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Chapter 3, Rings Definitions and Examples. We Now Have Several

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Chapter 3, Rings Definitions and Examples. We Now Have Several Chapter 3, Rings Definitions and examples. We now have several examples of algebraic systems with addition and multiplication: Z; Zn; R;Mn(R);2Z={2n|n∈Z}. We will write down a system of axioms which includes them all. Definition, p. 42. A ring is a nonempty set R with two binary operations (usually written as addition and multiplication) such that for all a; b; c ∈ R, (1) R is closed under addition: a + b ∈ R. (2) Addition is associative: (a + b)+c=a+(b+c). (3) Addition is commutative: a + b = b + a. (4) R contains an additive identity element, called zero and usually denoted by 0 or 0R: a +0=0+a=a. (5) Every element of R has an additive inverse: for each a, there exists an x ∈ R such that a + x =0=x+a. We write x = −a. (6) R is closed under multiplication: ab ∈ R. (7) Multiplication is associative: (ab)c = a(bc). (8) Multiplication distributes over addition: a(b + c)=ab + ac and (a + b)c = ac + bc. Other possible properties are captured by special types of rings. We will encounter many in this book; many more are studied as well by mathematicians. Theorem 3.3. Additive inverses are unique. Proof. Assume that x and y are both inverses of a.Thenx=x+0 = x+(a+y)= (x+a)+y=0+y=y. We don't have to add axioms about subtraction. We just define a − b to be a +(−b). Definition. A commutative ring is a ring R that satisfies the additional axiom that ab = ba for all a; b ∈ R. Examples are Z, R, Zn,2Z, but not Mn(R)ifn≥2. Definition. A ring with identity is a ring R that contains a multiplicative identity element 1R:1Ra=a=a1Rfor all a ∈ R. 10 Examples: 1 in the first three rings above, in M2(R). The set of even integers 01 1 2 is a ring without identity. The set of odd integers is not a ring. We can also work with matrices whose elements come from any ring we know about, such as Mn(Zr). Example. LetR = M2(Z2). This is a finite (16 elements) noncommutative ring with 10 00 identity 1R = and zero element 0R = . We give an example to show it 01 00 is noncommutative: 10 01 01 01 10 00 = but = 00 00 00 00 00 00 Example: rings of continuous functions. Let X be any topological space; if you don't know what that is, let it be R or any interval in R. We consider the set R = C(X; R), the set of all continuous functions from X to R. R becomes a ring with identity when we define addition and multiplication as in elementary calculus: (f + g)(x)=f(x)+g(x)and (fg)(x)=f(x)g(x). The identity element is the constant function 1. R is commutative because R is, but it does have zero divisors for almost all choices of X. There are many, many examples of this sort of ring. The functions don't have to be continuous. They can be restricted in many other ways, or not restricted at all. For example, you can look at polynomial functions or differentiable functions (for some choices of X). Definition, p. 46. An integral domain is a commutative ring R with identity 1R =06 R with no zero divisors; that is, ab =0R implies that a =0R or b =0R. Examples: Z, R, Zp for p prime. Nonexamples: C(R; R), Zn for n composite, the zero ring {0R}, the even integers 2Z. Definition, p. 47. A field is an integral domain in which every nonzero element a has a multiplicative inverse, denoted a−1. Examples: R, Q, C, Zp for p prime (Theorem 2.8). If an element of a ring has a multiplicative inverse, it is unique. The proof is the same as that given above for Theorem 3.3 if we replace addition by multiplication. (Note that we did not use the commutativity of addition.) This is also the proof from Math 311 that invertible matrices have unique inverses. Definition, p. 60. Any element a in a ring R with identity which has an inverse u (i.e., au =1r =ua)iscalledaunit. 