SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

32 I. R. Shafarevichthe sequence f 1 (a 0 ), ... , f k (a 0 ), as well as in the sequence f 1 (b 0 ), ... , f k (b 0 ), isdetermined by the Lemma. Thus, we obtain the wanted result:THEOREM 3. If polynomials f 1 and f 2 have no common roots, f 1 (a) 6= 0andf 1 (b) 6= 0, then the characteristics (f 1 f 2 ) b a is equal to the dierence between thenumbers of changes of sign in the sequence f 1 (a), ... , f k (a) and in the sequencef 1 (b), ... , f k (b), where f 1 (x), ... , f k (x) is Sturm's sequence corresponding to thepair of polynomials f 1 , f 2 .We shall show now that the Lemma is valid. Consider, for example, the valuex = a. Suppose that f i (a) =0for some i =1 ...k. By the assumption, i 6= 1since f 1 (a) 6= 0. Also, i 6= k since the polynomial f k (x) can dier from gcd(f 1 f 2 )only by sign and so it is a number distinct from 0. Note that then f i;1(a) 6= 0and f i+1 (a) 6= 0. Really, if we had,for example, f i (a) =0,f i+1 (a) =0, then itwould follow from formula (5) that f i;1(a) = 0. In exactly the same way, thiswould imply that f i;2(a) = 0 etc., and nally f 1 (a) = 0, which would contradictthe original assumption. But we can say even more|not only that the numbersf i;1(a) and f i+1 (a) are distinct from 0, but they have opposite signs|it followsimmediately by substituting x = a into equality (5) and taking into account theassumption that f i (a) =0.Compare now the sequences f 1 (a), ... , f k (a) and f 1 (a 0 ), ... , f k (a 0 ). Letf i (a) = 0. Then, as we have seen, f i;1(a) 6= 0 and f i+1 (a) 6= 0, and f i;1(a)and f i+1 (a) have opposite signs. But then f i;1(a 0 ) 6= 0 and f i+1 (a 0 ) 6= 0, andf i;1(a 0 ) has the same sign as f i;1(a), while f i+1 (a 0 ) has the same sign as f i+1 (a).This follows from the fact that the polynomials f i;1 and f i+1 have no roots in thesegment [a a 0 ], and so (by Bolzano's theorem) they can have novalues of oppositesigns. Write down the respective parts of our sequences. Suppose that f i;1(a) > 0.Then we obtain the following table:f i;1(x) f i (x) f i+1 (x)x = a + 0 ;x = a 0 + ? ;The characteristics (f 1 f 2 ) b0 a0 depends on the number of changes of sign in the lowestrow. But we see that it coincides with the number of changes of sign in the rowabove it|whatever the unkown sign, denoted by ?, is, there will be exactly onechange of sign in each of the rows. The case when f i;1(a) < 0 can be treatedexactly in the same way. The Lemma is proved.Combining Theorem 3 with Theorem 1 we obtain the basic result:THEOREM 4. (Sturm's Theorem) If a polynomial f(x) has no multiple rootsand does not vanish for x = a and x = b, then the number of its roots in the segment[a b] is equal to the dierence between the number of changes of sign of the valuesof polynomials in the Sturm's sequence, formed for the polynomials f(x) and f 0 (x)at x = a and x = b.