SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

Selected chapters from algebra 11Since by supposition 3 + a 2 + b + c = 0, then x 3 + ax 2 + bx + c =(x ; )(x 2 +(a + )x + b + a + 2 ). According to our denition the equationx 3 + ax 2 + bx + c =0hastwo roots equal to if is a root of the equation andalso if is a root of the polynomial x 2 +(a + )x + b + a + 2 . In other words, 2 +(a + ) + b + a + 2 =0,i.e.3 2 +2a + b =0. We see that a multipleroot of the equation x 3 + ax 2 + bx + c = 0 is the common root of the polynomialsx 3 + ax 2 + bx + c and 3x 2 +2ax + b. As we saw in Section 1, they are the roots ofthe polynomial g: c: d:(x 3 + ax 2 + bx + c 3x 2 +2ax + b) and the greatest commondivisor can be found by Euclid's algorithm.Wenow apply the same reasoning to the polynomial f(x) =a 0 +a 1 x++a n x nof arbitrary degree. When we divide f(x) by x ; we obtain as the quotient apolynomial g(x) of degree n ; 1 whose coecients depend on and so we shalldenote it by g(x ). We know (formula (3)) that the remainder is f():(12) f(x) =(x ; )g(x )+f():Putting x = into the polynomial g(x ) we obtain the polynomial in which iscalled the derivative of f(x) and is denoted by f 0 (). Hence, by denition,f(x)(13) f 0 ; f()() = ():x ; The above formula may cause some doubt, since after the substitution x = bothf(x) ; f()the numerator and the denominator in the expression become 0 and wex ; get 0 . This formula therefore needs to be explained: we rst (before substituting0x = ) divide the numerator by the denominator and we substitute x = intotheir quotient whichisapolynomial. For example, the meaning of the expressionx 2 ; 1x ; 1 (1) is: we rst get x2 ; 1= x + 1, and then (x + 1)(1) = 2.x ; 1Those of you who will continue to study mathematics will meet the derivativefor other functions, such asf(x) = sin x or f(x) =2 x . In essence they are denedby the same formula (13), but in general case it is more dicult to give the exactsense to the expression on right-hand side. In the case of polynomials everythingis cleared by applying Bezout's theorem to the polynomial f(x) ; f().If is a root of the polynomial f(x) in (12), i.e. if f() = 0, then we getf(x) =(x ; )g(x ) and by our denition is a multiple root of f(x) if is aroot of g(x ), i.e. if g( ) =0. But this means that f 0 () =0. We have provedthe assertion:THEOREM 5. Aroot of a polynomial f(x) is multiple if and only if it is alsoaroot of the derivative f 0 (x).We see that a multiple root is the common root of the polynomials f(x)and f 0 (x). In other words, isarootofg: c: d:(f(x)f 0 (x)) the greatest commondivisor can be found by Euclid's algorithm and it is, as a rule, a polynomial ofmuch smaller degree.