3 Making new rings. Theorem 3.1 (Product rings). Let R; S be rings and form the Cartesian product R × S. Define operations by (r; s)+(r0;s0)=(r+r0;s+s0) (r; s)(r0;s0)=(rr0;ss0): Then R × S is a ring. If R and S are both commutative, so is R × S.IfRand S both have an identity, then (1R; 1S) is the identity in R × S. Example. Let R be the ring Z × Z = { (n; m) | n; m ∈ Z }.Notethat(1;0)(0; 1) = (0; 0) = 0R,soRis not an integral domain. For the same reason, no product ring is an integral domain. Definition, p. 49. A subset of a ring which is itself a ring (using the same operations) is called a subring. A subset of a field which is itself a field is called a subfield. Q is a subfield of R, and both are subfields of C. Z is a subring of Q. Z3 is not a subring of Z. Its elements are not integers, but rather are congruence classes of integers. 2Z = { 2n | n ∈ Z } is a subring of Z, but the only subring of Z with identity is Z itself. The zero ring is a subring of every ring. As with subspaces of vector spaces, it is not hard to check that a subset is a subring as most axioms are inherited from the ring. Theorem 3.2. Let S be a subset of a ring R. S is a subring of R iff the following conditions all hold: (1) S is closed under addition and multiplication. (2) 0R ∈ S. (3) −s ∈ S for every element s ∈ S. Proof. Axioms 1, 4, 5 and 6 are in our hypotheses. Axioms 2, 3, 7 and 8 are inherited from R. As examples, do exercises from book, page 51{55: 7, 8, 9, 19, 34, discuss 35. 4 Basic properties of rings. Theorem 3.4{3.5. For all a; b; c in a ring R, (1) a + b = a + c implies b = c. (2) a · 0=0·a=0. (3) a(−b)=−(ab)=(−a)b. (4) −(−a)=a. (5) −(a + b)=(−a)+(−b). (6) (−a)(−b)=ab. (7) (−1)a = −a if R has an identity element. Proof. These make use of the definition of subtraction and negatives. For (1), add −a. (2) a · 0=a(0 + 0) = a · 0+a·0anduse(1). (3) Show that a(−b)and(−a)bare additive inverses for ab (we know there is a unique such element). (4) and (5) Definition of additive inverse. (6) Use (3) and (4). (7) Use (3) or add (−1)a to a. A quicker version of Theorem 3.2 can be obtained using subtraction. Theorem 3.6. Let R be a ring and ∅6=S⊆R. Sis a subring of R if S is closed under subtraction and multiplication. Proof. We need to show S is closed under addition, has 0 and has additive inverses. But S =6 ∅ implies there is some s ∈ S, hence 0 = s − s ∈ S. For any a ∈ S, −a =0−a∈S. And finally, for a; b ∈ S, a + b = a − (−b) ∈ S. Notation: For a in a ring R and n ∈ Z, write na for a sum of n copies of a and an 0 for a product of n copies of a. This includes the cases 0 · a =0R and a =1R if R has an identity element. Be careful: (a+b)2 = a2+ab+ba+b2 cannot be simplified unless the ring is commutative. Theorem 3.7. Let R be a ring and a; b ∈ R. The equation a + x = b has the unique solution x = b − a in R. Proof. b − a is a solution: check it! If z is another solution, then a + z = b = a +(b−a), so z = b − a by cancellation (Theorem 3.4). For multiplication, we need multiplicative inverses to get the cancellation in the proof. 5 Theorem 3.8. Let R be a ring with identity and a; b ∈ R.Ifais a unit, then the equations ax = b and ya = b have unique solutions in R. Proof. x = a−1b and y = ba−1 are solutions: check! Uniqueness works as in Theorem 3.7, using the inverse for cancellation: if z is another solution to ax = b,thenaz = b = a(a−1b). Multiply on the left by a−1 to get z = a−1az = a−1a(a−1b)=a−1b. A similar argument works for y. The solutions x = a−1b and y = ba−1 may not be the same. Exercise 4, p. 62 gives an example with 2 × 2 matrices. Sometimes multiplicative cancellation works without inverses (recall the integers). Theorem 3.10. Let R be an integral domain. If a =06 and ab = ac,thenb=c. Proof. ab = ac implies a(b − c) = 0. Since a =0and6 Ris an integral domain, we must have b − c =0,orb=c. If R is not an integral domain, this fails because of zero divisors: p. 62. Let a =6 0 in a ring R. a is a zero divisor if there exists an element b =0in6 R with either ab =0orba =0. Example: 2 · 3=0=2·0inZ6. Theorem 3.11. Every finite integral domain is a field. Combinatorial proof.
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