Section 18 Rings and fields Instructor: Yifan Yang Spring 2007
Motivation Many sets in mathematics have two binary operations (and thus two algebraic structures) For example, the sets Z, Q, R, M n (R) (the set of n n matrices) all have addition and multiplication. Note that the multiplication in Z, Q, and R are commutative, while that of M n (R) is not. Also, every non-zero element of Q and R has an multiplicative inverse, but this is not the case for Z and M n (R). We wish to give an axiomatic study on these subjects, and determine how various properties (such as commutativity of multiplication, the existence of multiplicative inverse and so on) affect the overall algebraic structures of the sets.
Rings and fields Definition A ring R, +, is a set R together with two binary operations + and, called addition and multiplication, such that the following axioms are satisfied: 1. R, + is an abelian group. 2. Multiplication is associative, i.e., a(bc) = (ab)c for all a, b, c R. 3. For all a, b, c R, the left distributive law a (b + c) = (a b) + (a c) and the right distributive law (a + b) c = (a c) + (b c) hold.
Examples The set Z, Q, R, and C are all rings with the usual addition and multiplication. The set Z n of all residue classes modulo n is a ring with ā + n b = a + b and ā n b = ab. The set M n (R) of all n n matrices is a ring. Let R 1 and R 2 be rings. Define + and on R 1 R 2 by (a 1, b 1 ) + (a 2, b 2 ) = (a 1 + a 2, b 1 + b 2 ) and (a 1, b 1 ) (a 2, b 2 ) = (a 1 a 2, b 1 b 2 ). Then R 1 R 2 is a ring, called the direct product of R 1 and R 2.
Examples Let F be the set of all continuous functions f : R R. Define f + g and f g to be f + g : x f (x) + g(x), f g : x f (x)g(x) Then F, +, is ring. Let F be the set of all linear transformations from R n to R n. Let addition and multiplication be defined by f + g : v f (v) + g(v), f g : v f (g(v)) Then F, +, is a ring. (Actually, this is just a different way to say that M n (R) is a ring.)
Notation For simplicity, we write ab in place of a b. The additive identity of a ring R is denoted by 0. In case some confusion may arise, we also write 0 R. For an element a of R, we let a denote the additive inverse of a. For a positive integer n and an element a of R, the notation n a refers to the sum a + + a having n summands. By convention we set 0 a = 0 R. Here the 0 on the left-hand side is the integer 0, while 0 R on the right is the additive identity element of R.
Basic properties Theorem (18.8) Let R be a ring. For any a, b R, we have 1. 0 R a = a0 R = 0 R. 2. a( b) = ( a)b = ab. 3. ( a)( b) = ab. Proof of (1). Since 0 R + 0 R = 0 R, we have (0 R + 0 R )a = 0 R a. Then by the distributive law, we have 0 R a + 0 R a = 0 R a. Then using the cancellation law for groups, we obtain 0 R a = 0 R. The proof of a0 R = 0 R is similar.
Proof of Theorem 18.8, continued Proof of a( b) = ab = ( a)b. We have, by definition of b, b + ( b) = 0 R. By (1), ab + a( b) = a0 R = 0 R. This means that a( b) is the additive inverse of ab. That is, a( b) = ab. The proof of ( a)b = ab is similar.
Proof of Theorem 18.8, continued Proof of ( a)( b) = ab. By (2) ( a)( b) = (a( b)). By (2) again, (a( b)) = ( ab). This means that ( a)( b) is the additive inverse of ab. But we know that the additive inverse of ab is ab. Thus, by the uniqueness of additive inverse, we conclude( a)( b) = ab.
In-class exercises Determine whether the following algebraic structures are rings. 1. The set Z[x] of all polynomials over Z. 2. The set GL n (R) of all n n invertible matrices under the usual matrix addition and multiplication. {( ) } a b 3. The set : a, b R under the usual matrix b a addition and multiplication. 4. The set 1 2Z = {n/2 : n Z} under the usual addition and multiplication. 5. The set Z[1/2] = { a 0 + a 1 2 + + a } n 2 n : a i Z under the usual addition and multiplication.
