Algorithms in Real Algebraic Geometry, Second Edition by Saugata Basu, Richard Pollack, Marie-Francoise Roy,

By Saugata Basu, Richard Pollack, Marie-Francoise Roy,

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17. 11 is nothing but an algebraic proof of the fundamental theorem of algebra. 18. [Modulus] If R is real closed, and R[i] = R[T ]/(T 2 + 1), we can identify R[i] with R2. For z = a + i b ∈ R[i], a ∈ R, b ∈ R, we define the conjugate of z by z = a − i b. The modulus of z = a + i b ∈ R[i] √ is |z | = a2 + b2 . 19. Let R be a real closed field, P ∈ R[X]. The irreducible factors of P are linear or have the form (X − c)2 + d2 = (X − c − i d)(X − c + i d), d 0 with c, d ∈ R. 11 and that the conjugate of a root of P is a root of P .

A0 and B = b qX q + + b0 be two normal Proof: Let A = a p X p + polynomials. e. that all the coefficients of A and B are positive. Let C = A B = c p+ q X p+ q + + c0. It is clear that all the coefficients of C are positive. It remains to prove that c2k ck−1 ck+1. 2 Real Root Counting Using the partition of (h, j) ∈ Z2 h > j and {(h, h − 1) h ∈ Z}. 49 in (j + 1, h − 1) ∈ Z2 h F F j F c2k − ck−1 ck+1 = ah a j bk−h bk− j + h j ah a j bk−h bk− j h>j − ah a j bk−h+1 bk− j −1 − h j = ah a j bk−h bk− j + h j a j +1 ah−1 bk− j −1 bk −h−1 h + ah a j bk −h+1 bk −j −1 h> j j ah ah−1 bk−h bk−h+1 − ah ah−1 bk−h+1bk−h h h − ah a j bk−h+1 bk− j −1 − h = j a j +1 ah−1bk −j bk−h h j (ah a j − ah−1 a j +1) (bk −j bk−h − bk− j −1 bk −h+1).

17, and using the corresponding notation, the elements of posgcd(P , P ) are (after removing obviously irrelevant factors), (S 4, (S 3, (Tru0(S 3), (S 2, (u, (Tru1(S 2), (Tru0(S 2), (P , a 0 ∧ s 0 ∧ δ 0), a 0 ∧ s 0 ∧ δ = 0), a 0 ∧ s = 0 ∧ t 0), a 0 ∧ s = t = 0), a = 0 ∧ b 0 ∧ u 0), a = 0 ∧ b 0 ∧ u = 0), a = b = 0 ∧ c 0), a = b = c = 0). e. if the degrees of the polynomials in the remainder sequence are 4, 3, 2, 1, 0), then gcd (P , P ) = S 4. e. if the degrees of the polynomials in the remainder sequence are 4, 3, 2, 1), then gcd(P , P ) = S 3.

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