Rings, Integral Domains, and Fields

Rings, Integral Domains, and Fields S. F. Ellermeyer September 26, 2006 Suppose that A is a set of objects endowed with two binary operations called addition (and denoted by + ) and multiplication (denoted by ). Let R fa; +; g. R is said to be a ring if the following properties are satis ed: 1. The associative laws of addition and multiplication hold. That is, for any elements a; b; and c 2 A, we have a + (b + c) (a + b) + c and a (bc) (ab) c. 2. The commutative law of addition holds. That is, for any elements a and b 2 C, we have a + b b + a. 3. An additive identity exists. That is, there exists an element 0 2 A such that a + 0 a for a 2 A. 4. Every element has an additive inverse. That is, for each a 2 A, there is an element a 2 A such that a + ( a) 0. 5. The left and right distributive properties holds. That is, for any elements a; b; and c 2 A, we have a (b + c) a b + a c and (b + c) a b a + c a. Remark 1 The de nition of ring does not require that the commutative law of multiplication holds or that a multiplicative identity exists or that each element of the ring has a multiplicative inverse. Remark 2 Some examples of rings that we have studied (either in this course or elsewhere) are Z (the set of all integers), Q (the set of all rational numbers), R (the set of all real numbers), C (the set of all complex 1

numbers), and (for any given integer m > 0) the ring of residues modulo m which we denote by Z m. Remark 3 When we have a ring, R (A; +; ), we will often abuse terminology and refer to the elements of A as elements of R. Thus when we say, for example, x 2 R, we really mean x 2 A. This can almost always be done without causing confusion. Exercise 4 Explain why the set of natural numbers, N, endowed with the usual addition and multiplication, is not a ring. If R (A; +; ) is a ring (meaning that properties 1 5 are satis ed), then R is called a commutative ring if it satis es the additional property 6. The commutative law of multiplication holds. That is, for any elements a and b 2 A, we have a b b a. Remark 5 All of the examples of rings mentioned in Remark 2 are commutative rings. If R (A; +; ) is a ring (meaning that properties 1 5 are satis ed), then R is called a ring with unity (or a ring with one) if it satis es the additional property 7. A multiplicative identity exists. That is, there exists an element 1 2 A such that a 1 a and 1 a a for all a 2 A. If R (A; +; ) satis es all of the properties 1 7, then R is called a commutative ring with unity (or a commutative ring with one). Remark 6 All of the examples of rings mentioned in Remark 2 are commutative rings with unity. De nition 7 An element, a, of a ring, R, is called a zero divisor if a 6 0 and there exists an element b 2 R such that b 6 0 and either a b 0 or b a 0. De nition 8 If R is a ring with unity and a is an element of R for which there exists an element b 2 R such that a b 1 and b a 1, then b is called a multiplicative inverse of a. 2

De nition 9 If a is an element of a ring, R, and a has a multiplicative inverse, then a is called a unit of R. If R is a commutative ring with unity (meaning that properties 1 7 are satis ed) and R has no zero divisors, then R is called an integral domain. Remark 10 All of the examples of rings given in Example 2 are integral domains with the exception of some of rings of residues. Recall that Z m has no zero divisors if and only if m is a prime number. Thus Z m is an integral domain if and only if m is a prime number. If R is an integral domain such that each non zero element of R is a unit, then R is called a eld. Remark 11 Z is not a eld. Q is a eld. R is a eld. C is a eld. Z m is a eld if and only if m is a prime number. (Recall that every non zero element of Z m is either a zero divisor or a unit, and if m is prime then Z m has no zero divisors.) De nition Suppose that R is a ring with unity and suppose that there exists an integer n > 0 such that n summands 0 but that if m is any positive integer less than n, then m summands 6 0. In this case, we call n the characteristic of R. If no such n exists, then we say that R has characteristic zero. Example 13 The eld of real numbers, R, has characteristic zero because for any positive integer n we have n summands 6 0. For any m > 0, the ring Z m has characteristic m because in Z m we have m summands 3 0

