Basic Properties of Rings A ring is an algebraic structure with an addition operation and a multiplication operation. These operations are required to satisfy many (but not all!) familiar properties. Some examples of rings: Each of the number systems Z, Q, R, and C is a ring; each has some additional properties. When viewed in the right context, the integers modulo n form a ring for any fixed value of n > 0. For any fixed n > 0, the set of n n matrices is a ring, using matrix addition and matrix multiplication as the two operations.
Ring Axioms for Addition For any a, b, c R, Properties 1-3 hold: 1. Closure: a + b R 2. Associativity: a + (b + c) = (a + b) + c 3. Commutativity: a + b = b + a 4. Identity: There exists 0 R R such that for every a R, a + 0 R = a. (the same element, 0 R, must work for every element in the ring). 5. Inverses: For any a R, there exists x R such that a + x = 0 R. (the value of x will depend on the given element a).
Ring Axioms for Multiplication For any a, b, c R, Properties 6-8 hold: 6. Closure: ab R 7. Associativity: a(bc) = (ab)c 8. Distributive Laws: a(b + c) = ab + ac and (a + b)c = ab + ac Note: ring multiplication still takes precedence over ring addition. More importantly, ring multiplication is NOT assumed to be Commutative!
Uniqueness of Zero, Cancellation Laws We did not assume that 0 R was unique, but this is easy to show: Suppose that z is a fixed element of R such that a + z = a for every a R. Then z = z + 0 R = 0 R + z = 0 R. Any ring satisfies the Cancellation Laws for Addition. If a + b = a + c, choose x to be a ring element such that a + x = 0 R. Then: b + a = c + a (b + a) + x = (c + a) + x b + (a + x) = c + (a + x) b + 0 R = c + 0 R b = c
Additive Inverses We do not need to assume that a + x = 0 R has a unique solution, but this follows easily from the Cancellation Laws. If y is another element of R such that a + y = 0 R, then a + x = 0 R = a + y, and thus x = y. Given a R, we can thus define a to be the unique x R such that a + x = 0 R. This means a + ( a) = ( a) + a = 0 R A general rule: To show that some expression (say x) is equal to a, you need to show that a + x = 0 R. Since additive inverses are unique, it then follows that x = a. Try proving the following: 0 R = 0 R. For any b R, ( b) = b.
Multiplication by Zero Before introducing ring subtraction and proving its familiar properties, we ll need the following result: For any a R, we have a 0 R = 0 R = 0 R a. Note: We do not assume ring multiplication is commutative, so there really are two things to prove here. We have: (a 0 R ) + (a 0 R ) = a (0 R + 0 R ) = a 0 R = (a 0 R ) + 0 R. By Cancellation, a 0 R = 0 R. The other equality is proven similarly. This is a useful proof to know it comes up surprisingly often! The two basic ingredients are the Additive Identity and the Distributive Laws.
Additive Inverses and Subtraction We can now define ring subtraction: a b = a + ( b). Prove the following familiar properties: a( b) = (ab) = ( a)b. commutative! Do not assume that multiplication is (a + b) = ( a) + ( b). Do not use the 1 symbol! (a b) = ( a) + b = b a. ( a)( b) = ab. [hint: Try using the results above].
Special Types of Rings The Ring Axioms do not include many familiar rules from ordinary algebra this is done with an eye towards generality. In some cases, we will want to assume some additional properties. Let R be a ring. If ab = ba for every a, b R, we say that R is a commutative ring. If there is an element 1 R R such that a 1 R = a = 1 R a for every a R, we say that R is a ring with identity/unity. Unless stated otherwise, we ll always assume 1 R 0 R. These two properties are independent of one another. See the next slide for examples.
Examples The integers (with ordinary addition/multiplication) are a commutative ring with identity [with some additional algebraic properties]. The even integers (with ordinary operations) are a commutative ring without identity. The set of 2 2 matrices with real entries are a non-commutative ring with identity (the ring operations are ordinary matrix addition and multiplication). The set of 2 2 matrices with even integer entries are a non-commutative ring without identity.
Units of a Ring, Multiplicative Inverses Let R be a ring with identity (with 1 R 0 R ). An element a R is a unit if there exist elements x, y R such that ax = 1 R = ya. Although we do not assume that x = y in our definition, it is easy to show that this must be true: x = 1 R x = (ya)x = y(ax) = y 1 R = y. It is also easy to show that if there exists x R such that ax = 1 R = xa, then x must be unique. We can thus define the multiplicative inverse of a to be this unique element x (if it exists), and write x = a 1 as usual. Question: What are the units in Z? In Q?
