Algebra 2. Rings and fields. Rings. A.M. Cohen, H. Cuypers, H. Sterk

Size: px

Start display at page:

Download "Algebra 2. Rings and fields. Rings. A.M. Cohen, H. Cuypers, H. Sterk"

Shanon Campbell
5 years ago
Views:

1 and fields 2 and fields A.M. Cohen, H. Cuypers, H. Sterk A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Multiplication turns each of the sets Z, Q, R, C, Z[X ], Q[X ], R[X ], C[X ] into monoids, whereas addition defines a group structure. These two structures are combined in the notion of a ring. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Definition A ring is a structure [R, +, 0,,, 1] consisting of a set R for which [R, +, 0, ] is a commutative group and [R,, 1] is a monoid, in such a way that the following laws hold for all x, y, z R: x (y + z)=x y + x z (left distributivity); (y + z) x=y x + z x (right distributivity). The ring is called commutative if the monoid [R,, 1] is commutative. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 1

2 and fields Notation and terminology: + is called the addition, is called the multiplication (the symbol is often omitted), 0 is the zero element, 1 is the identity element of the ring. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Usual arithmetic Each of Z, Q, R, C, with the usual addition and multiplication, is a commutative ring. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Modular arithmetic Addition and multiplication as defined in the chapter on modular arithmetic [] determine a commutative ring structure on Z/nZ. The zero element is the class of 0, the identity element is the class of 1. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 2

3 and fields Polynomial rings Let R be a commutative ring. Then R[X ], with the usual addition and multiplication [] is a commutative ring. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Residue class rings If R is a commutative ring and f is a polynomial in R[X ], then R[X ]/f R[X ], as defined in the chapter on modular polynomial arithmetic [], is a commutative ring. The zero element is 0 + f R[X ], the identity element is 1 + f R[X ]. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields The Gaussian integers The subset R=Z + Z i of the complex numbers is a ring with the usual addition and multiplication, with zero element 0=0 + 0 i and identity element 1=1 + 0 i. Most ring properties, like associativity of the multiplication, are inherited from the ring C: since they hold in the complex numbers they hold a fortiori in the subset R. A crucial issue is to verify if R is a ring, we have to check that R is closed with respect to the operations. For instance, (a + b i) (c + d i)=(a c b d) + (a d + b c) i shows that the set R is closed with respect to multiplication, since a c b d and a d + b c are integers if a, b, c, d are. The ring R is called the ring of Gaussian integers. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 3

4 and fields Matrix rings Let R be a ring. Then the following structure is a ring: S=[M(n, R), +, 0,,, 1], where M(n, R) is the set of n by n matrices with coefficients in R, where 0 is short for the zero matrix, 1 is short for the identity matrix, + denotes matrix addition and denotes matrix multiplication. If n > 1, it is easy, and left to the reader, to find matrices A, B such that A B and B A are distinct. Thus, S is not commutative for n > 1 even if R is commutative. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields The quaternions Take 1, i, j, k to be a set of four vectors (think of a standard basis) of the 4-dimensional real vector space H=R 1 + R i + R j + R k. On H we define the operations + and as follows. For x=a 1 + b i + c j + d k and x =a 1 + b i + c j + d k let x + x be the vector sum of x and x and set x x =p 1 + q i + r j + s k, where p=a a b b c c d d, q=a b + b a + c d d c, r=a c b d + c a + d b, and s=a d + b c c b + d a. Now H is a ring. (It is quite tedious to check associativity, etc.) Since i j = k = j i, the ring is not commutative. H is called the ring of real quaternions. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Here is an application of the ring of the Gaussian integers R=Z + Z i. Suppose that the integers k and l can both be written as sums of two squares of integers: k=a 2 + b 2 and l=c 2 + d 2. Then the product k l is also a sum of squares. You may find it hard to show this from scratch. Here is how the ring R comes into play: k=(a + b i) (a b i) and l=(c + d i) (c d i) so k l=(a + b i) (c + d i) (a b i) (c d i). Expanding the product of the first two factors gives a c b d +(a d +b c) i and the product of the last two factors is a c b d (a d + b c) i. This yields k l=(a c b d) 2 + (a d + b c) 2. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 4

5 and fields An argument similar to the Gaussian integers application 9, using the quaternions, can be used to show that if two integers can be written as sums of four squares of integers, then so can their product. The equality (a + b i + c j + d k) (a b i c j d k)=a 2 + b 2 + c 2 + d 2 plays a role in the proof. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Let R be a ring and let X be an indeterminate. By R[X ] we denote the set of all polynomials in X with coefficients in R, compare a previous section []. Let a=a 0 + a 1 X a n X n and b=b 0 + b 1 X b m X m be two elements of R[X ]. By adding, if necessary, some terms 0 X k we may assume n=m. The sum of these polynomials is a + b=a 0 + b 0 + (a 1 + b 1) X (a n + b n) X n. The product of these polynomials is a b=c 0 + c 1 X c n+m X n+m, where c k=a 0 b k + a 1 b k a k b 0. The symbol is often omitted. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Theorem The sum and product of polynomials define the structure of a commutative ring on the set R[X ] of all polynomials in X with coefficients in R. The zero element is the zero polynomial 0; the identity element is the polynomial 1. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 5

