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4.1. GCD IN ONE VARIABLE 117
This lemma is not very easy to use on its own, for it supposes that we know
the g.c.d. (or at least its leading coefficient) before we are able to check whether
the modular reduction has the same degree. But this leading coefficient has
to divide the two leading coefficients of A and B, and this gives a formulation
which is easier to use.
Corollary 10 If p does not divide the leading coefficients of A and of B (it
may divide one, but not both), then the degree of gcd(A p , B p ) is greater than or
equal to that of gcd(A, B).
As the g.c.d. is the only polynomial (to within an integer multiple) of its degree
which divides A and B, we can test the correctness of our calculations of the
g.c.d.: if the result has the degree of gcd(A p , B p ) (where p satisfies the hypothesis
of this corollary) and if it divides A and B, then it is the g.c.d. (to within an
integer multiple).
It is quite possible that we could find a gcd(A p , B p ) of too high a degree.
For example, in the case cited above, gcd(A 2 , B 2 ) = x + 1 (it is obvious that
x + 1 divides the two polynomials modulo 2, because the sum of the coefficients
of each polynomial is even). The following lemma shows that this possibility
can only arise for a finite number of p.
Lemma 6 Let C = gcd(A, B). If p satisfies the condition of the corollary above,
and if p does not divide Res x (A/C, B/C), then gcd(A p , B p ) = C p .
Proof. A/C and B/C are relatively prime, for otherwise C would not be the
g.c.d. of A and B. By the corollary, C p does not vanish. Therefore
gcd(A p , B p ) = C p gcd(A p /C p , B p /C p ).
For the lemma to be false, the last g.c.d. has to be non-trivial. This implies that
the resultant Res x (A p /C p , B p /C p ) vanishes, by proposition 58 of the Appendix.
This resultant is the determinant of a Sylvester matrix, and |M p | = (|M|) p , for
the determinant is only a sum of products of the coefficients. In the present case,
this amounts to saying that Res x (A/C, B/C) p vanishes, that is that p divides
Res x (A/C, B/C). But the hypotheses of the lemma exclude this possibility.
Definition 70 If gcd(A p , B p ) = gcd(A, B) p , we say that the reduction of this
problem modulo p is good, or that p is of good reduction. If not, we say that p
is of bad reduction.
This lemma implies, in particular, that there are only a finite number of values
of p such that gcd(A p , B p ) does not have the same degree as that of gcd(A, B),
that is the p which divide the g.c.d. of the leading coefficients and the p which
divide the resultant of the lemma (the resultant is non-zero, and therefore has
only a finite number of divisors). In particular, if A and B are relatively prime,
we can always find a p such that A p and B p are relatively prime. An alternative
proof that there are only finitely primes of bad reduction can be deduced from
Lemma 9.