Agoston M.K. Computer Graphics and Geometric Modelling: Mathematics

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10.3 More on Projective Space 689

10.3.6. Theorem. Let f Œ C[X

,...,X

] and V = V(f) Õ C

(1) The (projective) hypersurface V(H(f)) in P

H(V) of V.

(2) The projective completion H(V) of V is the topological closure of V in P

(C).

(3) Using the notation deﬁned by equations (10.23), j

-1

n+1

(V) = D(H(V)). (Basically,

this says that V is the afﬁne part of the projective completion of V.)

Proof. See [Kend77] or [Shaf94].

The next example shows that Theorem 10.3.6 is false if the ﬁeld is the reals. The

algebraic closure property of the complex numbers is essential.

10.3.7. Example. Consider

and let V = V(f) Õ R

. Figure 10.7(a) shows V. Note that the origin is an isolated

point of the graph. It is basically such an isolated point that will lead to our coun-

terexample but it will not be V directly because we need a variety that has its isolated

point at inﬁnity. To get this variety we simply move V. Consider the transformation

which moves the y-axis to the line at inﬁnity. This will transform

into

GXYZ YX Z ZX,,

()

=-+

232

Hf Y Z X XZ

()

=-+

232

TX Z

¢=

fXY Y X X,

()

=- -

()

V = V(f )

f(X,Y) = Y

– X

(X – 1) g(X,Y) = Y

X + X – 1

W = V(g)

(a) (b)

Figure 10.7. The varieties of Example 10.3.7.

690 10 Algebraic Geometry

with

Let W = V(g) Õ R

. Figure 10.7(b) shows W. The variety W is the counterexample we

are looking for. Note that G = H(g) and [1,0,0] Œ V(H(g)), but the topological closure

of W in P

is W » {[0,1,0]} (Exercise 10.3.4) and [1,0,0] is not in it. Although we do

not yet have the tools to prove this, it is the case that H(W) = V(H(g)) (Exercise 10.5.3).

We have found a counterexample to Theorem 10.3.6(2). Actually, W serves as our

counterexample no matter what might have happened because if it were the case that

H(W) π V(H(g)), then we would have violated Theorem 10.3.6(1) instead.

The reason that behavior as in Example 10.3.7 is impossible in complex projec-

tive space is that the algebraic closure of the complex numbers guarantees that equa-

tions have enough solutions to prevent the kind of isolated points that we found in

our example. Over the complex numbers the projective variety deﬁned by g(X,Y) will

be a pinched sphere and what was our isolated point over the reals will be the pinched

point which is no longer isolated (Exercise 10.3.5).

We shall have more to say about the projective completion of a variety at the end

of Section 10.8 once we have a little more algebra behind us.

10.4 Resultants

This section introduces an important tool for determining if polynomials in one vari-

able have a common factor. It will be needed later in several contexts.

Deﬁnition. Let

(10.24)

be two polynomials in D[X] of positive degrees m and n, respectively, where D is a

unique factorization domain. Deﬁne the Sylvester matrix SM(f,g) of f and g by

SM f g

aa a a

aa aa

aa a

bb b b

bb bb

bb b

()

110

LLL

OLOM

LLL

MO O LM

LLL

˜˜

fX a X a X a

gX b X b X b

()

=+ ++

()

=+ ++

... ,

...

gXY DG Y X X,.

()

=+-

(10.25)

10.4 Resultants 691

where we have n rows of a’s and m rows of b’s. The determinant of SM(f,g) is called

the (Sylvester) resultant of f and g and is denoted by R(f,g) (or R

(f,g) if we want to

emphasize the fact that f and g are polynomials in X in case a

and b

are themselves

polynomials in some other variables).

Note that the resultant is not a symmetric function, but it is easy to show from

basic properties of the determinant that R(g,f) = (-1)

R(f,g).

10.4.1. Lemma. If R = R(f,g) is the resultant of two polynomials f(X) and g(X) of

positive degree m and n, respectively, then

where F(X) and G(X) are polynomials with deg (F) < m and deg (G) < n.

Proof. For each i, 1 £ i £ m + n - 1, multiply the ith column of the Sylvester matrix

(10.25) by X

m+n-i

and add the result to the last column. This will produce the matrix

Let C

i,j

denote the cofactors of the matrix in (10.26). Since both of the matrices (10.25)

and (10.26) have the same determinant, expanding the determinant of (10.26) by the

last column gives that

Since the polynomials F(X) and G(X) have the desired properties, Lemma 10.4.1 is

proved.

