# g(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree. g(x) is an example of a power series.

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1 6 Polynomial Rings We introduce a class of rings called the polynomial rings, describing computation, factorization and divisibility in such rings For the case where the coefficients come from an integral domain, the corresponding polynomials also form an integral domain Definition 61 Let R be an integral domain A polynomial f(x) in indeterminate x with coefficients in R has the form f(x) = f 0 + f 1 x + + f n x n, where n is a nonnegative integer and f i R for each i If f n 0 we say that f(x) has degree n and write deg f = n We denote by R[x] the set of all polynomials with coefficients in R Example 61 Let f(x) = 3/2 x On the other hand 3 + 2ix Then f is a polynomial with complex coefficients, so f C[x] g(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree g(x) is an example of a power series We can use the addition and multiplication as defined on the ring R to define operations of addition and multiplication in R[x] As we ll soon see, with these operations, the set R[x] has the structure of a ring We ll mostly consider the case where R is a field, in which case R[x] is an integral domain Addition of polynomials is defined by and multiplication of polynomials is defined by f + g = (f 0 + g 0 ) + (f 1 + g 1 )x + + (f n + g n )x n (61) fg = f 0 g 0 + (f 1 g 0 + f 0 g 1 )x + (f 0 g 2 + f 1 g 1 + f 2 g 0 )x (f 0 g i + f 1 g i f i g 0 )x i + + f n g m x n+m (62 where f = f 0 +f 1 x+ +f n x n, g = g 0 +g 1 x+ +g m x m, and WLOG, n and m are a pair of non-negative integers satisfying n m (If i m then g i = 0) If f = f and g = g for some f, f, g, g R[x] then f i = f i, g i = g i for each i, so f + g = (f 0 + g 0 ) + (f 1 + g 1 )x + + (f n + g n )x n, = (f 0 + g 0) + (f 1 + g 1)x + + (f n + g n)x n, = f + g, since addition is well-defined on R It follows that addition is well-defined on R[x] Similarly, multiplication is well-defined on R[x] If f, g, h R[x] then (f + g) + h = ((f 0 + g 0 ) + h 0 ) + ((f 1 + g 1 ) + h 1 )x + + ((f n + g n ) + h n )x n, = (f 0 + (g 0 + h 0 )) + (f 1 + (g 1 + h 1 ))x + + ((f n + (g n + h n ))x n, = f + (g + h), since addition is associative in R Similarly, we can show that multiplication of polynomials over a ring is associative R[x] has an additive identity, namely 0, the additive identity of R, and the additive inverse of f = f 0 +f 1 x+ + f n x n is given by f = f 0 + ( f 1 )x + + ( f n )x n All this tells us the following: 19