but for any positive integer j < m, we have j summands 6 0. Proposition 14 If R is an integral domain with characteristic n > 0, then n is a prime number. Proof. Suppose R is an integral domain with characteristic n > 0. If n is a composite number, then there exist integers p and q with 1 < p < n and 1 < q < n such that n pq. Since n is the characteristic of R, then p summands 6 0 and 1 + 1 + {z + 1 } q summands 6 0 but This gives us 0 n summands 0. @1 + 1 + {z + 1A } @A 1 + 1 + {z + 1 } p summands q summands pq summands 1 n summands 0, which implies that the elements p summands and q summands are zero divisors of R. However, since R is an integral domain, then R has no zero divisors. This means that n must not be a composite number (and hence that n must be a prime number). Corollary 15 If F is a eld with only a nite number of elements, then the characteristic of F is a prime number. Proof. Let F be a eld with only a nite number of members. Then F is an integral domain. If we can prove that F has characteristic n for some integer n > 0, then the conclusion of this corollary will follow immediately from Proposition 14. Suppose, to the contrary, that F has characteristic 4

zero. Then 1 6 0. Furthermore, if we let j and k be any positive integers with j < k, then it cannot be true that j summands k summands for if this were true, then it would be true that (k j) summands which would mean that F does not have characteristic zero. Thus, if F does have characteristic zero, then the elements 1, 1 + 1, 1 + 1 + 1, : : : must all be di erent from each other, meaning that F must contain an in nite number of members. Since this is a contradiction to our hypotheses, we conclude that F must have nite characteristic and hence that the characteristic of F must be a prime number (by Proposition 14) 1 Examples of Rings We will now consider two examples of rings other than those mentioned in Remark 2. We will postpone a discussion of another very important class of examples, rings of polynomials, to Chapter 3. 0 1.1 The Ring of 2 2 Matrices with Real Entries Let M 2;2 (R) denote the set of all 2 2 matrices with real entries and with addition and multiplication de ned in the usual way. That is, a b M 2;2 (R) a; b; c; and d 2 R c d with addition de ned by b11 b + a 21 a 22 b 21 b 22 a11 + b 11 a + b a 21 + b 21 a 22 + b 22 and multiplication de ned by b11 b a11 b 11 + a b 21 a 11 b + a b 22 a 21 a 22 b 21 b 22 a 21 b 11 + a 22 b 21 a 21 b + a 22 b 22. 5

We will show that M 2;2 (R) is a ring with unity, but that this ring is not commutative. 1. First we show that the associative laws of addition and multiplication are satis ed. Let A a 21 a 22 b11 b ; B b 21 b 22 c11 c ; and C c 21 c 22 be elements of M 2;2 (R). Then A + (B + C) b11 + c + 11 b + c a 21 a 22 b 21 + c 21 b 22 + c 22 a11 + (b 11 + c 11 ) a + (b + c ) a 21 + (b 21 + c 21 ) a 22 + (b 22 + c 22 ) (a11 + b 11 ) + c 11 (a + b ) + c (a 21 + b 21 ) + c 21 (a 22 + b 22 ) + c 22 (by the associative law of addition for real numbers) a11 + b 11 a + b c11 c + a 21 + b 21 a 22 + b 22 c 21 c 22 (A + B) + C which shows that the associative law of addition holds in M 2;2 (R). Also, A (BC) b11 c 11 + b c 21 b 11 c + b c 22 a 21 a 22 b 21 c 11 + b 22 c 21 b 21 c + b 22 c 22 a11 (b 11 c 11 + b c 21 ) + a (b 21 c 11 + b 22 c 21 ) a 11 (b 11 c + b c 22 ) + a (b 21 c + b 22 c 22 ) a 21 (b 11 c 11 + b c 21 ) + a 22 (b 21 c 11 + b 22 c 21 ) a 21 (b 11 c + b c 22 ) + a 22 (b 21 c + b 22 c 22 ) a11 b 11 c 11 + a 11 b c 21 + a b 21 c 11 + a b 22 c 21 a 11 b 11 c + a 11 b c 22 + a b 21 c + a b 22 c 22 a 21 b 11 c 11 + a 21 b c 21 + a 22 b 21 c 11 + a 22 b 22 c 21 a 21 b 11 c + a 21 b c 22 + a 22 b 21 c + a 22 b 22 c 22 6