10.4.2. Lemma. If R = R(f,g) is the resultant of two polynomials f(X) and

g(X) of positive degree m and n, respectively, then R = 0 if and only if there

exist nonzero polynomials F(X) and G(X) with deg (F) < m and deg (G) < n, so

that

(10.27)

GXfX FXgX

()()

= 0.

Rfg X fXC X fXC fXC

XgXC XgXC gXC

GXfX FXg

mn nmn

nmn

nmn mnmn

, ...

...

,,,

()

()()

()

++ ++

()

aa a a XfX

aa aa XfX

aa fX

bb b b XgX

bb bb XgX

()

110

LLL

OLOM

LLL

MO O LM

()

RGXfX FXgX=

()()

(10.26)

692 10 Algebraic Geometry

Proof. If R = 0, then Lemma 10.4.1 shows that there are polynomials F(X) and G(X)

of degree less than m and n, respectively, satisfying equation (10.27), but we still need

to show that they are nonzero. Let

and

Collecting coefﬁcients of powers of X in (10.27) and setting them equal to 0 gives the

following system of equations:

(10.28)

This system is equivalent to the equation

(10.29)

Since the resultant R, which is the determinant of SM(f,g), is zero, there is a non-

trivial solution and we are done.

Next, we prove the converse. Assume therefore that the nonzero polynomials

F(X) and G(X) exist. We are again led to equation (10.29). Since we have a non-trivial

solution, the determinant of SM(f,g), namely, R, must be 0. Lemma 10.4.2 is

proved.

10.4.3. Theorem. Two nonconstant polynomials f(X) and g(X) have a nonconstant

common factor if and only if R(f,g) = 0.

Proof. Let d(X) be the greatest common divisor of f(X) and g(X). It sufﬁces to show

that R = R(f,g) = 0 if and only if d(X) is a nonconstant polynomial.

First, assume that d(X) is a nonconstant polynomial. Then f(X) = d(X)H(X) and

g(X) = d(X)G(X). It follows that F(X) =-H(X) and G(X) satisfy equation (10.27) in

Lemma 10.4.2, and hence R = 0. Conversely, assume that R = 0. By Lemma 10.4.2,

there exist nonzero polynomials F(X) and G(X) satisfying

having degree less than m and n, respectively. It follows that f(X) divides g(X)F(X).

Since f(X) is not a constant polynomial, deg F < deg f, and g(X) is not the zero poly-

nomial, some prime factor of f must divide g(X). In other words, f(X) and g(X) have

GXfX FXgX

()()

= 0

dd dc c cSMfg

nn mm

-- --

[]

()

12 0 1 2 0

0... ... , .

da cb

00 00

0+=.

da da cb cb

10 01 10 01

0+++ =

da da c b c b

nm n m mn m n-- - -- -

++ +=

11 2 11 2

da c b

nm mn--

GX d X d X d

()

=+++

... .

FX c X c X c

()

=+++

...

10.4 Resultants 693

a common factor and their greatest common divisor is not constant. Theorem 10.4.3

is proved.

10.4.4. Corollary. Two polynomials f(X) and g(X) deﬁned over an algebraically

closed ﬁeld have a common root if and only if R(f,g) = 0.

Proof. Obvious.

The resultant can be used to check for multiple roots of a polynomial.

10.4.5. Corollary. A polynomial f(X) has a nonconstant factor of multiplicity larger

than 1 if and only if R(f,f ¢) = 0. In particular, f(X) has a multiple root if and only if

R(f,f ¢) = 0.

Proof. By Theorem B.8.10, f(X) has a nonconstant factor of multiplicity larger than

1 if and only if that factor also divides f ¢(X). Now use Theorem 10.4.3.

Deﬁnition. Let f(X) be a nonconstant polynomial. The resultant R(f,f¢) is called the

resultant of f.

Here are two useful formulas for resultants that are worth stating explicitly.

10.4.6. Example. The resultant of the polynomials a

X + a

and b

X + b

10.4.7. Example. The resultant of the polynomials a

+ a

X + a

and b

+ b

10.4.8. Theorem. Let

be two polynomials where the a

and b

are homogeneous polynomials of degree

m - i and n - j, respectively, in the variables X

, X

,..., X

and a

π 0. Then the

resultant R(X

,...,X

) of f and g (with respect to X) is either identically equal to

0 or a homogeneous polynomial of degree mn.

gX b X b X b

()

=+ ++

...

fX a X a X a

()

=+ ++

... ,

aaa

bbb

aaa

bbb

aaa

bbb

210

=- = - .

