SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

30 I. R. Shafarevichand the formula of division with remainder (formula (2)) the formf 1 = f 2 q 1 ; f 3(we have denoted here q by q 1 ). Now itisclearhow to apply formula (4), reducingdegrees of polynomials considered. Starting from f 1 and f 2 dene polynomials f iby induction:(5) f i;1 = f i q i;1 ; f i+1 where the degree of f i+1 is smaller than the degree of f i (assuming that f i;1 andf i are already dened). Clearly, f i;1 are just those plynomials which appear as remaindersin the Euclid's algorithm, only with the changed signs. After several stepswe come to a polynomial f k ,dieringeventually only by sign with the gcd(f 1 f 2 ).Applying formula (4) to f 2 and f 3 instead to f 1 and f 2 , we obtain that(f 2 f 3 ) b a = (f 3f 4 ) b a +[f 2f 3 ] b a . Substituting this value for (f 2f 3 ) b a into formula(4), we get(f 1 f 2 ) b a =(f 3 f 4 ) b a +[f 1 f 2 ] b a +[f 2 f 3 ] b a:Repeating this process k times and noting that [f k f k+1 ] b a = 0 as a result we obtain:(6) (f 1 f 2 ) b a =[f 1f 2 ] b a +[f 2f 3 ] b a + +[f k;1f k ] b a :However, in order that we have the right to apply formula (4), we have to assumethat f i (a) 6= 0,f i (b) 6= 0 for all i =1 2 ...k.Consider carefully the expression [fg] b a which canbeevaluated using Table 1for F = fg. In our case it can be rewritten as8>< 0 if f(a)g(a) > 0andf(b)g(b) > 0, or f(a)g(a) < 0andf(b)g(b) < 0,[fg] b a = 1 if f(a)g(a) < 0andf(b)g(b) > 0,>:;1 if f(a)g(a) > 0andf(b)g(b) < 0.Table 2If two numbers A and B, distinct from 0, are given, then one says that in thepair (A B) there exists one change of sign if A and B are of opposite signs, andthat there is no change of sign if they are of the same sign. Using this terminology,one can reformulate information of Table 2, denoting by n the number of changesof sign in the pair (f(a)f(b)) and by m the number of changes of sign in the pair(f(b)g(b)). Table 2 obtains the form:[fg] b a m n0 0 00 1 11 1 0;1 0 1We see that in all the cases we have [fg] b a = m ; n.