A problem on associates

Question

Let $R$ be a commutative ring (not a domain) and $a, b \in R$. If $a | b, b | a$, then there exists a unit $c$ such that $a = bc$?

Answer 1

Consider the ring $R=\mathbb{R}[x,t]/\langle x(t^2+1) \rangle$.

Let $y=xt$. Then we have: $$y=xt,\, x=-yt.$$ However the units in $R$ are just the constant terms in $\mathbb{R}^*$, so there is no unit $c$ with $y=xc$.

To see this, note that every element of $R$ can be written uniquely in canonical form:$$ p(t)+xq(x),$$ for polynomials $p$ with coefficients in $\mathbb{R}$ and $q$ with coefficients in $\mathbb{C}$ (identifying $t$ with $i$ in the latter). That is, as vector spaces we have: $$R\cong \mathbb{R}[t]\oplus x\mathbb{C}[x].$$

If such an element were a unit, then it would remain a unit under the map $R\to \mathbb{C}[x]$, sending $t\mapsto i$. Thus $q=0$. We would then have that $p(t)r(t)=1$, for some polynomial $r(t)$ over $\mathbb{R}$, so we know $p(t)=\lambda$ for some $\lambda\in\mathbb{R} \backslash \{0\}$.