Suppose $R$ is a ring and let $I$ be a proper ideal of $R$. Which of the statements are true?:
a) $R$ has no units then $R/I$ has no units.
b) $R/I$ has no units then $R$ has no units.
I can prove that (b) is true but can't find counterexamples for (a). Any suggestions?
For part $(a)$, let $R = 2\mathbb{Z}$, and let $I=6\mathbb{Z}$.
Then $R$ has no units, and $I$ is a proper ideal of $R$, but the element $4+I$ is a multiplicative identity in $R/I$.
Note that every element of $R$ is congruent, mod $I$, to exactly one of $0,2,4$, hence $R/I$ has exactly $3$ elements, namely $0+I,\;2+I,\;4+I$.
To see that $4+I$ is a multiplicative identity in $R/I$, note that \begin{align*} (4+I)(0+I)&=0+I\\[4pt] (4+I)(2+I)&=8+I=2+I\\[4pt] (4+I)(4+I)&=16+I=4+I\\[4pt] \end{align*}
In fact, since $4+I$ is the multiplicative identity in $R/I$, and since $(2+I)(2+I)=4+I$, it follows that all nonzero elements of $R/I$ are units, hence $R/I$ is a field, isomorphic to $F_3$ (the field with $3$ elements).