Answers and Solutions to Selected Homework Problems From Section 2.5 S. F. Ellermeyer. and B =. 0 2

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1 Answers and Solutions to Selected Homework Problems From Section 2.5 S. F. Ellermeyer 5. Since gcd (2; 4) 6, then 2 is a zero divisor (and not a unit) in Z 4. In fact, we see that in Z 4. Thus 2x 2y in Z 4 does not imply that x y. In particular, we can write the equation 2x 2y as 2 (x ( y)) 0 and we know that any choice of x and y such that x ( y) 2 will make this equation be true. Thus suppose that x 3 and y (meaning that x ( y) 2 is true). Then 2x and 2y 2 2 so 2x 2y, but x 6 y. In Z 5, all non zero elements (including 2) are units. In particular, 3 2 in Z 5. Thus if we have 2x 2y, we can multiply both sides of this equation by 3 to obtain 3 (2x) 2 (2y) which gives (3 2) x (3 2) y which gives x y which gives x y. By similar reasoning, since 2 is a zero divisor in Z 20, we see that 2x 2y does not imply that x y. For example, suppose that x 6 and y 6. Then 2x and 2y so 2x 2y but x 6 y. In Z 5, 2 is a unit and 8 2. Thus 2x 2y does imply that x y. 6. In the ring M 2;2 (R) (which is a non commutative ring) suppose we take 0 A and B. Then and (A B) 2 A 2 2AB B

2 which shows that (A B) 2 6 A 2 2ABB 2. Using the ring properties, we see that if x and y are any two members of a ring then (x y) 2 (x y) (x y) (x y) x (x y) y x 2 yx xy y 2. (Note that yx need not be equal to xy so we can t replace the yx xy with 2xy.) Returning to our example with matrices, note that A 2 BA AB B which shows that (A B) 2 A 2 BA AB B Let us try plugging each member of Z 5 into the expression x 2 x 4: so we see that x 2 does satisfy the equation x 2 x 4 0 in Z 5. 2

3 Now we will try this in Z 7 : showing that x 2 x 4 0 does not have any solutions in Z The units of M 2;2 (R) are those matrices, A 2 M 2;2 (R) for which there exists a matrix, B, such that AB BA I. These are precisely the matrices that are invertible. In Linear Algebra, there are many criteria that can be used to determine whether or not a matrix is invertible. One such criteria is that a matrix, A, is invertible if and only if det (A) Suppose that x a b p 5 where a and b are rational numbers and suppose that x 6 0. (This implies that either a 6 0 or b 6 0.) If b 0, then x a 6 0 and it is obvious that the multiplicative inverse of x is a (which is also a rational number and also a member of Q p 5 ). Thus, let us suppose that b 6 0. If a 0, then x b p 5 and we can see that the multiplicative inverse of x is b p 5 p p 5 2 Q 5. 5b We are left to consider the case that b 6 0 and a 6 0. In this case, the multiplicative inverse of x is a b p 5 a b p 5 a b p 5 a b p 5 a bp 5 a 2 5b 2 a a 2 5b 2 b a 2 5b 2 p5. 3

4 The above computations are legitimate as long as we are not dividing by 0 in any of these computations. To see that we are not dividing by 0, note that, since we are assuming that a 6 0 and b 6 0, then it must be true that a b p because if a b p 5 0, then we would have p 5 ab which is impossible because p 5 is an irrational number and ab is a rational number. By similar reasoning, we see that a b p We can see that Z 3 [i] has exactly nine members. They are 0, i, 2i,, i, 2i, 2, 2 i, and 2 2i. To make sure that Z 3 [i] is closed under addition and multiplication, we take two members of Z 3 [i], x a bi and y c di (where a; b; c; and d are elements of Z 3 ) and observe that and x y (a bi) (c di) (a c) (b d) i 2 Z 3 [i] xy (a bi) (c di) ac adi bci bdi 2 (ac bd) (ad bc) i 2 Z 3 [i]. The additive identity of Z 3 [i] is 0 and the multiplicative identity is. It is clear that addition and multiplication in Z 3 [i] are both associative and commutative. Also, if x a bi; y c di, and z e fi are members of Z 3 [i], then x (y z) (a bi) ((c e) (d f) i) a (c e) a (d f) i b (c e) i b (d f) i 2 (ac ae bd bf) (ad af bc be) i (ac bd) (ad bc) i (ae bf) (af be) i (a bi) (c di) (a bi) (e fi) xy xz which shows that the distributive law holds. Thus Z 3 [i] is a commutative ring with unity. To see that every non zero member of Z 3 [i] is a unit, we observe that i 2i i 2i ( 2) i 2 2 4

5 so i and 2i are multiplicative inverses of each other. Likewise This shows that Z 3 [i] is a eld. ( i) (2 i) ( 2i) (2 2i) 2 2. Since the multiplicative inverse of 2i is i, we can solve the equation (2 i) x 2i by multiplying both sides of this equation by i to obtain which gives ( i) (2 i) x ( i) ( 2i) x 2. To see that this solution is correct, observe that (2 i) (2) 2i. 5. It is clear that Z 5 [i] is a commutative ring with unity. We ask "Does Z 5 [i] have any zero divisors?" The answer is yes because (2 i) (3 i) 0 in Z 5 [i]. Therefore Z 5 [i] is not an integral domain. 6. In an integral domain, the equation x 2 x can be written as x x ( x) 0 or as (x ( )) x 0. Since an integral domain has no zero divisors, it must be true that either x ( ) 0 or that x 0. The solution of x ( ) 0 is x. Thus x 0 and x are the only two solutions of x 2 x. Note: If 0, which is really not ruled out by the de nitions given in the textbook, then x 2 x has only one solution. However, this can happen only in the case of a one element eld. 5

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