Serge Ballif January 18, PDF Free Download

ballif@math.psu.edu The Pennsylvania State University January 18, 2008

Outline Rings Division Rings Noncommutative Rings s Roots of

Rings Definition A ring R is a set toger with two binary operations + such that following three properties hold. 1. (R, +) form an additive group. 2. Multiplication is associative (i.e. (ab)c = a(bc) for all a, b, c R). 3. The left right distributive laws hold: Rings Division Rings a(b + c) = ab + ac (a + b)c = ac + bc. Convention For purposes of this talk, we will assume that all rings have an identity element 1 such that 1r = r for all r R.

Examples of Rings Example The following rings are familiar to all mamaticians. R real numbers C complex numbers Z integers Mat n (R) ring of n n real-valued matrices Rings Division Rings

Division Rings Definition A ring R is a division ring iff every nonzero element of R has a two-sided inverse element iff (R {0}, ) is a group. Remark A field is a commutative division ring. Rings Division Rings Example R C are division rings. Z Mat n (R) are not division rings.

The Real Definition The ring of real quaternions H is a 4-dimensional real vector space with basis {1, i, j, k}. H is a division ring with multiplication structure determined by relations i 2 = j 2 = k 2 = 1 = ijk. Rings Division Rings An element q of H is of form q = a + bi + cj + dk, where a, b, c, d R. The real part of q is a, purely imaginary part of q is bi + cj + dk.

Quaternion Multiplication Remark Multiplication of quaternions is not commutative. We have ij = k, but ji = k. Mnemonic k i j Rings Division Rings Practice Multiply following quaternions: (1 + 2i + 3j + 4k)(4 + 3i + 2j + k)

s Definition For any ring R, we let R[x] denote polynomial ring with indeterminate x coefficients from ring R. Remark Each polynomial f(x) R[x] is of form f(x) = a i x i. Since indeterminate x commutes with coefficients, we also have f(x) = x i a i. s Roots of

Right Definition Let f(x) = n a ix i R[x]. For each element r R, we define (right) evaluation of f at r to be map R[x] R that maps f(x) to f(r) := a i r i. An element r is a (right) root of f(x) if f(r) = 0. s Roots of

Factoring Does Not Preserve Caution If R is noncommutative, n evaluation is not a ring homomorphism. That is, Example Consider f(x) = g(x)h(x) does not imply f(r) = g(r)h(r). f(x) = (x i)(x j) H[x]. We might expect i j to be roots of f(x). However, we note that f(x) = x 2 (i + j)x + k, so f(i) = i 2 (i + j)i + k = 1 i 2 ij + k = 2k 0 f(j) = j 2 (i + j)j + k = 1 ij j 2 + k = 0. Hence, j is a right root of f(x), but i is not. s Roots of

Remainder (Commutative Case) Let R be a commutative ring. An element r R is a root of a nonzero polynomial f(x) R[x] iff f(x) = g(x)(x r) for some polynomial g(x) R[x]. (Noncommutative Case) Let R be a noncommutative ring. An element r R is a right root of a nonzero polynomial f(x) R[x] iff... s Roots of

Proof of Remainder Proof. ( ) Let ( ) f(x) = a i x i (x r) = Then f(r) = a i (r) i+1 a i x i+1 a i rx i. a i r(r) i = 0, so r is a root of f(x). ( ) Suppose that f(r) = 0. Then by division algorithm (on right), we can write f(x) = g(x)(x r) + s for some g(x) R[x] some s R. The first part shows that r is a root of g(x)(x r). Thus f(r) = s = 0. s Roots of

Factor (Commutative Case) Let D be a field, let f(x) = g(x)h(x). An element d D is a root of f(x) iff d is a root of eir g(x) or h(x). (Noncommutative Case) Let D be a division ring, let f(x) = g(x)h(x). An element d D is a root of f(x) iff... s Roots of

Proof of Factor Proof. Let g(x) = n b ix i. Then ( ) f(x) = b i x i h(x) = b i h(x)x i. If h(d) = 0, n f(d) = 0. Suppose that h(d) = a 0 f(d) = 0. Then f(d) = b i h(d)d i = = b i ad i a 1 a ( b i ada 1 ) i a = g(ada 1 )h(d). Since f(d) = 0, but h(d) 0, we must have g(ada 1 ) = 0. s Roots of

The Number of Roots (Commutative Case) Let D be a field. A polynomial in D[x] of degree n has at most n distinct roots in D. (Noncommutative Case, ) Let D be a division ring. A polynomial in D[x] of degree n has at most... s Roots of

The Number of Roots (Commutative Case) Let D be a field. A polynomial in D[x] of degree n has at most n distinct roots in D. (Noncommutative Case, Gordon, Motzkin, 1965) Let D be a division ring. A polynomial in D[x] of degree n has at most roots. However, se roots will come from at most n distinct conjugacy classes of D. s Roots of

A Familiar Example In division ring H of real quaternions, i, j, k are each roots of second degree polynomial x 2 + 1. In fact, each of infinitely many conjugates of i is a root of x 2 + 1. s Roots of

Proof of Roots in at Most n Conjugacy Classes Proof. 1. We prove by induction. The case n = 1 is immediately true. 2. For n 2, let c be a root of f(x). Then f(x) = g(x)(x c). 3. Suppose that d c is anor root of f(x). Then a conjugate of d is a root of g(x). 4. By induction hyposis, d lies in at most one of (at most) n 1 conjugacy classes of roots of g(x). s Roots of

ically Closed Division Rings Definition A division ring is said to be right algebraically closed if every nonconstant polynomial has a right root. (Niven, Jacobson, 1941) The real quaternions H are right ( left) algebraically closed. s Roots of

Quaternionic Conjugates Definition The quaternionic conjugate of q = a + bi + cj + dk is q = a bi cj dk. Definition The conjugate polynomial of a polynomial f(x) = q i x i is f(x) = q i x i. Useful Facts For p, q H, pq = q p. Similarly, for f(x), g(x) H[x], fg = ḡ f. Also, q q R, while f f R[x]. s Roots of

Proof of Sketch of Proof. Let f(x) = n q ix i. 1. ff R[x] must have a root a in C. 2. Eir a is a root of f, or a conjugate b of a is a root of f. 3. Note that n q ib i = 0 n q i b i = 0, so b is a left root of f(x). 4. Thus, f(x) = (x b)g(x) for some g(x) D[x]. 5. Note that deg g < deg f, so by induction on degree of f, g(x) has a right root. 6. Therefore, f has a right root. s Roots of

Serge Ballif January 18, 2008