# 5.1 Commutative rings; Integral Domains

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1 5.1 J.A.Beachy Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following statements (listed in Section 5.1): (d) a 0 = 0 for all a R; Solution: Since 0 is the additive identity element, we have = 0. To connect this property to multiplication, we need to use the distributive law. We have a 0 = a (0+0) = a 0+a 0, and then we can subtract a 0 from both sides of the equation to get 0 = a 0. (e) ( a) = a for all a R; Solution: This holds in any group. (f) ( a) ( b) = ab for all a,b R. Solution: Because we are proving a relationship between addition (since a is defined to be the additive inverse of a) and multiplication, we need to use the distributive law. We have ab + a( b) = a(b b) = a 0 = 0, and so a( b) = (ab). Similarly, ( a)b = ab, and then ( a)( b) = (a( b)) = ( ab) = ab. 24. (a) Is Z 2 a subring of Z 6? Solution: No, because Z 2 is not even a subset of Z 6. (b) Is 3Z 6 a subring of Z 6? Solution: The set 3Z 6 = {[0] 6,[3] 6 } is closed under addition, since we know it is a subgroup of the underlying additive group of Z 6. Furthermore, 3Z 6 is easily checked to be closed under multiplication. But our definition of a subring requires that the subset must contain the identity element of the ring, and since [1] 6 3Z 6, we conclude that 3Z 6 is not a subring of Z 6. Comment: Be careful here, because some authors do not require that a subring of a commutative ring with 1 must have the same identity element. In fact, we changed our definition between the second and third editions of our text Abstract Algebra. 25. (a) Show that the ring of Gaussian integers is an integral domain. Solution: It is easy to check that the set Z[i] = {m+ni m,n Z} is closed under addition and multiplication and contains 1. It follows that Z[i] is a subring of C, and so Theorem implies that Z[i] is an integral domain. (b) Show that Z[ 2] = {m+n 2 m,n Z} is an integral domain. Solution: As in part (a), Z[ 2] is a subring of C and so it follows from Theorem that Z[ 2] is an integral domain. Comment: The details of the proofs that Z[i] and Z[ 2] are subrings of C can be found in Examples and of Abstract Algebra.

2 5.1 J.A.Beachy Let R be a commutative ring. Prove that the intersection of any collection of subrings of R is a subring of R. Solution: To use Proposition 5.1.4, let a and b beelements in the intersection. Then a and b each belong to every subringin the collection, so a+b, ab, and a also belong to every subring in the collection. Therefore a+b, ab, and a belong to the intersection. Finally, the multiplicative identity of R belongs to each subring, so it also belongs to the intersection. 27. Show that in an integral domain the only idempotent elements are 0 and 1. Solution: Let D be an integral domain, and suppose that e D is idempotent. Then e e = e 2 = e = e 1, and so if e 0, then e = 1 since the cancellation law holds in D. Since 0 and 1 are idempotent, they are the only idempotent elements in D. 28. Show that if n = p k (p prime), then in Z n the only idempotent elements are 0 and 1. Solution: If e 2 = e in Z n, then p k e(e 1), and since gcd(e,e 1) = 1, this implies the either p k e, and e = 0, or else p k (e 1) and e = In Z 24, (a) find all nilpotent elements; Solution: We have x n 0 (mod 24) if and only if 3 x and 2 x. Thus the list of nilpotent elements is {0,6,12,18} = 2 3. (b) find all idempotent elements. Solution: If e Z 24 is idempotent, then 24 e(e 1). Since gcd(e,e 1) = 1 this implies that 3 e or 3 e 1, or else 8 e or 8 e 1. The list of elements e with 8 e or 8 e 1 is {e e = 0,1,8,9,16,17}. Adding the condition that 3 e or 3 e 1 reduces the list to {0,1,9,16}. 30. Let R be the set of all 3 3 matrices of the form (a) Show that R is a commutative ring. b a 0 with a,b,c Z. Solution: From linear algebra we know that addition and multiplication of matrices satisfy all of the axioms of a commutative ring, except the commutative law. Therefore it is enough to show that the given set satisfies the commutative law, and satisfies the conditions of a subring given in Proposition First, it is clear that the set is closed under addition, is closed under forming additive inverses, and contains the identity matrix. The following calculations verify closure and commutativity of multiplication. b a 0 d 0 0 e d 0 f e d = ad 0 0 bd+ae ad 0 cd+bd+af bd+ae ad