6 and fields The ring R[X ] is called the polynomial ring over R in the indeterminate X. The ring R is called the coefficient ring of R[X ]. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Let R be a ring and take S to be the polynomial ring R[X ]. Then the polynomial ring S in the indeterminate Y is the same as the ring R [X, Y ] of polynomials in the two indeterminates X, Y. So its elements are of the form ij N N ai,j X i Y j, with a i,j R, nonzero for only a finite number of pairs (i, j). The element X Y is equal to the product Y X. This emphasizes that there are two ways to build this ring with indeterminates X and Y from R: As R[X ][Y ] and as R[Y ][X ]. To emphasize the symmetry in X and Y, we usually write R [X, Y ] for this ring. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Notions like degree are of course valid for all polynomial rings. But weird things may happen if the coefficient ring R is not a field: (2 X ) (2 X )=0 in Z/4Z[X ]. Here the degree of the product of two polynomials of degree 2 is not 4. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 6

7 and fields Let R be a ring. Addition defines a group structure on R. So every element a has an inverse with respect to the addition. This inverse is denoted a and is called the opposite of a. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Theorem The following properties hold for all a, b R. 1 a 0 = 0 a = 0; 2 a ( b) = (a b) = ( a) b; 3 ( a) ( b)=a b; 4 ( 1) a= a. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields The ring laws lead to rules for calculations which are known from the usual examples. For instance, if, in a product, one factor is 0, then the whole product is 0. Another example: ( a 1) ( a 2)... ( a n)=( 1) n a 1 a 2... a n. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 7

8 and fields Recall that a ring R is a monoid with respect to multiplication. It is not necessarily the case that every (nonzero) element has an inverse with respect to multiplication. Those elements of R that do have an inverse are called the invertible elements of R. The inverse of a in R is denoted by a 1. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Theorem The invertible elements of R form a multiplicative group (i.e., a group with respect to multiplication). This group is denoted by R. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields The Euler indicator [] can be expressed in terms of such a group: Φ(n)= (Z/nZ). A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 8

9 and fields Usual arithmetic Z ={1, 1}. Every nonzero element of Q, R, and C is invertible. (In other words, they are fields.) A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Modular arithmetic (Z/nZ) consists of the classes m+n Z of Z/nZ for which m is an integer such that gcd(m, n)=1. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Polynomials rings (Z[X ]) ={1, 1} and Q[X ] =Q\{0}. Similarly for R and C. To prove these statements you will need to involve the degree. We leave this to the reader. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 9

10 and fields Residue class rings If R=Q, R, or C, and f is a polynomial in R[X ] of positive degree, then (R[X ]/f R[X ]) consists of the residue classes of those polynomials g in R[X ] for which gcd(g, f )=1. If R=Z, then it is harder to describe the invertible elements of (R[X ]/f R[X ]) for general f. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Gaussian integers (Z + Z i) ={1, 1, i, i}. If a + b i is invertible, then there exists an element c + d i such that (a + b i) (c + d i)=1. Using the property z w = z w for the absolute value of complex numbers, we infer that (a 2 + b 2 ) (c 2 + d 2 )=1. Since a, b, c, d are integers, we find that the integer a 2 +b 2 divides 1. The conclusion is that a+b i must be one of the four elements 1, 1, i, i, as stated. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 and fields Matrix rings Let R be a commutative ring. The invertible elements of M(n, R) are those matrices whose determinant is invertible in R. This follows from Cramer s rule, which expresses the inverse of a matrix in terms of minors (elements of R) and the inverse of the determinant. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 10

11 and fields The Quaternions H ={x H N(x) R } where N(x)=a 2 + b 2 + c 2 + d 2 if x=a 1 + b i + c j + d k. As for the proof: write C(x)=a 1 (b i + c j d k). Then C(x) x=n(x). So, if N(x) is invertible, then (N(x)) 1 C(x) is the inverse of x. Also, if N(x) is not invertible, it is zero and so x is zero or a zero divisor; in particular, it is not invertible. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, / 31 11

Section 19 Integral domains

Section 19 Integral domains Instructor: Yifan Yang Spring 2007 Observation and motivation There are rings in which ab = 0 implies a = 0 or b = 0 For examples, Z, Q, R, C, and Z[x] are all such rings There