By Antonio Machì (auth.)
This ebook bargains with numerous themes in algebra priceless for desktop technology functions and the symbolic remedy of algebraic difficulties, stating and discussing their algorithmic nature. the subjects lined variety from classical effects akin to the Euclidean set of rules, the chinese language the rest theorem, and polynomial interpolation, to p-adic expansions of rational and algebraic numbers and rational services, to arrive the matter of the polynomial factorisation, particularly through Berlekamp’s approach, and the discrete Fourier rework. easy algebra techniques are revised in a sort fitted to implementation on a working laptop or computer algebra system.
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5 (Chinese remainder theorem for polynomials). Let m0 (x), m1 (x), . . , mn (x), be pairwise coprime polynomials, and let u0 (x), u1 (x), . . , un (x) be arbitrary polynomials. Then there exists a polynomial u(x) such that u(x) ≡ uk (x) mod mk (x), k = 0, 1, . . , n, and this u(x) is unique modulo the product m(x) = m0 (x)m1 (x) · · · mn (x). 3 Polynomials 23 Let us prove a lemma ﬁrst. 1. Let f , g and h be three polynomials with ∂h < ∂f + ∂g, and suppose that, for suitable polynomials a and b, the relation af + bg = h holds.
So we notice that all the coeﬃcients of the powers pi with i > k are equal to p − 1. ♦ Examples. 1. Determine the p-adic expansion of 1 . 1 we have: 1 · (1 − p) + 1 · p = 1, and hence: 1 1 =1+ p, 1−p 1−p and c0 = 1. From this follows 1 1 1 2 = 1 + (1 + p)p = 1 + p + p , 1−p 1−p 1−p and c1 = 1. Going on like this, or by induction, we ﬁnd: 1 = 1 + p + p2 + · · · . 1−p 2. More generally, let us expand 1 , 1 − pn this time by running the algorithm. We have: 1 ≡ (1 − pn )c0 = c0 − pn c0 mod p, and hence c0 = 1.
4. A series expansion modulo p in which the coeﬃcients from some point on are all equal to p − 1 represents a negative integer number, and conversely. 1 Expansions of rational numbers 45 Proof. c1 c2 . . ck p − 1 p − 1 . . be such an expansion. We may suppose that it is a reduced expansion, since a reduction procedure, if any, just moves rightwards the beginning of the repeating part. So, −x = −c0 − c1 p − · · ·−ck pk +(1−p)pk+1 +(1−p)pk+2 +· · · . Reducing, we get · · ·+(p−ck −1)pk + (1−p−1)pk+1 +· · · = · · ·+0·pk+1 +(1−p−1)pk+2 +· · · and so on, so ci = 0 for i > k.
Algebra for Symbolic Computation by Antonio Machì (auth.)