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1 Lecture 1 1.1 Definition of Rings and Examples Proof. Use the same proof that we used for groups! I.e. 1 = 1 · 10 = 10 and if b; b0 are both inverses to a; then b = b (ab0) = (ba) b0 = b0: A ring will be a set of elements, R; with both an addition and multiplication operation satisfying a number of \natural" axioms. Notation 1.6 (Subtraction) In any ring R, for a 2 R we write the additive inverse of a as (−a). So at a + (−a) = (−a) + a = 0 by definition. For any Axiom 1.1 (Axioms for a ring) Let R be a set with 2 binary operations a; b 2 R we abbreviate a + (−b) as a − b: called addition (written a + b) and multiplication (written ab). R is called a ring if for all a; b; c 2 R we have Let us now give a number of examples of rings. 1. (a + b) + c = a + (b + c) Example 1.7. Here are some examples of commutative rings that we are already 2. There exists an element 0 2 R which is an identity for +. familiar with. 3. There exists an element −a 2 R such that a + (−a) = 0: 1. Z = all integers with usual + and ·: 4. a + b = b + a: m 2. Q = all n such that m; n 2 Z with n 6= 0; usual + and ·: (We will generalize 5. (ab)c = a(bc). this later when we talk about “fields of fractions.") 6. a(b + c) = ab + ac and (b + c)a = ba + bc. 3. R = reals, usual + and ·: Items 1. { 4. are the axioms for an abelian group, (R; +) : Item 5. says mul- 4. C = all complex numbers, i.e. fa + ib : a; b 2 Rg ; usual + and · operations. tiplication is associative, and item 6. says that is both left and right distributive (We will explicitly verify this in Proposition 3.6 below.) over addition. Thus we could have stated the definition of a ring more succinctly Example 1.8. 2 = f:::; −4; −2; 0; 2; 4;::: g is a ring without identity. as follows. Z Example 1.9 (Integers modulo m). For m ≥ 2; = f0; 1; 2; : : : ; m − 1g with Definition 1.2. A ring R is a set with two binary operations \+" = addition Zm and \·"= multiplication, such that (R; +) is an abelian group (with identity + = addition mod m element we call 0), \·" is an associative multiplication on R which is both left and right distributive over addition. · = multiplication mod n: Remark 1.3. The multiplication operation might not be commutative, i.e., ab 6= Recall from last quarter that (Zm; +) is an abelian group and we showed, ba for some a; b 2 R: If we have ab = ba for all a; b 2 R, we say R is a [(ab) mod m · c] mod m = [abc] = [a (bc) mod m] mod m (associativity) commutative ring. Otherwise R is noncommutative. and Definition 1.4. If there exists and element 1 2 R such that a1 = 1a = a for all a 2 R, then we call 1 the identity element of R [the book calls it the unity.] [a · (b + c) mod m] mod m = [a · (b + c)] mod m Most of the rings that we study in this course will have an identity element. = [ab + ac] mod m = (ab) mod m + (ac) mod m Lemma 1.5. If R has an identity element 1; then 1 is unique. If an element which is the distributive property of multiplication mod m: Thus Zm is a ring a 2 R has a multiplicative inverse b, then b is unique, and we write b = a−1. with identity, 1: 10 1 Lecture 1 Example 1.10. M2(F ) = 2 × 2 matrices with entries from F , where F = Z; Q; Since 0 −1 0 −1 R; or C with binary operations; i2 = = I 1 0 1 0 a b a0 b0 a + a0 b + b0 + = (addition) c d c0 d0 c + c0 d + d0 it is straight forward to check that a b a0 b0 aa0 + bc0 ab0 + bd0 (aI+bi)(cI+di) = (ac − bd) I + (bc + ad) i and = : (multiplication) c d c0 d0 ca0 + dc0 cb0 + dd0 (aI+bi) + (cI+di) = (a + c) I + (b + d) i That is multiplication is the usual matrix product. You should have checked in which are the standard rules of complex arithmetic. The fact that C is a ring your linear