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.
- 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).]
- 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.
- 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.