Solutions to Assignment 4


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1 Solutions to Assignment 4 Math 412, Winter Define a new addition and multiplication on Z y a a + 1 and a a + a, where the operations on the righthand side off the equal signs are ordinary addition, sutraction, and multiplication. Prove that, with the new operations and, Z is an integral domain. OK, here goes: First we must show that with the new operations Z is a commutative ring. Then we must show that whenever a 0 for a, Z, then a 0 or 0. To keep things less confusing, we will write R to denote Z with these new and exciting operations. It is clear that R is a nonempty set. Let me tick off the other axioms. (a) If a, R, then a a + 1 is an integer, and hence is in R so that R is closed under. () If a,, c R, then (a ) c (a + 1) c a c 1 a + ( + c 1) 1 a ( c), so that is associative. (c) If a, R, then a a a 1 a, so that is commutative. (d) For all a R, note that a 1 a a, so that 0 R exists (it is the integer 1). (e) If a R, then a ( a + 2) a + ( a + 2) R so that inverses exist. (f) If a, R, then a a + a is an integer and hence is in R (that is, R is closed under ). (g) If a,, c R, then a ( c) a (+c c) a+(+c c) a(+c c) a++c c a ac+ac a+ a+c (a+ a)(c) (a+ a) c (a ) c, so that is associative. (h) If a,, c R, then a ( c) a ( + c 1) a + ( + c 1) a( + c 1) a + + c 1 a ac + a (a + a) + (a + c ac) 1 (a ) + (a c) 1 (a ) (a c). Furthermore, (a ) c (a + 1) c a c (a + 1)c a c ac c + c (a + c ac) + ( + c c) 1 (a c) + ( c) 1 (a c) ( c). We conclude that distriutivity holds. 1
2 (i) Note for all a R, 0 a 0 + a (0)(a) a, and thus 1 R exists (it is the integer 0). At this point we know that R is a ring. (j) Note that for all a, R, a a + a + a a a and thus R is commutative. Note that 0 R 1 R, and now suppose that a 0 R. We want to show that a 0 R or 0 R. Of course a 0 R implies that a + a 1 (an equation in Z). Because R is commutative it is enough to assume that 0 R (that is, that 1) and show that a 0 R follows. Rearranging things we get that a a 1, or a(1 ) (1 ). In turn this implies that a 1. Thus a 1 0 R and we have completed the proof Complete the tales for the following rings: + r s t r r s t s s t r t t r s r s t r r r r s r t t r Well, we want to compute st, ts, and tt. We know that st s(s + s) ss + ss t + t s. Also ts (s + s)s ss + ss t + t s. Finally tt t(s + s) ts + ts s + s t. So the diagrams should read: + r s t r r s t s s t r t t r s r s t r r r r s r t s t r s t Show that M(Z 2 ) is a 16element noncommutative ring with identity. Again, here goes: It is clear that M(Z 2 ) is a nonempty set. This set has 16 elements ecause there are 4 places in each matrix to fill, and for each spot we have two choices (either 0 or 1) thus there are possiilities. Let me tick off the other axioms. I will write M M(Z 2 ) for short. 2
3 a a (a) If, a a c d c d M, then + a + a + c d c d c + c d + d M so that M is closed under addition (I used that addition is closed in Z 2 ). a a () If, a c d c d, c d M, then ( ) a a + a c d c d + a + a + c d a c + c d + d + c d a + a + a + + a a c + c + c d + d + d + + a + c d c + c d + d ( ) a a + a c d c d + c d, so that addition is associative (I am using that addition is associative in Z 2 ). a a (c) If, a a c d c d M, then + a + a + c d c d c + c d + d a + a + a a c + c d + + d c d c d (using the fact that addition commutes in Z 2 ) so that addition is commutative. a a 0 0 a (d) For all M, note that +, so that a zero c d c d 0 0 c d element exists (using the fact that one exits in M). a a a 0 0 (e) If M, note that + 0 c d c d c d 0 0 M, so that multiplicative inverses exists in M (I am using the fact that a exists in Z 2 ). a a (f) If, a a c d c d M, then aa c d c d + c a + d ca + dc c + dd M so that M is closed under multiplication (using that Z 2 is closed under multiplication). a a (g) If, a c d c d, c d M, then ( ) a a a aa c d c d c d + c a + d a ca + dc c + dd c d aa a + c a + a c + d c aa + c + a d + d d ca a + dc a + c c + dd c ca + dc + c d + dd d () a a a + c a + d c d c a + d c c + d d ( ) a a a d c d c d + c d so that multiplication is associative (I used that multiplication is associative and distriutive in Z 2 ). 3
4 a a (h) If, a c d c d, c d M, then and ( ) a a a c d c d + a a c d + a + c d c + c d + d aa + aa + c + c a + a + d + d ca + ca + d + d c + c + dd + dd + c ) + + c ) (a + d ) + (a + d ) (ca + d ) + (ca + d ) (c + dd ) + (c + dd ) + c ) (a + d ) (ca + d ) (c + dd + + c ) (a + d ) ) (ca + d ) (c + dd ) a a a a c d c d + c d c d, ( ) a a + a a + a + c d c d c d a c + c d + d c d aa + a a + c + c a + a + d + d ca + c a + dc + d c c + c + dd + d d + c ) + (a a + c ) (a + d ) + (a + d ) (ca + dc ) + (c a + d c ) (c + dd ) + (c + d d ) + c ) (a + d ) (a (ca + dc ) (c + dd + a + c ) (a + d ) ) (c a + d c ) (c + d d ) a a a c d c d + a c d c d. We conclude that multiplication and addition distriute. a a 1 0 a 1 0 a (i) If M, then, so M has c d c d 0 1 c d 0 1 c d an identity (using that Z 2 has an identity) (j) Note that, so that multiplication is noncommutative. Thus M 2 (Z 2 ) is a noncommutative ring with 16 elements. Note that I actually proved that M 2 (Z n ) is a ring for all n > Let d e an integer that is not a perfect square. Show that Q( d) {r + s d r, s, Q} 4
