Show that if $I$ is an ideal contained in $J$ then $(R/I)\setminus (R/I)^{\times}$ is an ideal of $R/I$

Question

Question: Let $R\neq 0$ be a ring and $J= R\setminus R^{\times}$ an ideal of $R$.

Show that if $I\subset J$ is an ideal of $R$ then $(R/I)\setminus (R/I)^{\times}$ is an ideal of $R/I$

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Attempt: I have been able to show that $R/J$ is a division ring. Since $I\subset J$, all elements of $I$ are nonunits, but there may be nonunits of $R$ not in in $I$ (but they are in in $J$).

I believe I have shown that elements of $(R/I)^{\times}$ are of the form $r+I$, where $r\in R^{\times}$:

Let $r+I\in (R/I)^{\times}$. Then there exists some $s+I\in R/I$ such that $(r+I)(s+I)=(s+I)(r+I)=rs+I=1_R +I=1_{R/I}$ which implies that $rs=1_R$ and therefore $r\in R^{\times}$.

So then my intuition tells me that elements of $(R/I)\setminus (R/I)^{\times}$ are of the form $r'+I$ where $r'\notin R^{\times}$. From this point forward I believe it suffices to show that $(R/I)\setminus (R/I)^{\times}$ absorbs multiplication on the left and right, ie. for $r'+I \in (R/I)\setminus (R/I)^{\times}$ and $s+I \in R/I$:

$$ (s+I)(r'+I) = sr'+I \\ (r'+I)(s+I) = r's + I $$

Now $r's$ is not a unit (suppose it was, then $r'st=1_R$ for some $t\in R$. But then $r'$ must be a unit with inverse $st$, a contradiction).

Thus $(r'+I)(s+I)\in (R/I)\setminus (R/I)^{\times}$, if I can show that $sr'$ is a nonunit, then I can conclude that $(R/I)\setminus (R/I)^{\times}$ is indeed an ideal of $R/I$.

Answer 1

The condition that $J = R\backslash R^\times$ is an ideal means that $R$ is a local ring with $J$ the unique maximal ideal. To see this, note first that $J$ is proper, since $1$ is a unit. Let $\mathfrak{m}$ be some maximal ideal of $R$; then every element of $\mathfrak{m}$ is a nonunit since $\mathfrak{m}$ is proper, so that $\mathfrak{m}\subset J$ and thus $\mathfrak{m}=J$ since $\mathfrak{m}$ is maximal and $J$ is proper.

Now, consider the natural projection $\pi:R\to R/I$. By the lattice isomorphism theorem, maximal ideals of $R$ correspond to maximal ideals of $R/I$. Thus $R/I$ has a unique maximal ideal, which means that $R/I$ is also a local ring, so that $(R/I)\backslash (R/I)^\times$ is an ideal.