694 10 Algebraic Geometry

Proof. The resultant, or determinant of the Sylvester matrix, consists of a sum of

signed terms, each one of which is an (m + n)-fold product of factors. If the term is

nonzero, then n factors are a

’s and m factors are b

’s. Since the degree of a nonzero

(i,j)th element in the ﬁrst n rows is j - i and the degree of a nonzero (s,t)th element

in the last m rows is t - s, the total degree of the term is a sum of the form

where the sets of indices A and B are a partition of {1,2, ...,m + n}. This means that

the total degree is

10.4.9. Theorem. Let R be the resultant of the two polynomials

where m, n > 0. Then

(10.30)

(10.31)

(10.32)

Proof. Let us assume that a

, b

, r

, and s

are indeterminates. Expressing f(X) and

g(X) as in (10.24) we see that

(10.33)

where s

and t

are elementary symmetric polynomials in the r

’s and s

’s and of degree

n and m, respectively. From this we see that

Fact 1. a

divides R.

Next, we show

Fact 2. r

- s

divides R for 1 £ i £ m and 1 £ j £ n.

bb jn

()

££11t ,,

aa im

()

££11s ,,

() ()

’

bfs.

()

’

agr

Rab rs

()

’’

gX b X s X s X s

()

... ,

fX a X r X r X r

()

... ,

1mnmn mm nn mn+

()

= .

ji ts

jA s

()

=Œ=Œ

ÂÂÂÂ

10.5 More Polynomial Preliminaries 695

Observe that if we replace r

by s

then f(X) and g(X) have a common factor X - s

By Theorem 10.4.3, R vanishes after this substitution and hence Fact 2 is proved.

Fact 2 and the fact that the ring of polynomials k[a

,...,r

,...,s

] over

any ﬁeld k is a prime factorization domain imply that the right-hand side of equation

(10.30) divides R. On the other hand, (10.33) implies that R is homogeneous of degree

mn with respect to r

’s and s

’s. Therefore,

(10.34)

for some constant c. To determine c, set all the r

to 0. In that case, a

= a

= ...= a

= 0 and so the determinant of the Sylvester matrix (10.25), namely, R, is a

. This

fact and (10.34) implies that

(10.35)

The right-hand side of (10.35) is just

Therefore, c = 1 and (10.30) is proved. Equalities (10.31) and (10.32) follow easily, and

so the theorem is proved.

Theorem 10.4.8 leads to another useful formula for evaluating resultants.

10.4.10. Corollary. Given polynomials f(X), g(X), and h(X), then the resultants

satisfy the product identity

Proof. Exercise 10.4.2. For an alternate proof see [Seid68].

The resultant we have deﬁned above addresses the problem of ﬁnding the common

zeros of two polynomials. There are times when one is interested in the common zeros

of a larger collection of polynomials. For properties and applications of the corre-

sponding multipolynomial resultant see, for example, [CoLO98].

10.5 More Polynomial Preliminaries

Polynomials are the glue which holds the various aspects of algebraic geometry

together. This section summarizes some additional important basic facts about poly-

nomials and the hypersurfaces they deﬁne.

Now, we saw in Section 10.2 that if one wanted to understand curves it was impor-

tant to look at these in projective space (in fact, complex projective space) since one

Rfgh Rfh Rgh,,,.

()

()( )

()

ca b s ca b .

ab cab s

()

’

Rcab rs

()

’’

696 10 Algebraic Geometry

needs the points “at inﬁnity.” To do this one must express curves in homogeneous

coordinates and via homogeneous polynomials. This means that it is important that

one can relate properties of polynomials with their homogenized versions and vice

versa. The next two results basically say that the answers to questions about factor-

ization of polynomials for varieties are the same for projective varieties and their afﬁne

counterparts and so we may permit ourselves to not explicitly state which type of

variety we are talking about in such cases.

10.5.1. Proposition. Any factor of a homogeneous polynomial is homogeneous.

Proof. Exercise.

10.5.2. Theorem. Let F be a homogeneous polynomial and f = D(F) its

dehomogenization.

(1) Each factor of F dehomogenizes to a factor of f and conversely each factor of

f homogenizes to a factor of F.

(2) F is irreducible if and only if f is.

Proof. Exercise.

10.5.3. Proposition. Let k be an algebraically closed ﬁeld and let f(X,Y) be a homo-

geneous polynomial of degree d in k[X,Y]. Then f can be factored into linear factors,

that is, f has the form

Proof. Write f in the form

where g(Z) is a polynomial of degree d in k[Z]. Since k is algebraically closed, g (as a

polynomial in Z) factors into linear factors. Replacing Z by Y/X in this factorization

and simplifying gives the result.

For example,

Note that we could equally well have put things in terms of X/Y rather than Y/X,

because

XXYYX

YX YX

222

32 132

12 1

-+=-+

()

fXY X g

()

fXY aY bX a b k

ii ii

,,,.