3 5.1 J.A.Beachy 3 d 0 0 e d 0 f e d b a 0 (b) Find the units of R. = d ea+db da 0 fa+eb+dc ea+db da Solution: Because the entries are from Z, in which only 1 and 1 are invertible, a matrix in R is invertible if and only its determinant is ±1. Thus an element is invertible if and only if it has either 1 or 1 on the main diagonal. (c) Find the nilpotent elements of R. Solution: The nilpotent matrices in R are the strictly lower triangular matrices. 31. Let R and S be commutative rings. Prove or disprove the following statements. (a) An element (a,b) R S is idempotent if and only if a an idempotent in R and b is idempotent in S. Solution: Since (a,b) 2 = (a 2,b 2 ), it is clear that (a,b) 2 = (a,b) if and only if a 2 = a and b 2 = b. (b) An element (a,b) R S is nilpotent if and only if a nilpotent in R and b is nilpotent in S. Solution: This clearly holds since (a,b) n = (a n,b n ). Comment: If m and k are the smallest positive integers with a m = 0 and b k = 0, respectively, thenthesmallestpositiveinteger nwith(a,b) n = (0,0) isn = max{m,k}. (c) An element (a,b) R S is a zero divisor if and only if a is a zero divisor in R and b is a zero divisor in S. Solution: This part is false since (1,0) is a zero divisor in Z Z because (1,0)(0,1) = (0,0), but 1 is certainly not a zero divisor in Z. 32. In the ring Z Z, let R = {(x,y) x y (mod 2)}. (a) Show that R is a subring of Z Z. Comment: Although the definition of a direct sum was first given in Exercise of the text, there is also a formal definition in Section 5.2. Solution: To show that R is a subring we will use Proposition If (a,b) and (c,d) belong to R, then a b (mod 2) and c d (mod 2), so it follows that a + c b + d (mod 2) and ac bd (mod 2), and thus R is closed under addition and multiplication. Furthermore, a b (mod 2) implies (a,b) R, and it is clear that the identity (1,1) of Z Z is in R. (b) Is R an integral domain? Solution: No, because (2,0) (0,2) = (0,0). (c) Find all idempotent elements of R. Solution: An element of a direct sum of commutative rings is idempotent if and only if each component is idempotent. Thus the idempotent elements of Z Z are (0,0), (1,1), (1,0), (0,1), and of these only (0,0) and (1,1) belong to R.

4 5.1 J.A.Beachy Find a commutative ring with zero divisors a,b such that a+b is not a zero divisor and a+b 0. Solution: In Z 2 Z 2 we have (1,0) (0,1) = (0,0), so (1,0) and (0,1) are zero divisors, but their sum (1,0)+(0,1) = (1,1) is not a zero divisor. 34. Although the set Z 2 [x]/ x 2 +1 is not a field, Proposition shows that addition and multiplication of congruence classes of polynomials is well-defined. Show that in this set the cancellation law for multiplication does not hold. Solution: Since x , we have x Therefore we have x(1 + x) = x+x 2 x+1. The cancellation law does not hold since for the nonzero elements 1, x and 1+x we have x(1+x) = 1(1+x) but x Recall that Z 3 [x]/ x 2 +1 is a field (this follows from Theorem since x 2 +1 is irreducible over Z 3 ). Find the order of each element in its group of units. Solution: The nonzero elements are 1, 1, x, x+1, x 1, x, x+1 = (x 1), and x 1 = (x+1). Of course, 1 has order 2. Since x 2 1, we have x 3 x and x 4 1. It follows that x and x have order 4, while x 2 has order 2. We have (x+1) 2 = x 2 x+1 x, so x+1 has order 8, making it a generator of the group of units. Starting over again and calculating powers of x+1, we have the following. (x+1) 2 x (x+1) 3 (x+1)( x) x 2 x x+1 (x+1) 4 ( x) 2 1 (x+1) 5 (x+1) (x+1) 6 (x+1) 2 x (x+1) 7 x(x+1) x 1 (x+1) 8 ( 1) 2 1 Thus the congruence classes of ±x±1 have order 8, while those of ±x have order 4. Finally 1 has order 2 and 1 has order 1. Comment: Review Proposition if you have questions about finding the order of an element of a cyclic group. 36. Let p be an odd prime number that is not congruent to 1 modulo 4. Prove that the ring Z p [x]/ x 2 +1 is a field. Hint: A root of x 2 = 1 leads to an element of order 4 in Z p. Solution: We must show that x 2 +1is irreducibleover Z p, or, equivalently, that x 2 +1 has no root in Z p. Supposethataisarootofx 2 +1inZ p. Thena 2 1 (mod p), andsoa 4 1 (mod p). The element a cannot be a root of x 2 1, so it does not have order 2, and thus it must have order 4. By Lagrange s theorem, this means that 4 is a divisor of the order of Z p, which is p 1. Therefore p = 4q +1 for some q Z, contradicting the assumption.

