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114 CHAPTER 4. MODULAR METHODS
as being in R and check that it is indeed the right result. The key questions are
then the following.
1. Are there domains R i for which the behaviour of the computation in R i
is sufficiently close to that in R for us to make deductions about the computation
in R from that in R i ? Such R i will generally be called “good”.
2. Can we tell, either immediately or with hindsight, whether an R i is
“good”? It will often turn out that we can’t, but may be able to say
that, given R i and R j , one of them is definitely “bad”, and preferably
which.
3. How many reductions should we take? In practice we will only count
“good” reductions, so this question is bound up with the previous one.
4. How do we combine the results from the various R i ? The answer to this
will often turn out to be a variant of the Chinese Remainder Theorem.
5. How do we check the result? Can we be absolutely certain that this result
is in fact the answer to the original question? In category speak, does
Figure 4.1 commute?
4.1 Gcd in one variable
Let us consider Brown’s example (from which page 46 was modified):
A(x) = x 8 + x 6 − 3x 4 − 3x 3 + 8x 2 + 2x − 5; (4.1)
B(x) = 3x 6 + 5x 4 − 4x 2 − 9x + 21. (4.2)
Let us suppose that these two polynomials have a common factor, that is a
polynomial P (of non-zero degree) which divides A and B. Then there is a
polynomial Q such that A = P Q. This equation still holds if we take each
coefficient as an integer modulo 5. If we write P 5 to signify the polynomial P
considered as a polynomial with coefficients modulo 5, this equation implies that
P 5 divides A 5 . Similarly, P 5 divides B 5 , and therefore it is a common factor 1 of
A 5 and B 5 . But calculating the g.c.d. of A 5 and B 5 is fairly easy:
A 5 (x) = x 8 + x 6 + 2x 4 + 2x 3 + 3x 2 + 2x;
B 5 (x) = 3x 6 + x 2 + x + 1;
C 5 (x) = remainder(A 5 (x), B 5 (x)) = A 5 (x) + 3(x 2 + 1)B 5 (x) = 4x 2 + 3;
D 5 (x) = remainder(B 5 (x), C 5 (x)) = B 5 (x) + (x 4 + 4x 2 + 3)C 5 (x) = x;
E 5 (x) = remainder(C 5 (x), D 5 (x)) = C 5 (x) + xD 5 (x) = 3.
Thus A 5 and B 5 are relatively prime, which implies that P 5 = 1. As the leading
coefficient of P has to be one, we deduce that P = 1.
1 Note that we cannot deduce that P 5 = gcd(A 5 , B 5 ): a counter-example is A = x − 3,
B = x + 2, where P = 1, but A 5 = B 5 = x + 2, and so gcd(A 5 , B 5 ) = x + 2, whereas P 5 = 1.