5 is a sufield of C. Denote Q( d) y S, and suppose a d and a2 + 2 d are elements of S. Then (a d) + (a2 + 2 d) (a1 + a 2 ) + ( ) d, which is an element of S (a 1 + a 2 and are in Q), and (a d)(a2 + 2 d) a 1 a 2 + a 1 2 d + a2 1 d d (a 1 a d) + (a a 2 1 ) d, which also elongs to S (each of these coefficients in again in Q). We conclude that S is closed under multiplication and addition. Note also that d is in S. Finally, for (a + d) S, we know that a + ( ) d S, and (a + d) + ( a + ( ) d) 0, so that the solution to s + x 0 is in S for all s S. We can now conclude y theorem 3.2, that S will e a suring of C. To demonstrate that S is a sufield, it remains to show that if s 0 is in S, then s 1 S. So consider a + d S. If a 0 then 0, and an inverse is 1 d d which is an element of S. If 0 then a 0, and an inverse is 1/a + 0 d which is also an element of S. If a 0, then consider a a d + d. a d a I claim that this is an element of S. The quantities and will e a d a d in Q as long as a d is not zero. If it were zero, then a 2 2 d or (a/) 2 d, so that a/ d. Because d is not a perfect square, we know that a. But 2 a 2. This gives a contradiction. To show this we use induction. Suppose that p 1 p n is a prime factorization of and let P (n) e the statement that if 2 a 2 then a. For n 1, this says that if p 2 1 a 2 then p 1 a, which is clearly true ecause p 1 is prime. So suppose that P (t) holds for some t > 1. We are required to prove that P (t + 1) must e true. If (p 1... p t+1 ) 2 a 2 this implies that p t+1 a 2, and thus that p t+1 a (again ecause p t+1 is prime). Thus there exists a 1 Z such that p t+1 a 1 a. Now we have that (p 1 p t ) 2 (a 1 ) 2. By induction this means that (p 1, p t ) a 1. Thus there is a m Z such that p 1 p t m a 1, and we conclude that p 1 p t p t+1 m a 1 p t a, that is, that p 1 p t+1 a and hence that P (t + 1) is true. Thus if 2 a 2, it must e the case that a, a contradiction ecause d a/ is not an integer. The product (a + ( d) a + a d a d ) d is equal to 1, so that the inverse of a + d is in S, and this completes the proof. (I found this equation y multiplying out (a + d)(c + e d) and solving for c and e in terms of a and ) Let R and S e nonzero rings. Show that R S contains zero divisors. Consider the elements (0, s) and (r, 0) where r R, s S, and r 0 s. Such 5
6 r and s can e found ecause R and S are nonzero rings. Then (0, s) (r, 0) (0 r, s 0) (0, 0), so that oth (0, s) and (r, 0) are zero divisors Assume that R {0 R, 1 R, a, } is a ring and that a and are units. Write out the multiplication tale of R. The first two columns and rows are easy ecause we know how 0 and 1 ehave a a a 0 a 0 Now we know that units have unique inverses (see page 60) and are not zero divisors (done in class). If aa 1, then a 1 (ecause a ), so 1. Then a 0 and a 1, so a a or a. By theorem 3.10 (cancellation), the first of these options implies that 1, and the second that a 1, oth contradictions. (Note that this argument is symmetric in a and, that is, if we started with 1 we would get the same conclusion). So it must e the case that a 1, and hence that a 1 as well. Clearly aa a (again y theorem 3.10), so it must e that aa. Then a and we are finished a a a 0 a a Let R e a nonzero finite commutative ring with no zero divisors. Prove that R is a field. If we can show that R must contain the identity, then R will e a finite integral domain (ecause it will contain no elements for which a 0, ut a 0 ). Then we will simply use theorem Let {0.r 1,..., r n } e the elements of R and consider the set R i {r i r 1, r i r 2,..., r i r n }, 1 i n. For each i, if r i r j r i r k, then r j r k y cancellation, and thus the elements in R i are distinct. It follows that for each i there is an t i such that 1 t i n and r i r ti r i. Note that r i r ti r i r ti r tti r i r tti 6
7 so y cancellation, r ti r tti. That is, r ti r i r i r i r ti. Now we claim that r t1 is the identity in R. It is enough to show that for each r i R, r t1 r i r i. Given i, there exists a k such that r t1 r k r i, and thus r 1 r k r 1 r t1 r k r 1 r i. By cancellation, r k r i and thus r t1 r i r i as required. We conclude that R contains an identity. As mentioned aove, theorem 3.11 now completes the proof. 7
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