3.1 Do the following problem together with those of Section 3.2:

Transcription

1 Chapter First check if the subset S is really a subring. The checklist for that is spelled out in Theorem 3.2. If S is a subring, try to find an element 1 S in S that acts as a multiplicative identity. 15. Multiplication and addition is understood to be component by component. See Theorem Begin by finding an integer b such that a b = a for all integers a. That element (which is not the number zero) will be your 0 R. Then try to find an element b in Z such that a b = a for all a in Z. That is your 1 R, even though it is not the number 1. After you have checked that (Z,, ) is a commutative ring with identity, show that a b = 0 R implies that either a = 0 R or b = 0 R 26. It might help to realize that a log b = e (log a)(log b). (Here, the logarithm is the natural logarithm with base e.) Which numbers play the role of 0 R and 1 R? Do not be misled, 0 R will not be the number 0; it couldn t be, since only positive real numbers are considered. Similarly, 1 R will not be the number 1. You need: a 0 R = a and a 1 R = a for all positive real numbers a. For a 0 R can you find a positive real number b such that a b = 1 R? 29. For example to find out what s t should equal, try to multiply out s t = s (s + s) = Carefully investigate the multiplication table for Z 2 Z 3, which is printed in the back of your textbook as the solution to Problem 15(a). Notice that both Z 2 are Z 3 are fields. (Why?) 44. Before you do this problem, you should have a look at Problem 19, whose solution is in the back of the text. For 0 R and 1 R try the sets and S, respectively. 45. Notice that this is exactly the arithmetic of the complex numbers: (a + bi)(c + di) = (ac bd) + (ad + bc)i Therefore, 0 R = (0, 0) and 1 R = (1, 0). Moreover, given (a, b) (0, 0), the element (a/(a 2 + b 2 ), b/(a 2 + b 2 )) will serve as a multiplicative inverse, like in Example 11 on page 49 about the complex numbers.

2 46. This problem shows you a systematic way of creating rings and subrings that have different multiplicative identities. If you want to work out some examples first, notice that all you need to get started, is two relatively prime integers r and s with r 2 and s 2. Since gcd(r, s) = 1, there are integers α and β with the property that αr + βs = 1. So, αr = ( β)s + 1. Notice that β will not be a multiple of r, since otherwise r would have to divide 1, which it cannot. If β is in the range from 1 to (r 1), take k = β and you are set: r divides ks + 1 and 1 k < r. Otherwise, notice that you can always change β by any multiple of r, so as to force it to be in that range, because we have (α s)r + (β ± r)s = 1. For example, for r = 2 and s = 5, we get α = 3, β = 1 and k = 1 yielding Problem 21, namely the subring S = {[0], [2], [4], [6], [8]} of Z 10 with a multiplicative identity 1 S = [ks + 1] = [6] which is different from 1 R = [1]. The point of the problem is that the element [ks + 1] is in the set S = {[0], [r], [2r], [3r],, [(s 1)r]} (why?) and acts like a multiplicative identity in that set: [xr] [ks + 1] = = [xr] in the ring Z rs (which you should check). Carefully explain why 1 S = [ks + 1] [1] = 1 R, where 1 R denotes the multiplicative identity element of the larger ring R = Z rs. Do not forget to show that the set S is a subring of R = Z rs. 3.1 Do the following problem together with those of Section 3.2: 38. Use Theorem 3.6 of Section 3.2 with S = A R as defined in the problem: First, verify that 0 R S so that S is not empty. Then let x, y S (i.e. assume that a x = 0 R and a y = 0 R ) and show that x y S (i.e. a (x y) = 0 R ) and that x y S (i.e. a (x y) = 0 R ). [Caution: only verifying that S is closed under addition and multiplication does not make S a subring! Remember: Z + is not a subring of Z.] (a) Yes, S T will always be a subring of R provided S and T are subrings of R. To prove this, use Theorem 3.6: Find an element, which lies in both S and T to show that S T is not

3 empty. Then let a, b S T and explain why a b S T and a b S T. (b) Examine R = Z with the subrings S = 2Z = {2n n Z} and T = 3Z = {3n n Z}. 14. If e 2 = e, then e(e 1 R ) = 0 R. 15. (b) Try some 2 2 matrices. 16. Where is the zero 0 R? 20. Pick r R with r 0 R and s S with s 0 S. Multiply (r, 0) and (0, s). 25. (a) You have already seen such an example. (b) Work out the expression 1 S (1 S 1 R ). What do you conclude? Justify, very carefully, every step of your analysis. [In fact, if 1 S 1 R, then every non-zero element s of S is a zero divisor of R, as can be seen by working out s(1 S 1 R ).] 28. First recall that a unit is never a zero divisor. Now use Problem 21 to explain why the row labeled a, in the multiplication table for R must feature every one of the four elements of R. The same is true for the column labeled a and the row/column labeled b. 31. Use the hint in the back of your textbook. 32. (a) It is obvious that addition satisfies all those axioms, which feature addition alone. There is no need to write those out. Clearly, 0 T = (0 R, 0 Z ). However, do check associativity of multiplication and at least one of the two distributive laws. The fun part, of course, is to hunt down the identity element: 1 T = (?,?) (b) After you have shown that R = {(r, 0) r R} is a subring of T, lean back and think about what this whole problem proves: this subring R is actually identical to the ring R along with its arithmetic. The only difference between R and R is that the elements of R keep carrying a second zero-coordinate with them. In summary, you are proving that every ring R without identity can be placed inside of a ring T with identity! This trick will be used later in the course. 33. Be careful, this problem is trickier than it may seem. Start by choosing elements u and v such that (ab)u = 1 R = u(ab) and