5 5.1 J.A.Beachy Let S be a commutative ring. (a) Let R = {n 1 n Z}. Show that R is a subring of S. Solution: If m,n Z, we have m 1+n 1 = (m+n) 1 by the associative law for addition, and (m 1) (n 1) = (mn) 1 by the distributive law. Finally, 1 = 1 1, and so R is subring of S. (b) Find the subring defined in part (a) for the ring given in Problem 30 and for the field given in Problem 35. Solution: In the ring b a 0 with a,b,c Z defined in Problem 27, the identity element is the identity matrix. The subring defined in part (a) is therefore the set of all scalar matrices. In the field Z 3 [x]/ x 2 +1 defined in Problem 30, the elements are the congruence classes represented by linear polynomials. The identity element is the constant polynomial [1] 3, and so the subringdefined in part (a) is the set of all constant polynomials. 38. (a) Let S be a set that satisfies all of the axioms of commutative ring, with the possible exception of the commutative law for multiplication. Show that the set R = {r S rs = sr for all s S} is a commutative ring. Note: The ring R is called the center of S. Solution: Since all of the axioms except the commutative law hold, we can use the proof of Proposition to simplify the calculations. If r 1,r 2 R, then (r 1 +r 2 )s = r 1 s + r 2 s = sr 1 + sr 2 = s(r 1 + r 2 ), and (r 1 r 2 )s = r 1 (r 2 s) = r 1 (sr 2 ) = (r 1 s)r 2 = (sr 1 )r 2 = s(r 1 r 2 ). Furthermore, ( r 1 )s = (r 1 s) = (sr 1 ) = s( r 1 ). Finally, by definition 1 R, and since it is clear that elements in R commute with each other, we are justified in concluding that R is a commutative ring. (b) Find the center of the set M 2 (R) of all 2 2 matrices with real entries. Comment: I need to comment on the following calculations, so that they don t just look like a trick. It would be natural to start out by just writing down an element in the center and seeing what you get by letting it commute with a general element of M 2 (R). In fact (you can try it), you would get four equations in four unknowns, with four general (undetermined) constants, and the situation is quite out of hand. The point is to take these equations and carefully choose some of the constants so that you can get some meaningful equations. The best way to do this seems me to be to pick (carefully) a couple of elements of M 2 (R) and see what specific information you get. My choice of elements is based on trial and error. [ ] a b Solution: Let belong to the center. In particular, it must commute with c d [ ] [ ][ ] [ ][ ] 1 0 a b a b the element, and so we must have =. 1 1 c d c d This yields four equations, and the useful ones are a+b = a, which shows that b has to be zero, and c+d = a+c, which shows that a = d. Next, any such element must

6 5.1 J.A.Beachy 6 [ ] [ ][ ] [ ][ ] 1 1 a a 0 also commute with, so we get =. Now 0 1 c a c a we get a = a+c, and so c = 0. We conclude that any element in the center of M 2 (R) must be a scalar matrix. On the other hand, it is clear that any scalar matrix is in the center, and so the center is precisely the set of scalar matrices. ANSWERS AND HINTS 42. Let R and S be commutative rings. Show that (R S) = R S (as groups). Hint: Define φ : R S (R S) by φ((u,v)) = (u,v) and verify that (u,v) is a unit of R S if and only if u is a unit in R and v is a unit in S. 44. Let R be a Boolean ring with 16 elements. Find R. Answer: R = {1}. Note: the number 16 is a red herring. 45. Let R be the ring of continuous functions from R to R. (Remember that R is a ring under multiplication of functions, not composition of functions.) (b) Characterize the units of R. Answer: Either f(x) > 0 for all x R, or f(x) < 0 for all x R. (d) Characterize the nilpotent elements of R. Answer: The only nilpotent element is the zero function. 48. Let R be a commutative ring. (a) Can R[x] have nilpotent elements of positive degree? Hint: Look at Z 4 [x]. (b) Can R[x] have idempotent elements of positive degree? Hint: Look at Z 6 [x].

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