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132 CHAPTER 4. MODULAR METHODS
4. that we can use single-word primes;
the running time of the ‘modular first’ (Figure 4.10) algorithm is
O ( l 2 (d + 1) n + (d + 1) n+1) .
This contrasts with the subresultant algorithm’s bound [op. cit. (80)] (assuming
the p.r.s. are normal) of
(
O l 2 (d + 1) 4n 2 2n2 3 n) .
The real challenge comes with the potential sparsity of the polynomials.
The factor (d + 1) n in the running time comes simply from the fact that a dense
polynomial of degree d in n variables has that many terms, and we will need
(d + 1) n−1 univariate g.c.d.s (therefore (d + 1) n coefficients) to deduce these
potential terms.
4.3.1 A worked example
Let f be
x 5 y 7 z 4 + x 4 y 8 z 4 + x 4 y 7 z 4 + x 3 y 7 z 4 + x 2 y 8 z 4 + x 5 y 3 z 5 + x 4 y 4 z 5 +
x 3 y 4 z 6 + x 2 y 7 z 4 + x 2 y 5 z 6 − x 5 y 7 + x 5 yz 6 − x 4 y 8 + x 4 y 3 z 5 +
x 4 y 2 z 6 − x 3 y 4 z 5 − x 2 y 5 z 5 + x 2 y 4 z 6 − x 4 y 7 + x 4 yz 6 − x 2 y 4 z 5 − x 3 y 7 −
x 2 y 8 − x 2 y 7 − x 5 y 3 − x 4 y 4 − x 4 y 3 − x 5 y − x 4 y 2 − x 4 y + x 3 z+
x 2 yz − x 3 − x 2 y + x 2 z − x 2 + xz + yz − x − y + z − 1
(4.6)
and g be
x 6 y 7 z 4 − x 4 y 7 z 4 + x 6 y 3 z 5 + x 4 y 7 z 3 + x 4 y 4 z 6 − x 6 y 7 + x 6 yz 6 + x 4 y 7 z 2
−x 4 y 4 z 5 − 2 x 2 y 7 z 4 + x 4 y 7 z − 2 x 4 y 3 z 5 + x 2 y 7 z 3 − 2 x 2 y 4 z 6 + 2 x 4 y 7 +
x 4 y 3 z 4 − 2 x 4 yz 6 + x 2 y 7 z 2 + 3 x 2 y 4 z 5 + x 4 y 3 z 3 + x 4 yz 5 + x 2 y 7 z−
x 6 y 3 + x 4 y 3 z 2 + x 4 yz 4 + 3 x 2 y 7 + x 4 y 3 z + x 4 yz 3 − x 6 y + 3 x 4 y 3 +
x 4 yz 2 + x 4 yz + 3 x 4 y + x 4 z − x 4 − x 2 z + 2 x 2 − 2 z + 3
(4.7)
polynomials with 42 and 39 terms respectively (as against the 378 and 392 they
would have if they were dense of the same degree).
Let us regard x as the variable to be preserved throughout. If we compute
gcd(f| z=2 , g| z=2 ) (which would require at least 4 8 y values) we get (assuming
that z = 2 is a good reduction)
gcd(f, g)| z=2 = ( 15 y 7 + 31 y 3 + 63 y ) x 4 + ( 15 y 7 + 32 y 4 + 1 ) x 2 + 1. (4.8)
We note that each coefficient (with respect to x) has at most three terms.
A dense algorithm would compute five more equivalents of (4.8), at different
4 In fact, with y = 0 both f and g drop in x-degree, whereas with y = −2 we get a g.c.d.
with x-degree 5, as y = −2 is a root of the resultant in Lemma 8.