Past exam of the algebra 2 course of the University of Oxford / 2015 / 1 / a / i / Solution

If $a=ub$ with $u$ a unit in a ring, then $a$ lies in the principal ideal generated by $b$. This gives $(a)\subseteq (b)$. Since $u$ has a multiplicative inverse, $b=u^{-1}a$, giving the reverse inclusion. ThereforeThus associate elements always generate the same principal ideal; the associator ring condition asks whether equality of these ideals also forces the generators to differ by a unit.