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.
• 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).]
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.
• Assume that [c,d] divides b. Show that h is injective if and only if [c,d]=b.
• Assume that [c,d] divides b. Show that h is surjective if and only if (c,d) = 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) = ([5]c,[12]d). Explain how you obtain your answer.