SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

20 I. R. Shafarevichgreatest degree of all monomials which comprise a polynomial is the degree of thepolynomial. For example, the degree of the polynomial 2x 3 ; 3x + 7 is 3 and itscoecients are a 0 =7,a 1 = ;3, a 2 =0,a 3 =2. Thus, a polynomial of degree ncan be written in the formf(x) =a 0 + a 1 x + a 2 x 2 + + a n x n where some of a k 's can be zero, but a n 6=0,because otherwise the degree of thepolynomial would be less than n. The term a 0 is called the constant term of thepolynomial, a n is its leading coecient. The equation f(x) =0iscalledanalgebraicequation with one unknown. Anumber is called its root if f() =0. Arootofthe equation f(x) =0isalsocalledaroot of the polynomial f(x). Degree of thepolynomial f(x) is the degree of the equation. Obviuosly, the equations f(x) =0and cf(x) = 0, where c is a number, distinct from 0, are equivalent.Now we shall treat such equations f(x) = 0 whose coecients a 0 , a 1 , ... ,a n are rational numbers, some of which may be equal to 0 or negative. If c isthe common denominator of all, distinct from 0, coecients, we can pass from theequation f(x) = 0 to the equation cf(x) = 0 having integer coecients. In thesequel we shall treat only such equations. In this connection we shall have to dealwith the divisibility of (not only nonnegative) integers. Recall that an integer a is,by denition, divisible by aninteger b if a = bc for some integer c.THEOREM 10. Let f(x) be a polynomial with integer coecients and withleading coecient equal to 1. If the equation f(x) =0has arational root , then is an integer and it is a divisor of the constant term of the polynomial f(x).Let us represent in the form = a b , where the fraction a is irreducible,bi.e. positive integers a and b are relatively prime. By the condition, the polynomialf(x) =a 0 + a 1 x + a 2 x 2 + + a n;1 x n;1 + x n has integer coecients a i . Let ussubstitute into the equation f(x) =0. By the assumption,(16) a 0 + a a 1 + + a a n;1 a nn;1 + =0:bbbMultiply the equation by b n and transfer (a) n to the right-hand side. All theterms remaining on the left-hand side will be divisible by b:(a 0 b n;1 + a 1 (a)b n;2 + + a n;1 (a) n;1 b n;2 )b =(1) n;1 a n :We see that b divides a n . If were not an integer, b would be > 1. Let p be someof its prime divisors. Then it has to divide a n ,andby Theorem 5, p divides a, too.However, by the assumption, a and b are relatively prime and we have obtained acontradiction. Hence, b =1and = a.In order to obtain the second assertion of the theorem, let us leave justa 0 onthe right-hand side, and transfer all the other terms to the right-hand side (recallthat b = 1). All the terms on the right-hand side are divisible by a:a 0 = a(a 1 ; a 2 (a) ;;a n;1 (a) n;2 ; (a) n;1 ):Obviously, it follows that a devides a 0 .