# Abstract Algebra: Chapters 16 and 17

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1 Study polynomials, their factorization, and the construction of fields.

2 Chapter 16 Polynomial Rings Notation Let R be a commutative ring. The ring of polynomials over R in the indeterminate x is the set R[x] = {a a nx n : n N, a 0,..., a n R}. We can consider equality, addition, multiplication and degree of a polynomial f(x) R[x]. Theorem 16.1 If D is an integral domain, then D[x] is an integral domain. Theorem 16.2 If F is a field, and f(x), g(x) F [x] with g(x) 0, then there exist unique polynomials q(x), r(x) such that f(x) = g(x)q(x) + r(x) with deg(r(x)) deg(g(x)).

3 Corollary Let F be a field, f(x) F[x], a F. Then the following holds. (a) f(x) = (x a)q(x) + f(a), i.e., f(a) is the remainder. (b) (x a) is a factor of f(x) if and only if f(a) = 0. (c) If deg(f(x)) = n, then f(x) has at most n zeros, counting multiplicities. Theorem If F is a finite field, then the nonzero elements in F is a cyclic group under multiplication. Proof. Suppose F = n. By the FTFAG, (F, ) is isomorphic to Z m p 1 Z m 1 p k such that p m 1 1 p m k k = n 1. k We show that m = lcm(p m 1 1,..., pm k k ) = n 1. Then p 1,..., p k are all distinct, and the conclusion will follow. If m < n 1, then every element in F is a zero of f(x) = x m 1 F[x], which is absurd.

4 Definition A principal ideal domain is an integral domain D in which every ideal has the form a = {ra : r D} for some a D. Theorem Let F be a field. Then F[x] is a principal ideal domain. In fact, for any ideal A of F [x], A = g(x), where g(x) is a nonzero monic polynomial in A with minimum degree. Example 1 Suppose f(x) = x 2 2 Q[x] and A = x 2 2. Then F = Q[x]/A = {ax + b + A : a, b Q} is a field. For every nonzero ax + b + A F, the multiplicative inverse is (ax b)/(2a 2 b 2 ) + A as (ax + b + A)((ax b)/(2a 2 b 2 ) + A) = (a 2 x 2 b 2 )/(2a 2 b 2 ) + A = (2a 2 b 2 )/(2a 2 b 2 ) + A = 1 + A. Here note that 2a 2 b 0 because a, b Q. Note that by factor theorem, f(x) has no zeros in Q. But x + A F is a solution of the equation y 2 2 = 0, where 2 = 2(1 + A) = 2 + A, as (x + A) 2 (2 + A) = (x 2 2) + A = 0 + A.

5 Examples Example 2 Suppose f(x) = x R[x] and A = x Then F = R[x]/A = {a + bx + A : a, b R} is a field. For every nonzero a + bx + A F, the multiplicative inverse is (a bx)/(a 2 + b 2 ) + A as (a + bx + A)((a bx)/(a 2 + b 2 ) + A) = (a 2 b 2 x 2 )/(a 2 + b 2 ) + A = (a 2 + b 2 )/(a 2 + b 2 ) + A = 1 + A. Note that f(x) has no zeros in R. But x + A F is a solution of the equation y = 0. Example 3 Suppose f(x) = x 2 + x + 1 Z 2[x] and A = x 2 + x + 1. Then F = Z 2[x]/A = {ax + b + A : a, b Z 2} is a field. For every nonzero ax + b + A F, one can find the inverse. Here are the inverse pairs: (1 + A, 1 + A), (x + A, 1 + x + A). Note that f(x) has no zeros in R. But x + A F is the solution of the equation y 2 + y + 1 = 0.

6 Chapter 17 Factorization of Polynomials Definition Let D be an integral domain. A polynomial f(x) in D[x] is reducible if f(x) = g(x)h(x) for some polynomials g(x), h(x) D[x] such that both g(x), h(x) have degrees smaller than f(x). If f(x) has degree at least 2 and not reducible, then it is irreducible. Theorem 17.1 Let F be a field, f(x) F[x] with degree 2 or 3. Then f(x) is reducible over F if and only if f(x) has a zero in F. Proof. If f(x) is a product of two polynomial of smaller positive degrees, then one of them must be linear. The result follows from the factor theorem.

7 Theorem 17.2 Let f(x) Z[x]. Then f(x) is reducible over Q if and only if it is reducible over Z. Proof. The content of f(x) = a a nx n Z[x] is gcd(a 0,..., a n). If the content of f(x) is 1, then f(x) is primitive. Assertion 1. Suppose u(x), v(x) Z[x] are primitive. We claim that u(x)v(x) is primitive. If not... Return to the proof of the theorem. Suppose f(x) Z[x]. We may divide f(x) by its content and assume that it is primitive. Suppose f(x) = g(x)h(x) so that g(x), h(x) Q[x] have lower degrees. Then abf(x) = ag(x)bh(x) so that a, b N are the smallest integers such that ag(x), bh(x) Z[x]. Suppose c and d are the contents of ag(x) and bh(x), then abf(x) has content ab and abf(x) = ag(x)bh(x) = (c g(x))(d h(x)) with has content cd. Thus, ad = cd and f(x) = g(x) h(x). Clearly, if f(x) is reducible in Z[x], then it is reducible in Q[x].