2 Lemma 61 R[x] is a ring with the operations of polynomial addition and multiplication as defined in Equations 61, 62 If R is unital and commutative, then so is R[x] If F is a field then since F has no zero divisors, it is straightforward to show that degree (fg) = degree f+ degree g for all f, g F[x] Since F is unital and commutative, it can be shown that Lemma 62 If F is a field then F[x] is an integral domain Given nonzero polynomials a, b F[x], we say that a is a divisor of b in F[x], expressed a b, if b = ac for some c F[x] If c / F we say that a is a proper divisor of b Since F[x] is an ID, if f, g F[x] then f g and g f if and only if f = ug for some u F Definition 62 Let f, g F[x] A common divisor of f, g in F[x] is an polynomial d F[x] such that d f and d g A common divisor d of f, g is called a greatest common divisor (gcd) of f and g if any other common divisor of f and g is also a divisor of d If d is a gcd of f and g then we write d = gcd(f, g), which is unique in F[x], up to multiplication by nonzero elements of F (the units of F[x]) If gcd(f, g) = 1 we say that f and g are coprime The polynomial f = f 0 + f 1 x + f n x n chosen to represent the gcd of a pair of polynomials has f n = 1 Such a polynomial is called monic A non-constant polynomial f F[x] is called irreducible if it has no proper divisors Irreducibles in F[x] play a role similar to the prime numbers in Z If f gh for some g, h F[x] and f is irreducible, then f g or f h Theorem 61 (Division Algorithm in F[x]) Let f, g F[x], g 0 Then there exist m, r F[x] satisfying f = mg + r with 0 degree r < degree g, or r = 0 Proof Let S = {f mg : m F[x]} If 0 S then r = 0 gives the required element, so assume otherwise Let r = f m 0 g 0 have minimal degree in S Let degree r = l, degree g = k and suppose that l k = d 0 Then r ( r l x d )g = f (m 0 + r l x d )g S, g k g k where r i, g j are the ith and jth coefficients of r and g, respectively, and degree(r ( r l g k x d )g) < l, unless r ( r l g k x d )g = 0 The former contradicts the minimality of r in S, and the latter the fact that 0 / S The result follows The Euclidean algorithm is a technique for computing the gcd of a pair of polynomials If f, g F[x], f, g 0 and degree f degree g, then repeated applications of the division algorithm yields f = m 1 g + r 1, g = m 2 r 1 + r 2, r 1 = m 3 r 2 + r 3, r i = m i+2 r i 1 + r i+2, r k 2 = m k r k 1 + r k, r k 1 = m k+1 r k + 0, with degree r k < degree r k 1 < < degree r 1 < degree g and r k 0 Since we have generated a sequence of strictly decreasing positive integers, the process must terminate in a finite number of steps (in fact in at 20

3 most t steps where t is the difference in the degrees of f and g) Note that r k divides r k 1 in F[x], and hence r k 2 = m k r k 1 + r k Repeat this argument to see that r k divides each remainder r i, and finally f and g in F[x] If d is a common divisor of f and g then d divides r 1, r 2,, r k, so r k is the greatest common divisor of f and g Example 62 Let f = x 6 + 2x 4 + x 3 + 2x 2 + x + 1, g = x 3 + x 2 + x + 1 be a pair of rational polynomials Then, implementing the Euclidean algorithm we obtain the equations: so x is the gcd of f and g in Q[x] f = xg + x 2 + 1, g = (x + 1)r 1 + 0, Theorem 62 Let f, g F[x] Then f and g have a well-defined gcd, say d = gcd(f, g), and there exist s, t F[x] such that fs + gt = d Moreover, d is unique up to multiplication by units in F Proof Let S = {fs + gt : s, t F[x]} Clearly f, g S, so S φ Let d S have minimal degree and suppose d = fs 0 + gt 0 for some s 0, t 0 F[x] We claim that d =gcd(f, g) Let w = fs 1 + gt 1 S From the DA in F[x], there exist m, r R such that w = md + r and r = 0 or degree r < degree d But then r = w md = (s 1 ms 0 )f + (t 1 mt 0 )g S It follows by the minimality of d in S that r = 0 Thus d divides every element of S, so, in particular, d f, g On the other hand, if c is a common divisor of f and g then c (fs + gt) for all s, t F[x], so any common divisor of f, g also divides d Thus d =gcd(f, g) To see that gcd(f, g) is unique, suppose that e is also a gcd of f and g in F[x] Then d e and e d in F[x], and in particular degree d = degree e It follows that e is a unit multiple of d in F[x] Definition 63 An ID R is called a unique factorization domain (UFD) if every non-zero non-unit in R can be expressed uniquely (up to multiplication by a unit) as a finite product of irreducibles in R Example 63 The ring of integers forms a UFD - every number can be uniquely expressed as a finite product of prime numbers, and up to multiplication by ±1 (the units of Z), this can be done in only one way Theorem 63 F[x] is a UFD By convention, if f F[x] we express its factorization as f = δf 1 f 2 f t, where each f i is monic and irreducible in F[x] and δ is a constant in F Definition 64 If f F[x] and f(α) = 0 for some α in an extension field of F, we say that α is a root of f Theorem 64 (The Factor Theorem) Let f F[x] and let α F Then α is a root of f, if and only if (x α) f in F[x] Proof From the DA, there exist g, r F[x] such that f = (x α)g + r, and either degree r < degree (x α) = 1, or r = 0 Then r is constant, since and f(α) = (α α)g(α) + r(α) = r(α) = 0 Then r = 0 and x α divides f in F[x] This gives us a criterion for irreducibility Corollary 61 (Root Test) Let f F[x] and let degree f {2, 3} Then f has no roots in F if and only if f is irreducible in F[x] There are a number of classical results which help us to determine whether or not an integer polynomial is reducible over the rationals 21

