Graphics Reference
In-Depth Information
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
1
f
2
, where f
i
Œ
k(Y)[X]. The coefficients of the f
i
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
1
h
2
, h
i
Πk[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
1
, u
2
Πk(Y)[X], so that u
1
f + u
2
g = 1. Again, multiplying
through by an appropriate polynomial in Y, this equation can be transformed into
an equation v
1
f + v
2
g = w, where v
1
, v
2
Πk[X,Y] and w(Y) Πk[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
1
,...,X
n
], and let g be an arbitrary polynomial in k[X
1
,...,X
n
]. 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
1
,...,X
n
] of the form
nn
n
k
fff
=
12
...
f
,
(10.36)
12
k
where the f
i
are irreducible and nonassociates and n
i
> 0. If V(f) = V(g) for some poly-
nomial g in
C
[X
1
,...,X
n
], then g has the form
mm
m
k
gcff
=
12
...
f
,
12
k
where m
i
> 0 and c is a constant.
Proof.
Write
