## Math 310: Problem set 8

Instructions: This problem set is due Thursday, November 9, 2006. Your goal is not only to give correct answers but to communicate your ideas well. Make sure you use good English.
1. Let R be the ring whose elements are continuous functions f:[0,1] -> R (i.e., real valued functions y=f(x) defined on the interval 0 <= x <= 1. Thus x has to be between 0 and 1 inclusive, and for each such x, the value f(x) of f at x has to be real, and f:[0,1] -> R has to be continuous).
• If f:[0,1] -> R is a unit in R, show that there is no x with 0 <= x <= 1 such that f(x)=0.
• Answer: For f(x) to be a unit there must be an element g(x) in R such that f(x)g(x) = 1 for all x from 0 to 1. But if f(x) = 0 for some x, then f(x)g(x) = 1 is impossible. [Note that the converse of this problem is also true: if f(x) is never 0 on the interval [0,1], then f(x) is a unit. This uses the fact from calculus that 1/f(x) is continuous at each x such that f(x) is not 0.]
• If there are reals c, d and e in the interval [0,1] such that 0 <= c < d <= 1, f(e) is not 0 and such that f(x) = 0 for all c <= x <= d, show that f(x) is a zero divisor. [Hint: show that there is a function y=g(x) defined on the interval [0,1] such that g(x) is not identically 0, but f(x)g(x) = 0 for all x from 0 to 1. You can do this by drawing the graph of g(x).]
• Answer: Define g(x) such that the graph of g(x) consists of two line segments. Since f(x) is not zero at e, it can't be true that c = 0 and d = 1. Thus either 0 < c or d < 1. If d < 1, take the graph of g(x) to consist of the segment from the point (0,0) to the point (d,0), and the segment from (d,0) to (1,1). This is continuous since the segments connect together. The function g(x) is not the 0 element of R since g(1) is not 0. If 0 < c, take the graph of g(x) to be the segment from (0,1) to (c,0), and the segment from (c,0) to (1,0). This is continuous since the segments connect together. The function g(x) is not the 0 element of R since this time g(0) is not 0. Thus f(x) is not the 0 element of R (since f(e) is not 0), and neither is g(x) (since either g(0) is not 0 or g(1) is not 0), but f(x)g(x) is 0 everywhere, hence f(x)g(x) is the 0 element of R, so f(x) is a zero divisor. [Note: The converse of this problem is also true. I.e., if f(x) is a zero divisor, then there is an interval [c,d] contained in but not equal to [0,1] such that f(x) is 0 on [c,d]. The idea is that if f(x) is a zero divisor, then there is a continuous function g(x) such that neither f(x) nor g(x) is 0 everywhere on [0,1], but f(x)g(x) is 0 everywhere. To prove this, pick a value t of x such that g(t) is not 0. Since g is continuous, there is an interval [c,d] containing t with c < d such that g(x) is not zero for all x in [c,d]. Since f(x)g(x) = 0 for all x in [0,1], it must be true that f(x) = 0 for all x in [c,d].]
2. Let b, c and d be positive integers bigger than 1. Suppose we try to define a map h:Z/bZ -> Z/cZ x Z/dZ by specifying that h([x]b) = ([x]c,[x]d).
• Show that h is well-defined if and only if [c,d] divides b.
• Answer: The map h is well defined if and only if whenever [x]b = [y]b, we also have ([x]c,[x]d) = ([y]c,[y]d). I.e., if and only if whenever b | x-y we also have c | x-y and d | x-y. But if [c,d] divides b, then c | x-y and d | x-y whenever b | x-y since b is a multiple of both c and d. Conversely, assume that whenever b | x-y we also have c | x-y and d | x-y. Let x = b and y = 0. Then b | x-y, so by assumption, c | x-y=b and d | x-y=b. Thus b is a common multiple of c and d so [c,d] | b.
• Assume that [c,d] divides b. Show that h is injective if and only if [c,d]=b.
• Answer: Recall h is injective iff ker(h) = {b}. Let m = [c,d] and note that [m]b is in ker(h). First say h is injective. Then ker(h) = {b}, so, since [m]b is in ker(h), we know [m]b = b; i.e., b | m. But m | b by assumption (in order for h to be well-defined), so m=b. Conversely, suppose that m=b. Let [x]b be in ker(h); this means that ([x]c,[x]d) = (c,d). We must show that [x]b = b. But we know that [x]c=c and that [x]d=d. Thus c | x and d | x, so it follows that b=[c,d] | x. Thus [x]b = b, so ker(h) = {b}.
• Assume that [c,d] divides b. Show that h is surjective if and only if (c,d) = 1.
• Answer: First suppose that (c,d) = 1. Consider any element ([u]c,[v]d) of Z/cZ x Z/dZ. As shown in class, since (c,d) = 1, we can solve:
```x = u (mod c)
x = v (mod d)
```
If x = n is a solution, then h([n]b) = ([u]c,[v]d), so h is surjective. Conversely, assume that h is surjective. Let g = (c,d). Then there is an s such that h([s]b) = (c,d), so d | s-1 (hence g | s-1) and c | s-0=s (hence g | s-0=s). Thus g | s-(s-1) = 1, so g = 1.
3. Let b = 210, c = 14 and d = 15. Let h:Z/bZ -> Z/cZ x Z/dZ be defined as in the previous problem. Find an integer n such that h([n]b) = (c,d). Explain how you obtain your answer.
4. Answer: From the previous problem or from the Chinese Remainder Theorem stated in class, we know h is bijective. Thus there is a solution. To find it, we must solve
```x =  5 (mod 14)
x = 12 (mod 15)
```
simultaneously. But from x = 5 (mod 14) we have x = 5 + 14q for some integer q. From x = 12 (mod 15) we have x = 12 + 15r for some integer r. Thus 5 + 14q = x = 12 + 15r or 7 = 14q - 15r. Solve this using Euclid's algorithm. (Alternatively, note that 7 = (14-15)(-7) so we can take q = -7 and r = 7.) We get x = 5 - 7*14 = -93 (which we check with x = 12 - 7*15 = -93). Thus n = -93 works, as does -93 + m210 for any integer m.