4 av = 1 R = va. You need to find an element w such that bw = 1 R = wb. Try w = ua. Watch it, though, you also need to prove that bw = 1 R not only that wb = 1 R. The problem here is that the ring R need not be commutative. Trick: first prove that a(bw) = a and then solve this equation for bw Under the given assignment, the tables do not correspond. Find elements a and b such that either f(a+b) f(a)+f(b) or f(a b) f(a) f(b). Notice, however, that the two rings are isomorphic. This was shown on Page 67. It s just that the function suggested in this problem is not an isomorphism. 15. Careful: f(r) might be 0 S, in which case we would not call it a zero divisor. 19. First you should verify that the function f is well-defined. (After all it is defined using representatives!) That is, if x x mod 7, explain why [8x] 28 = [8 x] 28 in S. Since S = {[0], [8], [2 8], [3 8],, [6 8]} is a complete list of all multiples of 8, the function f is clearly onto. What makes f a homomorphism is the fact that [8] [8] = [8] in Z 28. Counting elements will show that f must be one-to-one also. Alternatively, you can show directly that f is one-to-one: If f([x 1 ]) = f([x 2 ]), then [8x 1 ] = [8x 2 ] in Z 28. So, 8x 1 8x 2 mod 28, and therefore also mod 7. Then why x 1 x 2 mod 7? This time around, f will automatically be onto, by counting elements. An aside: If you are interested in how they keep finding these examples, reread Problem 46 of Section 3.1. If r and s are two integers with r 2, s 2, and gcd(r, s) = 1, we can find an integer k with 1 k < r such that r divides ks + 1. In this example, r = 4, s = 7, and k = 1. (See the discussion for 3.1 #46 above.) Consider the subring S = {[0], [r], [2r], [3r],, [(s 1)r]} of Z rs. The element e S = [ks + 1] of Z rs is an element of S and it has the property that e S e S = e S in Z rs. (Indeed, e S = 1 S.) Hence, f : Z s S given by f([t]) = [t]e S = [(ks + 1)t] is a homomorphism. (The function f is well-defined because r divides ks + 1, so that if t 1 t 2 mod s, then (ks + 1)t 1 (ks + 1)t 2 mod rs.) We see that f is injective, because

5 f([t 1 ]) = f([t 2 ]) will imply that [(ks + 1)t 1 ] = [(ks + 1)t 2 ] in Z rs. So, (ks+1)t 1 (ks+1)t 2 mod rs and also mod s. But (ks+1) 1 mod s, so that t 1 t 2 mod s. Now, counting elements makes f also onto. Investigating this further: If we interchange the role of r and s in the above, we can compile a function g : Z r Z rs given by g([t]) = [t]e R, where e R = [lr+1] is chosen so that s divides lr+1 and 1 l < s. We can combine all this to obtain a homomorphism ϕ : Z r Z s Z rs given by ϕ([t 1 ], [t 2 ]) = [t 1 ]e R + [t 2 ]e S. The reason why this is a homomorphism lies in the following rules that hold in Z rs : e S e S = e S e R e R = e R e S e R = 0 (These elements are called orthogonal idempotents.) Looking back at how we constructed e R and e S ( 3.1 #46 above) we see that e R + e S = [αr] + [βs] = 1 in Z rs. This makes ϕ onto: if [a] Z rs, then ϕ([a], [a]) = [a]e R + [a]e S = [a](e R + e S ) = [a]. Hence, counting elements, ϕ is also one-to-one and therefore an isomorphism. Indeed, we see that ϕ is the inverse of the isomorphism ψ : Z rs Z r Z s that we found in class as ψ([a]) = ([a], [a]). We have come full circle! 21. Follow the hint in the back of your textbook. 35. (b) Recall that M(R) denotes all 2 2 matrices with real-number entries. Are both rings commutative? (d) How many solutions can you find to the equation x 2 = 1 R +1 R in each of the two rings? For isomorphic rings that needs to be the same number. (f) Count units.

6 40. Think about the following example and generalize the idea: 3Z is not isomorphic to 5Z. Proof: Suppose to the contrary that there is an isomorphism f : 3Z 5Z. Then f(9) = f(3 3) = f(3) f(3) = (5q)f(3) for some q Z. On the other hand, f(9) = f(3+3+3) = f(3)+f(3)+f(3) = 3f(3). Conclude that f(3) = 0. But f(0) = 0, also. So, f was not injective after all. Contradiction.