algebra course that M2 (F ) is a non-commutative ring with identity, now easily follows from the fact that M2 (R) is a ring. 1 0 I = : In this last example, the reader may wonder how did we come up with the 0 1 0 −1 matrix i := to represent i: The answer is as follows. If we view as 2 1 0 C R For example let us check that left distributive law in M2(Z); in disguise, then multiplication by i on C becomes, a b e f p q + c d g h r s (a; b) ∼ a + ib ! i (a + ib) = −b + ai ∼ (−b; a) a b p + e f + q = while c d g + r h + s a 0 −1 a −b i = = : b (g + r) + a (p + e) a (f + q) + b (h + s) b 1 0 b a = d (g + r) + c (p + e) c (f + q) + d (h + s) Thus i is the 2 × 2 real matrix which implements multiplication by i on C: bg + ap + br + ae af + bh + aq + bs = Theorem 1.12 (Matrix Rings). Suppose that R is a ring and n 2 Z+: Let dg + cp + dr + ce cf + dh + cq + ds n Mn (R) denote the n × n { matrices A = (Aij)i;j=1 with entries from R: Then while Mn (R) is a ring using the addition and multiplication operations given by, a b e f a b p q + (A + B) = Aij + Bij and c d g h c d r s ij X (AB) = A B : bg + ae af + bh ap + br aq + bs ij ik kj = + dg + ce cf + dh cp + dr cq + ds k bg + ap + br + ae af + bh + aq + bs Moreover if 1 2 R; then = 2 3 dg + cp + dr + ce cf + dh + cq + ds 1 0 0 . I := 6 .. 7 which is the same result as the previous equation. 4 0 0 5 0 0 1 Example 1.11. We may realize C as a sub-ring of M2 (R) as follows. Let is the identity of Mn (R) : 1 0 0 −1 I = 2 M ( ) and i := 0 1 2 R 1 0 Proof. I will only check associativity and left distributivity of multiplication here. The rest of the proof is similar if not easier. In doing this we will make and then identify z = a + ib with use of the results about sums in the Appendix 1.2 at the end of this lecture. Let A; B; and C be n × n { matrices with entries from R: Then 1 0 0 −1 a −b aI+bi := a + b = : 0 1 1 0 b a Page: 10 job: 103bs macro: svmonob.cls date/time: 2-Apr-2009/8:32 1.2 Appendix: Facts about finite sums 11 ! X X X 1.2 Appendix: Facts about finite sums [A (BC)]ij = Aik (BC)kj = Aik BklClj k k l Throughout this section, suppose that (R; +) is an abelian group, Λ is any set, X = AikBklClj and Λ 3 λ ! rλ 2 R is a given function. k;l Theorem 1.15. Let F := fA ⊂ Λ : jAj < 1g : Then there is a unique function, while S : F! R such that; ! 1. S (;) = 0; X X X [(AB) C]ij = (AB)il Clj = AikBkl Clj 2. S (fλg) = rλ for all λ 2 Λ. l l k 3. S (A [ B) = S (A) + S (B) for all A; B 2 F with A \ B = ;: X = AikBklClj: Moreover, for any A 2 F;S (A) only depends on frλgλ2A : k;l Proof. Suppose that n ≥ 2 and that S (A) has been defined for all A 2 F Similarly, with jAj < n in such a way that S satisfies items 1. { 3. provided that jA [ Bj < X X n: Then if jAj = n and λ 2 A; we must define, [A (B + C)]ij = Aik (Bkj + Ckj) = (AikBkj + AikCkj) k k S (A) = S (A n fλg) + S (fλg) = S (A n fλg) + rλ: X X = AikBkj + AikCkj = [AB]ij + [AC]ij : We should verify that this definition is independent of the choice of λ 2 A: To k k see this is the case, suppose that λ0 2 A with λ0 6= λ, then by the induction hypothesis we know, S (A n fλg) = S ([A n fλ, λ0g] [ fλ0g) Example 1.13. In Z6, 1 is an identity for multiplication, but 2 has no multi- 0 0 0 plicative inverse. While in M2(R), a matrix A has a multiplicative inverse if = S (A n fλ, λ g) + S (fλ g) = S (A n fλ, λ g) + rλ0 and only if det(A) 6= 0: so that Example 1.14 (Another ring without identity). Let 0 S (A n fλg) + rλ = [S (A n fλ, λ g) + rλ0 ] + rλ 0 0 a = S (A n fλ, λ g) + (rλ0 + rλ) R = : a 2 R 0 0 0 = S (A n fλ, λ g) + (rλ + rλ0 ) 0 = [S (A n fλ, λ g) + r ] + r 0 with the usual addition and multiplication of matrices.
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