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126 CHAPTER 4. MODULAR METHODS
This notation is analogous to the previous section, where A p was the remainder
on dividing A by p.
Observation 8 Clearly R[x, y] ≡ R[x][y] also, and the definition of g.c.d. is
independent of this choice. Algorithmically, though, it seems as if we must make
such a choice. Some systems may already have imposed a default choice, but if
we have a free hand it is usual to choose as the main variable (x in Notation
19) the one which minimises min(deg(A), deg(B)).
4.2.2 The evaluation–interpolation relationship
In this sub-section, we answer a question analogous to that in section 4.1.2:
what do we do if the g.c.d. of the evaluations is not the image under evaluation
of the g.c.d. calculated before evaluation?
Lemma 7 If y−v does not divide the leading coefficient of gcd(A, B), the degree
of gcd(A y−v , B y−v ) is greater than or equal to that of gcd(A, B).
Proof. Since gcd(A, B) divides A, then (gcd(A, B)) y−v divides A y−v . Similarly,
it divides B y−v , and therefore it divides gcd(A y−v , B y−v ). This implies that the
degree of gcd(A y−v , B y−v ) is greater than or equal to that of gcd(A, B) y−v .
But the degree of gcd(A, B) y−v is equal to that of gcd(A, B), for the leading
coefficient of gcd(A, B) does not cancel when it is evaluated at v.
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 11 If y − v does not divide the leading coefficients of A and of B
(it may divide one, but not both), then the degree of gcd(A y−v , B y−v ) is greater
than or equal to that of gcd(A, B).
As the g.c.d. is the only polynomial (to within a multiple from R[y]) of its degree
(in x) 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 y−v , B y−v ) (where v satisfies the
hypothesis of this corollary) and if it divides A and B, then it is the g.c.d. (to
within a multiple from R[y]).
As in section 4.1.2, it is quite possible that we could find a gcd(A y−v , B y−v )
of too high a degree: consider A = x − 1, B = x − y and the evaluation y ↦→ 1.
The following lemma shows that this possibility can only arise for a finite number
of v.
Lemma 8 Let C = gcd(A, B). If v satisfies the condition of the corollary above,
and if y − v does not divide Res x (A/C, B/C), then gcd(A y−v , B y−v ) = C y−v .
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 y−v does not vanish. Therefore
gcd(A y−v , B y−v ) = C y−v gcd(A y−v /C y−v , B y−v /C y−v ).