Zero divisors dividing each other In problem 7.15 of Shoup's number theory text, we were to consider the ring of functions from $\mathbb{R}$ to itself with addition and multiplication defined pointwise, i.e. $(f+g)=f(x)+g(x)$ and $(fg)(x)=f(x)g(x)$. In part (d) it is asked to find examples of two continuous functions, $a,b$ which divide each other in the subring of continuous functions, $C$, but with no function $c\in C^\times$ such that $a$ and $b$ differ by this unit, i.e. $a\vert b$ and $b\vert a$ but $ac\neq b$ for any $c\in C^\times$. However, 4 pages earlier theorem 7.4(a) states that two nonzero divisors $x,y$ in ring $R$ divide each other if and only if they differ by a unit, and it seems to me that these two statements contradict each other.  The problem does not explicitly say that $a,b$ are not zero divisors, so I would assume that the answer lies there but I am unsure if this is possible to begin with. An interpretation of the problem I heard was that the two functions divide each other in the larger ring, but the unit does not come from $C$ which I do not believe to be correct as the problem does explicitly say "in the ring $C$ we have $a\vert b$ and $b\vert a$..." which leads me to believe that the former interpretation is what was meant but I could be wrong.
 A: The interpretation in the problem is correct: you can find $a,b\in\mathcal{C}$ such that $a\mid b$ and $b\mid a$ in the ring $\mathcal{C},$ and such that there does not exist $c\in\mathcal{C}^\times$ such that $ac = b.$ As you note, this will require $a$ and $b$ to be zero divisors in $\mathcal{C}.$
Hint 1: It might help to think about what zero divisors in $\mathcal{C}$ must look like.
Hint 2: Use part (c)!
Spoilers below:

 Let $a$ be a function which is identically $0$ on some $[\alpha,\beta]$, and nonzero outside of the interval (you might take $a(x) = -x - 1$ when $x \leq 1,$ $a(x) = 0$ when $-1\leq x\leq 1,$ and $a(x) = x - 1$ when $x\geq 1$). Take $b$ to be identically $0$ on $[\alpha,\beta],$ with $b(x) = -a(x)$ when $x\leq \alpha$ and $b(x) = a(x)$ when $x\geq \beta.$


Then set $s(x) = -1$ for $x\leq \alpha,$ $s(x) = 1$ for $x\geq \beta,$ and $s(x) = \frac{2x}{\beta - \alpha} - \frac{\beta + \alpha}{\beta - \alpha}.$ Less explicitly, $s$ is piecewise linear, and equal to $-1$ when $b(x) = -a(x),$ and equal to $1$ when $b(x) = a(x).$


 You can  check that $b(x)s(x) = a(x)$ and $a(x) s(x) = b(x),$ so that $a\mid b$ and $b\mid a.$ However, there cannot be a $c\in\mathcal{C}^\times$ such that $ac = b$ by part (c): such a $c$ is either always positive or always negative, but if $c$ is always positive, we will not have $b(x) = a(x)c(x)$ when $x\leq \alpha,$ and if $c$ is always negative, we will not have $b(x) = a(x) c(x)$ when $x\geq\beta.$