4 Theorem 65 (Rational Root Test) Let f = f 0 + f 1 x + + f n x n Z[x] If α = r/s Q, (r, s) = 1 is a root of f then r f 0 and s f n Proof If r/s Q with gcd(r, s) = 1 and r/s is a root of f then and hence f(r/s) = f 0 + f 1 (r/s) + f n (r/s) n = 0, f 0 s n + f 1 rs n 1 + f n 1 r n 1 s = f n r n Since r, s have no factors in common, s divides f n Similarly, r divides f 0 Definition 65 Let f = f 0 + f 1 x + + f n x n Z[x] Then we say that f is primitive if the gcd of f 0, f 1,, f n in Z is 1 Lemma 63 If f, g Z[x] are primitive, then so is f g Theorem 66 (Gauss Lemma) If the primitive polynomial f Z[x] can be factored as f = gh, g, h Q[x], then there exist g 1, h 1 Z[x] such that f = g 1 h 1 We can state Gauss Lemma in another way Corollary 62 A polynomial f in Z[x] is irreducible in Z[x] if and only if it is irreducible in Q[x] Example 64 Let f = x 3 + 2x + 1 Q[x] From the rational root test (RRT), if f has a rational root it is one of ±1 From the root test, since f has degree 3, it is irreducible if and only if f has no rational roots Since f(1), f( 1) 0, we deduce that f is irreducible over Q Theorem 67 (Eisenstein s Criterion) Let f = f 0 + f 1 x + + f n x n Z[x] If there exists a prime p Z such that p f n, p f n 1, p f n 2,, p f 0, p 2 f 0 Then f is irreducible Proof Let p Z, f Z[x] be as in the statement of the theorem We may assume that f is primitive Suppose f = gh for some g, h Z[x] Then f 0 = g 0 h 0 We may assume that p g 0, which implies that p does not divide h 0 Since f is primitive, p does not divide all the coefficients of g Let 0 < j < n be the least integer such that p g j Then f j = g 0 h j + g 1 h j g j h 0, so p f j, g 0, g 1,, g j 1, and, since p does not divide g j, p h 0, giving a contradiction We deduce that f is irreducible over Q Let f F[x] Then f(x) = g(x)h(x) for some g, h F[x] if and only if f(x+a) = g(x+a)h(x+a) for some a F, in which case g(x + a), h(x + a) F[x] It follows that f is irreducible in F[x] if and only if f(x + a) is irreducible in F[x] for all a F It is sometimes convenient to use this substitution method to change the form of a polynomial to one which can be more easily identified as irreducible or reducible Example 65 Let f = x Q[x] Then f(x) is irreducible in Q[x] if and only if f(x + 1) is irreducible in Q[x] Now f(x + 1) = x 4 + 4x 3 + 6x 2 + 4x + 2, which is irreducible in Q[x] by Eisenstein s irreducibility criterion with p = 2 Theorem 68 Let f R[x] and let α C be a root of f Then ᾱ is also a root of f Proof f(ᾱ) = f 0 + f 1 ᾱ + f 1 ᾱ f n ᾱ n = f 0 + f 1 α + f 1 α f n α n = 0 22

5 This means that no odd degree polynomial f is irreducible in R[x]: if f has no real roots and α is a root of f then x 2 + 2Re(α) + α 2 divides f in R[x] It turns out that every polynomial of degree f over the complex numbers factorizes as a product of n linear factors over C A field with this property is called algebraically closed Theorem 69 (The Fundamental Theorem of Algebra) Every polynomial of degree n in C[x] has exactly n (not necessarily distinct) roots in C 23

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