Principal ideals and units

Question

Let $R$ be a commutative ring with identity. Suppose $(a)=(b)$ with $a,b$ in $R$. Then $a=bu$, with $u$ a unit in $R$. Is this true?

Answer 1

No. Let $R = \{f: [0, 3] \to \mathbb{R} \mid f \text{ is continuous}\}$. For $x \in [0, 1]$, let $f(x) = 1 - x$ and $g(x) = 1 - x$. For $x \in [1, 2]$, let $g(x) = f(x) = 0$. For $x \in [2, 3]$, let $f(x) = x - 2$, and $g(x) = 2 - x$. (I recommend drawing the picture).

Now $f \mid g$ and $g \mid f$, let for example $h(x) = 1$ for $x \in [0, 2]$ and $h(x) = -1$ for $x \in [2, 3]$. Then $h \cdot f = g$ and $h \cdot g = f$. So $(f) = (g)$.

Let's assume there's a unit $u$ so that $u \cdot f = g$. Then $u(\frac{1}{2}) = 1$ and $u(\frac{5}{2}) = -1$. Since $u$ is continous, $u(y) = 0$ for some $y \in [0, 3]$. But then $u$ can't be a unit, since for any $u^{-1}$ we have $(u \cdot u^{-1})(y) = 0 \neq 1$.