# No units in quotient ring equivalent to no units in original ring?

Definition: Let $R$ be a ring with $1$. $r\in R$ is a unit if and only if $r \neq 1$ and there exists $s\in R, s \neq 1$ such that $rs=1=sr$.

Let $R$ be a ring with $1$ and let $I$ be a proper ideal of $R$.

$R/I$ has no units $\Rightarrow$ $R$ has no units?

$R$ has no units $\Rightarrow$ $R/I$ has no units?

• Different here. This is about "units". The other question is about "zero divisors". Feb 23 '14 at 8:53
• Right, terribly sorry about that. Should've paid more attention. Feb 23 '14 at 8:54
• Hrm. What precisely do you mean by "no units"? That the only units are roots of unity? That you're using the variation on the definition of "ring" that doesn't require a ring to have 1, and "no unit" insists that it, in fact, does not?
– user14972
Feb 23 '14 at 8:55
• I've added my definition of unit now. Feb 23 '14 at 8:58
• If by "no units" you mean no element in the ring is a unit, then your implication is vacuously true, since $R/I$ and $R$ always hav a unit: e.g. $1$.
– user14972
Feb 23 '14 at 9:11

The natural map $R\to R/I$ maps invertible elements to invertible elements. If $R/I$ has no invertible elements except $1$, then all invertible elements of $R$ are congruent modulo $I$ with $1$.
This can well happen: for example, take $R=\mathbb Z_2[x]/(x^2)$ and $I=(x)$. Then $R/I$ (which is isomorphic to $\mathbb Z_2$) has no invertible element except $1$, but $R$ does have more than $1$ invertible element.\
The converse implication is also false: if $R=\mathbb Z_2[x,y]$ is a polynomial ring in two variables over the field of two elements, $R$ has exactly one invertible element, the unit. Yet $R/(xy-1)$ has many invertible elements.