()

’

10.5 More Polynomial Preliminaries 697

10.5.4. Proposition. Let (a

,...,a

) Œ k

. Every polynomial f Œ k[X

,...,X

]

can be written in the form

where b Œ k, f

Œ k[X

,...,X

Proof. The proposition follows from the fact that

so that k[X

,...,X

] can be thought of as a polynomial ring in X

- a

,...,

- a

10.5.5. Theorem. If k is an algebraically closed ﬁeld, then any nonconstant poly-

nomial f in k[X

,...,X

], n > 1, has an inﬁnite number of zeros.

Proof. We shall sketch a proof for the case n = 2 and leave the general case to the

reader. Since f is not constant, we may assume without loss of generality that f(X,Y)

is of the form

where r > 0 and a

(Y) π 0. If the degree of a

(Y) is s, then a

(Y) has at most s roots.

Since k is inﬁnite, there are inﬁnitely many c in k so that a

This equation has a root b because k is algebraically closed. By deﬁnition, f(b,c) = 0.

To ﬁnish the proof one simply needs to show that the inﬁnite number of choices for

c lead to an inﬁnite number of solutions to f(X,Y) = 0.

10.5.6. Theorem. Let k be an arbitrary ﬁeld, f an irreducible polynomial in

k[X

,...,X

], and g an arbitrary polynomial in k[X

,...,X

]. If f does not divide g,

then there are only a ﬁnite number of solutions to the equations

Proof. Again, we shall only give a proof for the case n = 2 and leave the general case

to the reader. By hypothesis f cannot be constant and so we may assume without loss

fX X gX

0,..., ,. ...,

()

acX a cX ac

()

0... .

fXY a YX a YX Y Y Y

aak, ... , ,

()

() ()

[]

+Œ

kX X X kX a X a X a

nnn12 1 12 2

, ,..., , ,... ,

[]

=- - -

[]

fbX af X af

nnn

=+ -

()

++ -

()

111

... ,

XXYYY

XYX Y

222

32 32

-+=

()

698 10 Algebraic Geometry

of generality that f(X,Y) contains a positive power of X. Consider f to belong to

k(Y)[X].

Claim. f is irreducible in k(Y)[X].

To prove the claim, suppose that f factors in k(Y)[X], that is, f = f

, where f

k(Y)[X]. The coefﬁcients of the f

are rational functions in Y. That means that there is

some polynomial a(Y) (we could use the product of all the denominators) so that

af = h

, h

Œ k[Y][X] = k[X,Y]. This easily contradicts the irreducibility of f, and so

the claim is proved.

Now, as polynomials in X over the ﬁeld k(Y), f and g are relatively prime. It follows

that there are polynomials u

, u

Œ k(Y)[X], so that u

f + u

g = 1. Again, multiplying

through by an appropriate polynomial in Y, this equation can be transformed into

an equation v

f + v

g = w, where v

, v

Œ k[X,Y] and w(Y) Œ k[Y]. If f(a,b) = g(a,b) =

0, then w(b) = 0. But w(Y) has only a ﬁnite number of roots. For each root b of

w, consider f(X,b) = 0. This is either identically equal to zero or has only a ﬁnite

number of roots. It cannot be identically equal to zero, because if it were, then Y - b

would divide f, which is impossible. This clearly proves that f and g have only a

ﬁnite number of roots in common and ﬁnishes the proof of Theorem 10.5.6 for the

case n = 2.

Theorem 10.5.6 implies a special case of the Hilbert Nullstellensatz, which is

proved in Section 10.8.

10.5.7. Corollary. Let k be an algebraically closed ﬁeld, let f be an irreducible

polynomial in k[X

,...,X

], and let g be an arbitrary polynomial in k[X

,...,X

]. If

g vanishes wherever f does, then f divides g.

Proof. First, assume that f has degree 0, that is, f Œ k. If f π 0, then the result is

obvious since f is invertible. If f = 0, then g vanishes everywhere and by Theorem

B.11.12 must be 0.

Now assume that f has degree r, r > 0. Since k is algebraically closed, f and g have

an inﬁnite number of common zeros by Theorem 10.5.5. If f did not divide g, then we

would have a contradiction to Theorem 10.5.6.

10.5.8. Theorem. Consider a hypersurface V(f) (either afﬁne or projective) deﬁned

by a polynomial f in C[X

,...,X

] of the form

(10.36)

where the f

are irreducible and nonassociates and n

> 0. If V(f) = V(g) for some poly-

nomial g in C[X

,...,X

], then g has the form

where m

> 0 and c is a constant.

Proof. Write

gcff f

... ,

fff f

... ,