and (AB) C a11 b 11 + a b 21 a 11 b + a b 22 c11 c a 21 b 11 + a 22 b 21 a 21 b + a 22 b 22 c 21 c 22 (a11 b 11 + a b 21 ) c 11 + (a 11 b + a b 22 ) c 21 (a 11 b 11 + a b 21 ) c + (a 11 b + a b 22 ) c 22 (a 21 b 11 + a 22 b 21 ) c 11 + (a 21 b + a 22 b 22 ) c 21 (a 21 b 11 + a 22 b 21 ) c + (a 21 b + a 22 b 22 ) c 22 a11 b 11 c 11 + a 11 b c 21 + a b 21 c 11 + a b 22 c 21 a 11 b 11 c + a 11 b c 22 + a b 21 c + a b 22 c 22 a 21 b 11 c 11 + a 21 b c 21 + a 22 b 21 c 11 + a 22 b 22 c 21 a 21 b 11 c + a 21 b c 22 + a 22 b 21 c + a 22 b 22 c 22 A (BC) which shows that the associative law of multiplication holds in M 2;2 (R). 2. We now show that the commutative law of addition holds in M 2;2 (R). Let A a 21 a 22 be elements of M 2;2 (R). Then A + B b11 b + a 21 a 22 b 21 b 22 a11 + b 11 a + b a 21 + b 21 a 22 + b 22 b11 + a 11 b + a b 21 + a 21 b 22 + a 22 b11 b and B b 21 b 22 (by the commutative law of addition for real numbers) b11 b + b 21 b 22 a 21 a 22 B + A. 3. The additive identity element for M 2;2 (R) is O because if A is any member of M 2;2 (R), then it is easy to see that A + O A. 7

4. Every element of M 2;2 (R) has an additive inverse. It is easy to check that the additive inverse of A a 21 a 22 is A a 11 a a 21 a 22. 5. We now show that the left and right distributive properties hold. (Actually we will just show that the left distributive property holds and it can be considered a homework assignment to show that the right distributive property holds.) Let A a 21 a 22 b11 b ; B b 21 b 22 c11 c ; and C c 21 c 22 be elements of M 2;2 (R). Then A (B + C) b11 + c 11 b + c a 21 a 22 b 21 + c 21 b 22 + c 22 a11 (b 11 + c 11 ) + a (b 21 + c 21 ) a 11 (b + c ) + a (b 22 + c 22 ) a 21 (b 11 + c 11 ) + a 22 (b 21 + c 21 ) a 21 (b + c ) + a 22 (b 22 + c 22 ) a11 b 11 + a 11 c 11 + a b 21 + a c 21 a 11 b + a 11 c + a b 22 + a c 22 a 21 b 11 + a 21 c 11 + a 22 b 21 + a 22 c 21 a 21 b + a 21 c + a 22 b 22 + a 22 c 22 (by the distributive law for real numbers) a11 b 11 + a b 21 a 11 b + a b 22 a11 c + 11 + a c 21 a 11 c + a c 22 a 21 b 11 + a 22 b 21 a 21 b + a 22 b 22 a 21 c 11 + a 22 c 21 a 21 c + a 22 c 22 AB + AC. We have now proved that M 2;2 (R) is a ring. To see that it is a ring with unity, we observe that the element I 0 1 8

is the multiplicative identity element of M 2;2 (R) because if A a 21 a 22 is any element of M 2;2 (R), then AI a 21 a 22 0 1 a 21 a 22 A and IA 0 1 A. a 21 a 22 a 21 a 22 We can see that M 2;2 (R) is not a commutative ring because if we let A and B be A ; B, 6 0 then and AB BA 6 0 6 0 6 6 5 1 so AB 6 BA. To conclude our study of M 2;2 (R), we will show that M 2;2 (R) has both zero divisors and units. For we see that A AB ; B O even though neither A nor B is equal to O. Thus A and B are both zero divisors. For A 0 1 ; B,, 9