Homomorphisms Definition For rings R and R, a function φ : R R is a (ring) homomorphism if 1. φ(a + b) = φ(a) + φ(b), 2. φ(ab) = φ(a)φ(b), for all a, b R. The kernel of φ is the set Ker(φ) = {a R : φ(a) = 0}.
Examples Let n be a positive integer. Let R = Z and R = Z n. Then the function φ : Z Z n defined by φ(a) = a mod n (or φ(a) = a + nz in our notation from the last semester) is a ring homomorphism. The function ψ : Z 2Z defined by ψ(a) = 2a is not a ring homomorphism since ψ(ab) = 2ab, but ψ(a)ψ(b) = 4ab. Note that if we consider Z and 2Z as additive groups, then ψ is a group homomorphism.
Example Let R = R[x] be the set of all polynomials over R. Given a R, define φ a : R[x] R by φ a (f (x)) = f (a). Then φ a is an evaluation homomorphism.
Isomorphisms Definition Let R and R be two rings. A function φ : R R is an isomorphism if 1. φ is a ring homomorphism, 2. φ is one-to-one, or equivalently, Ker(φ) = {0}, 3. φ is onto. We then say R and R are isomorphic.
Example {( ) a b The map φ : C b a } : a, b R defined by ( ) a b φ(a + bi) = b a is a ring isomorphism. The map ( ) a b ψ(a + bi) = b a is another isomorphism.
Multiplicative definitions Definition A ring in which the multiplication is commutative is a commutative ring. If an element a of R satisfies ra = ar = r for all r R, then a is the multiplicative identity or the unity, and is denoted by 1 or 1 R. A ring with a multiplicative identity is a ring with unity. A multiplicative inverse of an element a in a ring with unity is an element a 1 of R such that a 1 a = aa 1 = 1.
Multiplicative definitions Definition Let R be a ring with unity 1 0. An element u of R is a unit if it has a multiplicative inverse. If every nonzero element of R is a unit, then R is a division ring (or skew field). A commutative division ring is a field. A noncommutative division ring is a strictly skew field.
Remarks If a ring R has a multiplicative identity element, it is unique. (See the proof of Theorem 3.13.) If an element a of a ring with unity has a multiplicative inverse, the inverse is unique. The assumption 1 0 in the definition of a field is to avoid the trivial case where the ring consists of just one single element 0. (In this case 0 is both the additive identity and the multiplicative identity.) Conversely, if 1 = 0, then for all a R, a = 1a = 0a = 0, and R = {0} is the trivial ring.
Examples The sets Z, Q, R, and C are all commutative rings with unity under usual addition and multiplication. The rings Q, R, and C are fields, while Z is not since only ±1 are units in Z. The set M n (R) of all n n matrices is a noncommutative ring. The set of units in M n (R) is GL n (R). The set M n (R) is not a division ring since there exist nonzero matrices that are not invertible. The set Z 2, Z 3, and Z 5 are fields. (Take Z 5 for example, the multiplicative inverses of 1, 2, 3, 4 are 1, 3, 2, 4, respectively.) If gcd(a, n) > 1, then ā never has a multiplicative inverse in Z n, since ā b will be a multiple of gcd(a, n) for any b. This shows that if n is composite, then Z n is never a field.
In-class exercises 1. Find the units in Z 12. 2. Find the units in Z 15. 3. Let F be the ring of all continuous functions f : R R with addition and multiplication given by f + g : x f (x) + g(x), f g : x f (x)g(x). What are the units in F?
Subrings and subfields Definition A subset S of a ring R is a subring of R if S is a ring under the induced addition and multiplication from R. We denote this relation by S < R. A subfield is similarly defined.
Homework Problems 8, 10, 12, 18, 19, 24, 25, 37, 40, 48, 50 of Section 18.