + 4 w x1 y2 − 4 w 2 x1 y2−
+
4 w x3 y2 + 4 w 2 x3 y2 − 4 w x1 y3 + 4 w 2 x1 y3 + 4 w x2 y3 − 4 w 2 x2 y3
y − (x1 y1) + 2 w x2 y1 + x3 y1 − 2 w x3 y1 − 2 w x1 y2 + 4 w 2 x1 y2 − 4 w 2 x2 y2 + 2 w x3 y2 + x1 y3+
+
2 w x1 y3 − 4 w 2 x1 y3 − 2 w x2 y3 + 4 w 2 x2 y3 − x3 y3
x y1 2 − 4 w 2 y1 y2 + 4 w 2 y2 2 − 2 y1 y3 + 4 w 2 y1 y3−
4 w 2 y2 y3 + y3 2 + t 2 w x2 y1 2 − 2 w x3 y1 2 −
2 w x1 y1 y2 − 4 w 2 x2 y1 y2 + 2 w x3 y1 y2 + 4 w 2 x3 y1 y2+
4 w 2 x1 y2 2 − 4 w 2 x3 y2 2 + 2 w x1 y1 y3 − 2 w x3 y1 y3−
2 w x1 y2 y3 − 4 w 2 x1 y2 y3 + 4 w 2 x2 y2 y3 + 2 w x3 y2 y3 + 2 w x1 y3 2 − 2 w x2 y3 2 ,
x1 2 − 4 w 2 x1 x2 + 4 w 2 x2 2 − 2 x1 x3 + 4 w 2 x1 x3 − 4 w 2 x2 x3+
x3 2 y + 2 w x1 x2 y1 − 4 w 2 x2 2 y1 + x1 x3 y1 − 2 w x1 x3 y1+
4 w 2 x2 x3 y1 − x3 2 y1 − 2 w x1 2 y2+
4 w 2 x1 x2 y2 + 2 w x1 x3 y2 − 4 w 2 x1 x3 y2 − x1 2 y3+
2 w x1 2 y3 − 2 w x1 x2 y3 + x1 x3 y3 + t x 4 w x2 y1 − 4 w 2 x2 y1−
4 w x3 y1 + 4 w 2 x3 y1 − 4 w x1 y2 + 4 w 2 x1 y2 + 4 w x3 y2 − 4 w 2 x3 y2+
4 w x1 y3 − 4 w 2 x1 y3 − 4 w x2 y3 + 4 w 2 x2 y3
+
x − (x1 y1) − 2 w x2 y1 + 4 w 2 x2 y1 + x3 y1 + 2 w x3 y1 − 4 w 2 x3 y1 + 2 w x1 y2 − 4 w 2 x2 y2−
2 w x3 y2 + 4 w 2
x3 y2 + x1 y3 − 2 w x1 y3 + 2 w x2 y3 − x3 y3 +
t −2 w x1 x2 y1 + 4 w 2 x2 2 y1 + 2 w x1 x3 y1 − 2 w x2 x3 y1 − 4 w 2 x2 x3 y1 + 2 w x3 2 y1+
2 w x1 2 y2 − 4 w 2 x1 x2 y2 + 4 w 2 x2 x3 y2 − 2 w x3 2 y2 − 2 w x1 2 y3+
2 w x1 x2 y3 + 4 w 2 x1 x2 y3 − 4 w 2 x2 2 y3 − 2 w x1 x3 y3 + 2 w x2 x3 y3
I’ve blocked this output into three sections each representing a polynomial. Setting the first of these polynomials
equal to zero gives the equation of the variety obtained by eliminating t and the reader can check
without too much difficulty that it is a conic.
22
}