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140 CHAPTER 4. MODULAR METHODS
Proposition 46 If M is a n × n matrix whose entries are polynomials in
x of degree at most d, then the degree of det(M) is at most nd.
In the integer case, the fact that we are adding n! products makes the
estimate more subtle: see Proposition 66.
4. How do we combine the results — one of Algorithms 39 (integer) or 40
(polynomial).
5. How do we check the result: irrelevant.
4.4.2 Resultants and Discriminants
The resultant of two polynomials f = ∑ n
i=0 a ix i and g = ∑ m
i=0 b ix i is defined
(Definition 89) as the determinant of a certain (m+n)×(m+n) matrix Syl(f, g).
Hence if Syl(f, g)| p = Syl(f| p , g| p ) (or the equivalent for | y=v ), the theory of the
previous section holds. But the only way this can fail to hold is if Syl(f| p , g| p ) is
no longer an (m + n) × (m + n) matrix, i.e. if a n or b m evaluate to zero. Hence
again we have simple answer to the questions on page 114.
1. Are there ”good” reductions from R: yes — all evaluations which do not
reduce a n b m to zero,
2. How can we tell if R i is good?: in advance by checking that a n b m does
not reduce to zero.
3-5 As above.
4.4.3 Linear Systems
We consider the matrix equation
M.x = a, (3.13)
and the case of M and a having integer entries, and reduction modulo a prime.
Conceptually, this has the well-known solution
x = M −1 .a. (3.14)
This formulation shows the fundamental problem: M −1 , and x itself, might not
have integer entries. In fact, if p divides det(M), (M| p ) −1 does not exist. There
are two solutions to this problem.
4.4.3.1 Clear Fractions
If we compute det(M) first, by the method of section 4.4.1, we can clear denominators
in (3.14), and get
̂x := det(M)x = det(M)M −1 .a, (4.16)