we see that and AB BA 0 1 0 1 I I from which we conclude that A and B are both units. It is in fact true that every element of M 2;2 (R) is either a zero divisor or a unit (and not both). The theory of linear algebra is used to prove this. 1.2 The Ring of Rational Numbers with the Number p 2 Adjoined Let Q p 2 be the set of all real numbers of the form a+b p 2 where a and b are rational numbers and with addition and multiplication de ned in the usual way. Then Q p 2 is closed under addition (meaning that if x 2 Q p 2 and y 2 Q p 2, then x + y 2 Q p 2 ) and closed under multiplication (meaning that if x 2 Q p 2 and y 2 Q p 2, then xy 2 Q p 2 ). To see that Q p 2 p p is closed under addition, let x a 1 + b 1 2 2 Q 2 and p p y a 2 +b 2 2 2 Q 2. Then x+y (a1 + a 2 )+(b 1 + b 2 ) p 2 2 Q p 2. To see that Q p 2 p p is closed under multiplication, let x a 1 + b 1 2 2 Q 2 p p and y a 2 + b 2 2 2 Q 2. Then p p xy a 1 + b 1 2 a 2 + b 2 2 a 1 a 2 + a 1 b 2 p 2 + b1 a 2 p 2 + 2b1 b 2 (a 1 a 2 + 2b 1 b 2 ) + (a 1 b 2 + b 1 a 2 ) p 2 shows that xy 2 Q p 2. Hence, both addition and multiplication are binary operations on Q p 2. We will show that Q p 2 is a eld. 1. The associative laws of addition and multiplication obviously hold in Q p 2 because Q p 2 is a subset of R (the set of all real numbers) and these laws hold in R. 2. The commutative law of addition holds in Q p 2 because it holds in R. 10

3. 0 0 + 0 p 2 2 Q p 2 is the additive identity element of Q p 2. 4. If a + b p 2 2 Q p 2, then a b p 2 2 Q p 2 and the latter element if the additive inverse of the former. 5. The distributive laws hold in Q p 2 because they hold in R. 6. Multiplication is commutative in Q p 2 because it is commutative in R. 7. The number 1 1 + 0 p 2 is a member of Q p 2 and this number is the multiplicative identity for Q p 2 because it is the multiplicative identity for R. The above observations show that Q p 2 is a commutative ring with unity. It is also clearly an integral domain because, in R, if ab 0, then either a 0 or b 0, and so the same must be true in Q p 2. (Thus Q p 2 contains no zero divisors.) Finally, to show that Q p 2 is a eld, we must show that every non zero member of Q p 2 has a multiplicative inverse in Q p 2. To do this let x a + b p 2 2 Q p 2 and suppose that x 6 0. Then either a 6 0 or b 6 0 (because if a and b were both zero, then x would be zero). Also, a 2 2b 2 is a rational number and we claim that, in addition, a 2 2b 2 6 0. This claim is true because if it were true that a 2 2b 2 0, then we would have a 2 2b 2 which would give us jaj p 2 jbj, which can t be true because jaj is a rational number and p 2 jbj is an irrational number (or zero if b 0, but this would also mean that a 0 and a and b can t both be zero). Since a 2 2b 2 is a rational number which is not equal to zero, we see that a a 2 2b + b p2 2 a 2 2b 2 is a member of Q p 2. The following calculation shows that the above element is the multiplicative inverse of a + b p 2: a + b p a 2 a 2 2b + b p2 1 a + b p 2 a b p 2 2 a 2 2b 2 a 2 2b 2 1 a 2 2b 2 a 2 2b 2 1. We have now proved that Q p 2 is a eld. 11

2 Some Algebraic Properties of Rings Henceforth, we will often use juxtaposition (no symbol) for the operation of multiplication in a ring. This means that we will write ab instead of a b to denote multiplication. The following propositions give some basic algebraic facts about rings. Proposition 16 If R is a ring, then R contains only one additive identity element (which we denote by 0), and if R is a ring with unity, then R contains only one multiplicative identity element (which we denote by 1). Also, if R is a ring with unity and a is a unit in R, then a has only one multiplicative inverse (which we denote by a 1 ). Proof. Suppose that 0 1 and 0 2 are additive identity elements for R. Then, since 0 1 is an additive identity element, we have 0 2 + 0 1 0 2 and since 0 2 is an additive identity element, we have 0 2 + 0 1 0 1. This shows that 0 1 0 2. Therefore R has exactly one additive identity element. Suppose that and 1 2 are multiplicative identity elements for R (a ring with unity). Then, since is a multiplicative identity element, we have 1 2 1 2 and since 1 2 is a multiplicative identity element, we have 1 2. This shows that 1 2. Therefore R has exactly one multiplicative identity element. Now let R be a ring with unity and let a be a unit in R. Suppose that b 1 and b 2 are elements of R such that ab 1 b 1 a 1 and ab 2 b 2 a 1.