8 Theorem 17.3 Let p be a prime number, and suppose f(x) = a a nx n Z[x] with n 2. Suppose f(x) = [a 0] p + + [a n] px n has degree n. If f(x) is irreducible then f(x) is irreducible over Z (or Q). Proof. If f(x) = g(x)h(x) then f(x) = g(x) h(x) has degree n implies that g(x) and g(x) have the same degree and also h(x) and h(x) have the same degree. So, f(x) is reducible. Example Consider 21x 3 3x 2 + 2x + 9 Q[x]. Try x = m/n for m = 1, 3, 7, 21 and n = ±1, 3, 9. Send it to Z p[x] for p = 2, 3, 5. Example Consider (3/7)x 4 (2/7)x 2 + (9/35)x + 3/5. Send 35f(x) = 15x 4 10x 2 + 9x + 21 to Z 2[x] and check irreducibility.

9 Theorem 17.4 Suppose f(x) = a a nx n Z[x] with n 2. If there is a prime p such that p does not divide a n and p 2 does not divide a 0, but p a n 1,..., p a 0, then f(x) is irreducible over Z. Proof. Assume f(x) = g(x)h(x) with g(x) = b b rx r and h(x) = c c sx s. We may assume that p b 0 and p does not divide c 0. Note that p does not divide b rc s so that p does not divide b r. Let t be the smallest integer such that p does not divide b t. Then p (b ta 0 + b t 1a b 0a t) so that p b ta 0, a contradiction. Example Show that 3x x 4 20x x + 20 is irreducible over Q.

10 Corollary For any prime p, the pth cyclotomic polynomial is irreducible over Q. Proof. Φ(y + 1) = p j=k ( p k) y k... Φ p(x) = xp 1 x 1 = xp 1 + x p

11 Theorem 17.5 Let F be a field, and p(x) F[x]. Then p(x) is maximal if and only if p(x) is irreducible. Proof. If p(x) = g(x)h(x) then p(x) g(x). If A is an ideal not equal to F[x] and not equal to p(x) such that p(x) A, then A = g(x) and p(x) = g(x)h(x) such that g(x) has degree less than p(x). Corollary Let F be a field. Suppose p(x) is irreducible. (a) Then F[x]/ p(x) is a field. (b) If u(x), v(x) F[x] and f(x) u(x)v(x), then p(x) u(x) or p(x) v(x). Proof. (a) By the fact that D/A is a field if and only if A is a maximal. (b) A = p(x) is maximal, and hence is prime...

12 Theorem 17.6 Every f(x) F[x] can be written as a product of irreducible polynomials. The factorization is unique up to a rearrangement of the factors and multiples of the factors by the field elements. Proof. By induction on degree. f(x) = f i(x) such that every f i(x) is irreducible. If f i(x) = g j(x), then f i(x) divides some g j... Extra Credit Homework 1 If r R such that r + 1/r Z \ {2, 2}, than r is irrational. 2 Let f(x) = a a nx n Z[x]. If f(r/s) = 0, where r/s Q is in its lowest form. Show that r a 0 and s a n. 3 Show that the ideal x is prime in Z[x], but it is not a maximal ideal. [Hint: Consider Z[x]/ x ]

13 Final Examination (Take home part) Due: Dec. 16, 17:00 1 Show that if G is a group with no proper non-trivial subgroup and has at least two elements, then G is isomorphic to Z p for some prime number p. 2 Let F be a field. Show that F has a subfield isomorphic to Z or Z p for a prime number p. [Hint: If F has characteristic 0, then D = {n 1 : n Z} is a subring (integral domain) of F, and the filed of quotient of D is... If F has characteristic m, then...] 3 (a) Show that x 4 + x + 1 Z 2[x] is irreducible. (b) Let F = Z 2[x]/ x 4 + x + 1. Pair up all the elements that are multiplicative inverse of each other in F, and determine all the generators of the group F under multiplication. 4 Let p > 2 be a prime number. (a) Show that there is a Z p such that a 2 { 1, 2, 2}. [Hint: Consider φ : Z p Z p defined by φ(x) = x2. (a.1) Show that ker(φ) = { 1, 1} and H is isomorphic to Z p /ker(φ) has index 2. (a.2) Show that if 1, 2 / H = φ(z p ), then H = 2H H and H = ( H)( H) = ( 2)H so that 2 H.] (b) Show that x is reducible over Z p. [Hint: If there is a 2 = 1, then x = (x 2 + a)(x 2 a). If there is a 2 = 2, then x = (x 2 + ax + 1)(x 2 ax + 1). If there is a 2 = 2, then...] (c) Show that x is reducible over Z 2.

14 Coda Division rings/skew-fields A non-commutative ring with unity such that every nonzero element has an (multiplicative) inverse is a division ring. Example Real quaternions. H = {a 0 + a 1i + a 2j + a 3k : a 0, a 1, a 2, a 3 R} with i 2 = j 2 = k 2 = 1, ij = k = ji, jk = i = kj, ki = j = ik. Define addition and multiplication by: See Wiki for the interesting history. Factoring / finding roots of a polynomial in an extension field Note that R is a subring of R[x]. If f(x) = a a nx n is irreducible in F[x], then E = F[x]/A with A = f(x) is a field containing F. Moreover, y = x + A is a zero for f(y).

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