4
$\begingroup$

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?

What about the converse,

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

$\endgroup$
13
  • $\begingroup$ Different here. This is about "units". The other question is about "zero divisors". $\endgroup$
    – user130893
    Feb 23 '14 at 8:53
  • $\begingroup$ Right, terribly sorry about that. Should've paid more attention. $\endgroup$ Feb 23 '14 at 8:54
  • 1
    $\begingroup$ 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? $\endgroup$
    – user14972
    Feb 23 '14 at 8:55
  • $\begingroup$ I've added my definition of unit now. $\endgroup$
    – user130893
    Feb 23 '14 at 8:58
  • 1
    $\begingroup$ 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. $1$. $\endgroup$
    – user14972
    Feb 23 '14 at 9:11
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.