(That is, suppose that b 1 and b 2 both serve as multiplicative inverses for a.) Then, since ab 1 ab 2, we obtain b 1 (ab 1 ) b 1 (ab 2 ) and the associative law of multiplication then gives us and this gives us (b 1 a) b 1 (b 1 a) b 2 1b 1 1b 2 from which we obtain b 1 b 2. This proves that a has a unique multiplicative inverse. Proposition 17 Let R be a ring and let a and b be elements of R. Then: 1. a0 0 and 0a 0. 2. a ( b) (ab) and ( a) b (ab). 3. ( a) a 4. ( a) ( b) ab. 5. ( 1) a a and a ( 1) a (in rings with unity). Proof. 1. Because 0 is the additive identity element of R, we know that 0+0 0. Multiplying both sides of this equation by a, we obtain a (0 + 0) a0 which, by the left distributive property, gives us a0 + a0 a0. Since a0 is an element of R, we know that a0 has an additive inverse, which we denote by (a0). Adding this element to both sides of the above equation, we obtain (a0) + (a0 + a0) (a0) + a0. 13

We now use the associative property of addition to obtain and this gives us ( (a0) + a0) + a0 (a0) + a0 0 + a0 0 which, since 0 + a0 a0, gives us a0 0. The proof that 0a 0 is similar and is left as homework. 1. Since (ab) is the additive inverse of ab, we know that (ab)+ab 0. Now observe that a ( b) + ab a ( b + b) (by the left distributive property) and since a ( b) + ab a ( b + b) a0 0. b + b 0, we see that Since the additive inverse of ab is unique (by Proposition 16), then it must be the case that a ( b) (ab). The proof that ( a) b (ab) is similar and is left as homework. 2. Since a + a 0, we see that a is the additive inverse of a. In other words, ( a) a. 3. By using parts 2 and 3 of this Proposition, we obtain ( a) ( b) (a ( b)) ( (ab)) ab. 4. Using part 2 of this Proposition, we obtain ( 1) a (1a) a and a ( 1) (a1) a. Proposition 18 If R is a ring with unity, then no element of R can be both a zero divisor and a unit. 14

Proof. Suppose that a is a unit in R (meaning that a 1 exists). If a 0, then a is not a zero divisor. Thus suppose that a 6 0. Suppose also that either ab 0 or ba 0 for some element b 2 R. If ab 0, then we obtain a 1 (ab) a which gives us (a 1 a) b 0 which gives us 1b 0 which gives us b 0. If ba 0, then we also obtain the conclusion that b 0 (by similar reasoning). This shows that a cannot be a zero divisor. Exercise 19 1. Prove or disprove: If R is a ring and x and y are elements of R such that xy 0, then yx 0. 2. Prove or disprove: If R is a ring and x and y are elements of R such that y 6 0 and xy y, then x 1. 3. Let c be a symbol and let F ffcg ; +; g with addition and multiplication de ned by c + c c and c c c. (Note that the underlying set here contains only one element.) Prove that F is a eld. 4. Prove that if R is a ring in which 1 0, then R must have only one element and that this element is a unit (and not a zero divisor). 5. Prove that if R is a ring with unity, y is a unit in R, and x is an element of R such that xy 1, then x y 1. 6. Prove that if R is a ring with unity and y is a unit in R, then y 1 is also a unit in R and (y 1 ) 1 y. 7. Suppose that R is a ring and that a; b; and c are elements of R such that ba 1 and ac 1. Prove that a is a unit and that a 1 b c. Exercise 20 In the textbook, Section 2.5 (page 89), do problems 5, 6, 11, (requires some recall of linear algebra), 13, 14, 15, and 16. 15