Division Rings and Fields Let R be a ring with identity. Note that 0 R cannot be a unit. Suppose that for each a 0 R, there exist elements x, y R such that ax = 1 R = ya. Then R is called a division ring. As before, the condition for a division ring implies that x = y and that x is unique. A commutative division ring is called a field. Examples: Q, R, C.
Zero Divisors Let R be any ring. A zero divisor is an element a 0 R such that for some b R, ab = 0 R and/or ba = 0 R. Example: In the ring of 2 2 matrices, we have: ( ) 1 0 0 0 ( ) 0 0 0 1 = ( ) 0 0 0 0 Example: In the ring of integers (mod 6), 2 and 3 are zero divisors, since 2 3 0 (mod 6).
Integral Domains The familiar rings (Z, Q, R, C) have no zero divisors. This is usually stated in the following equivalent form: If ab = 0 R, then a = 0 R or b = 0 R (or possibly both). Using induction and associativity, this extends to: If a product (of finitely many factors) is equal to 0 R, then at least one factor equals 0 R. A commutative ring with identity (with 1 R 0 R ) having no zero divisors is called an integral domain.
Cancellation Laws for Multiplication Suppose that R is an integral domain. If a 0 R and ab = ac, then ab ac = ac ac ab ac = 0 R a(b c) = 0 R b c = 0 R b = c Cancellation on the right can be proven similarly, and also follows from that fact than multiplication is commutative in an integral domain.
Cancellation Laws for Multiplication Conversely, suppose that R is a commutative ring with identity, 1 R 0 R, and the multiplicative Cancellation Laws hold in R. If ab = 0 R and a 0 R, then ab = 0 R = a 0 R. By Cancellation, b = 0 R. Thus R is an integral domain. In particular, note that any field is an integral domain. If ab = 0 R and a 0 R, then we can multiply both sides by a 1 and simplify to get b = 0 R. Z is an integral domain that is not a field (2x = 1 has no solution in Z).
Repeated Ring Addition Let R be a ring, a R, and n Z. Define na (in that order!) as follows: For n = 0, 0a = 0 R [note that 0 Z, but 0 R R]. For n > 0, na = (n 1)a + a [note that is done in Z, + is done in R, and the multiplication is what we re defining with this notation]. For n < 0, na = ( n)( a) [ n > 0; a is the additive inverse in R]. Practically speaking, if n is a positive integer, then we have: na = a + a +... + a [this is n copies of a]. ( n)a = ( a) + ( a) +... + ( a) [this is n copies of a].
Be VERY CAREFUL with this notation! Although na is written as multiplication, it is not the same thing as ring multiplication. In particular, we always have 1a = a (1 Z), even when R does not have an identity 1 R. It follows immediately from the definition that ( 1)a = a for any a R. We also have the following distributive/associative-like laws: For any n, m Z and a R, (n ± m)a = na ± ma. For any n Z and a, b R, n(a ± b) = na ± nb. For any n Z and a, b R, n(ab) = (na)b = a(nb). For any n, m Z and a, b R, (nm)(ab) = (na)(mb). Note that the same notation is used for different operations above. Note also that multiplication is commutative in Z, but might not be in R.
Repeated Ring Multiplication We use the ordinary power notation to denote repeated multiplication. For any a R, we define a n, where n is a positive integer, recursively: a 1 = a. For n > 1, a n = a(a n 1 ). If R is a ring with identity, we define a 0 = 1 R for all a 0 R. If a has a multiplicative inverse, we define a n = (a 1 ) n for n > 0. Question: Can you simplify (ab) 2, or, more generally, (ab) n? What about (a + b) 2?
Since ring multiplication might not be commutative... Unless R is a Commutative Ring, you cannot assume (ab) n = a n b n. In particular (ab) 1 does not always equal a 1 b 1. However, it is easy to check that if a and b have multiplicative inverses, then ab has the multiplicative inverse b 1 a 1 ; note the the order has been reversed! Unless R is a Commutative Ring, you cannot assume (a+b) 2 = a 2 +2ab+b 2. Similar considerations hold for higher powers of binomials. In general, (a + b) 2 = a 2 + ab + ba + b 2. This is a good exercise to try on your own; use the distributive laws for rings very carefully! Fortunately, a n+m = (a n )(a m ), with appropriate restrictions on zero or negative exponents.
Exercises Let R be a ring, a, b R. If R is a ring with identity, show that ( 1 R )a = a. Is a 2 b 2 = (a b)(a + b)? If not, what conditions are required? Does a 2 = 0 R imply a = 0 R? If not, what conditions are required? If n, m Z, does na = ma imply n = m? m = 0. For an easier case, take If a and b are units, is a + b a unit? Is ab a unit?