# Chapter 3, Rings. Definitions and examples.

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2 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 M n (Z r ). Example. Let ( R = ) M 2 (Z 2 ). This is a finite ((16 elements) ) noncommutative ring with identity 1 R = and zero element R =. We give an example to show it 0 0 is noncommutative: ( )( ) ( ) ( )( ) ( ) = but = 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 1 R 0 R with no zero divisors; that is, ab =0 R implies that a =0 R or b =0 R. Examples: Z, R, Z p for p prime. Nonexamples: C(R, R), Z n for n composite, the zero ring {0 R }, 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, Z p 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 =1 r =ua) iscalledaunit.

3 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)+(r,s )=(r+r,s+s ) (r, s)(r,s )=(rr,ss ). Then R S is a ring. If R and S are both commutative, so is R S. If R and S both have an identity, then (1 R, 1 S ) is the identity in R S. Example. Let R be the ring Z Z = { (n, m) n, m Z }. Note that (1,0)(0, 1) = (0, 0) = 0 R,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. Z 3 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) 0 R 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 4 Basic properties of rings. Theorem 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 S R. S is 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 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 a n for a product of n copies of a. This includes the cases 0 a =0 R and a 0 =1 R if R has an identity element. Be careful: (a+b) 2 = a 2 +ab+ba+b 2 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. 5 Proof. x = a 1 b 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 1 b). Multiply on the left by a 1 to get z = a 1 az = a 1 a(a 1 b)=a 1 b. A similar argument works for y. The solutions x = a 1 b 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 Let R be an integral domain. If a 0and ab = ac, then b =c. Proof. ab = ac implies a(b c) = 0. Since a 0andRis 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 0 in a ring R. a is a zero divisor if there exists an element b 0inR with either ab =0orba =0. Example: 2 3=0=2 0inZ 6. Theorem Every finite integral domain is a field. Combinatorial proof. Let a 0 in the integral domain R. The set ar = { ar r R } is a permutation of the elements of R: ax = ay implies x = y by Theorem 3.10 and there are only finitely many elements. Therefore some ar must be 1 and a has an inverse. Examples. Discuss exercises 8, 11, 17, 25 on pages

6 6 Isomorphisms and homomorphisms. Recall from linear algebra that a linear transformation is a function between vector spaces that preserves the operations on the vector space. For rings we only want to consider the functions that preserve their two operations. Definition, p. 71. Let R and S be rings. A function f : R S (which means the domain is R and f takes values in S) iscalledahomomorphism if f(a + b) =f(a)+f(b)and f(ab) =f(a)f(b) for all a, b R. Examples. Z Q defined by n n is the natural embedding of the integers into the rational numbers. Z Z n defined by n [n]. This is a homomorphism by the definition of addition and multiplication in Z n. (Theorem 2.6) ι R : R R, the identity map for any ring R. f : C C defined by f(a + bi) =a bi (complex conjugation). Check the definition. The last two examples are special in that they are one-to-one (injective) andonto (surjective). In this case, we say the homomorphism is an isomorphism. Iff: R Sis a ring isomorphism, we say R and S are isomorphic. An interesting example in the book is given on p. ( 69: the ) complex numbers are isomorphic to the ring of real 2 2 matrices a b of the form, which corresponds to the complex number a + bi. b a Besides the identity, there is one other rather trivial example: the zero mapping z : R S defined by z(r) =0 S for all r R is a ring homomorphism. Example. Recall Exer. 7, p. 51: R = { (r, r) r R }. Define f : R R by f (r) =(r, r), Check that f is an isomorphism. Theorem Let f : R S be a ring homomorphism. (1) f(0 R )=0 S. (2) f( a) = f(a)for all a R. (3) f(a b) =f(a) f(b)for all a, b R. If R is a ring with identity and f is surjective, then (4) S is a ring with identity and f(1 R )=1 S. (5) Whenever u R is a unit, then f(u) is a unit in S and f(u) 1 = f(u 1 ). Proof. (1) f(0) = f(0 + 0) = f(0) + f(0) implies f(0 R )=0 S. (2) f( a)+f(a)=f( a+a)=f(0) = 0 S by (1), so f( a) = f(a). (3) f(a b) =f(a+( b)) = f(a)+f( b)=f(a) f(b)by(2).

7 (4) Let s S. We must show sf(1) = f(1)s = s. Sincefis surjective, there is some r R with f(r) =s.thensf(1) = f(r)f(1) = f(r 1) = f(r) =sand similarly for f(1)s. (5) f(u 1 )f(u) =f(u 1 u)=f(1) = 1 by (4). Note that we really need surjectivity in (4) and (5). If S = R R and we define f : R S by f (r) =(r, 0), then f(1) = (1, 0) is not the identity in S. Furthermore, 1 R is a unit, but f(1) is not a unit. When the homomorphism f : R S is not surjective, it is handy to have a name for the subset of S that it maps onto. We write im(f) ={f(r) r R}and call this set the image of f. (The book does not have a symbol for the image or for the kernel which we shall define later.) Now we can say that f : R S is surjective if S =im(f). 7 Corollary Given a ring homomorphism f : R S, the image of f is a subring of S. Proof. im(f) since f(0) = 0 im(f). Closure under subtraction is Theorem 3.12(3). Closure under multiplication follows from the definition of homomorphism: f(a)f(b) = f(ab) im(f). Students should read pages for hints on how to decide (or prove) whether two rings are isomorphic or not. Examples. 1. Z 6 is isomorphic to Z 2 Z 3. They have the same number of elements. We must find a bijection which preserves the operations. Define f : Z 6 Z 2 Z 3 by f ([n]) = ([n] 2, [n] 3 ). Check that f is a homomorphism. Since both sets have 6 elements, f will be a bijection if it is either injective OR surjective. Surjectivity is an example of the Chinese Remainder Theorem (page 408). We show injectivity. Assume f([n]) = f([m]). Then f([m n]) = f([m]) f([n]) = 0, so it suffices to prove only [0] maps to ([0], [0]). (Recall the situation for vector spaces!) If [a] ([0] 2, [0] 3 ), then a is divisible by both 2 and 3, hence by 6 since they are relatively prime. Therefore [a] =0inZ Z 4 is not isomorphic to Z 2 Z 2. What fails in trying to use the previous proof is that 2 and 2 are not relatively prime. What is really different about them? Any isomorphism must take 1 Z 4 to (1, 1) Z 2 Z 2.ButinZ 4, we must add 1 to itself four times to get zero, while (1, 1) + (1, 1) = (0, 0). Thus any homomorphism f : Z 4 Z 2 Z 2 will have f(2) = f(0) = 0 and will not be injective. These two examples can be generalized. See exercises 39, 40 on page Pages 76 79, exercises 18, 25, 33.

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