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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
()
=
m
-
1
m
-
2
FX
c
X
+
c
X
+
...
+
c
m
-
1
m
-
2
0
and
()
=
n
-
1
n
-
2
GX
d
X
+
d
X
+
...
+
d
.
n
-
1
n
-
2
0
Collecting coefficients of powers of X in (10.27) and setting them equal to 0 gives the
following system of equations:
da c b
+
=
0
nm mn
-
1
-
1
da
+
d a c b
+
+
c b
=
0
nm
-
1
-
1
n m mn
-
2
-
1
-
1
m n
-
2
L
da
+
d a
+
cb
+
c b
=
0
10
01
10
01
da
+
cb
=
0
.
(10.28)
00
00
This system is equivalent to the equation
T
[
]
()
=
dd
...
dc c
...
cSMf g
,
0
.
(10.29)
nn
-
1
-
2
0
mm
-
1
-
2
0
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
()()
+
()()
= 0
GXfX
FXgX
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
