Maximal right ideal

Question

Let $R$ be a ring with unit, $I$ be a maximal right ideal, and $a\notin I$ is invertible. Define $$ J=\{r\in R ; ar\in I \}.$$ Prove that $J$ is a maximal right ideal.

Answer 1

Consider the map $$ f\colon R\to R/I,\qquad f(r)=ar+I $$ This is clearly a homomorphism of right modules and $$ \ker f=\{r\in R:ar\in I\}=J $$ In particular $J$ is a right ideal. Note that up to now we have not used the fact that $I$ is maximal, nor that $a$ is invertible.

Let's assume $a$ invertible; then $f(a^{-1})=1+I$, so $f$ is surjective. If moreover $I$ is maximal, $R/I$ is a simple module. Therefore $R/J=R/\ker f\cong R/I$ is a simple module too, which is the same as saying that $J$ is a maximal right ideal.

Answer 2

Let us prove that $J$ is a right ideal first. If $x$ and $x'$ are elements in $J$, then $ax$, $ax'$ and hence $a(x+x')=ax+ax'$ are elements in $I$. Therefore $x+x'$ is in $J$. If $x$ is $J$ and $r$ is $R$, then $ax$ is in $I$ and hence since it is a right ideal so is $a(xr)=(ax)r$. Therefore $xr$ is in $J$.

To prove that $J$ is proper note that if $J= R$, then $1 \in J$ and hence $a \in I$.

Now lets prove $J$ is maximal. Suppose that $J\subseteq J'$ and let $I'=\{ab\,|\,b\in J'\}$. I leave it as an exercise to check that $I'$ is a right ideal. It follows that $I+I'$ is a right ideal such that $I \subseteq I+I'$. Since $I$ is maximal it follows that either $I=I+I'$ or $I+I'=R$. Now let us show that $J' = \{r\in R\,|\,ar\in I+I'\}$. Clearly $J' \subseteq \{r \in R\,|\,ar \in I'\} \subseteq \{r\in R\,|\,ar\in I+I'\}$ and so we must show that for $r \in R$ if $ar \in I + I'$ then $r$ is in $J'$. If $ar \in I + I'$ then $ar = x +ab$ where $x$ is in $I$ and $b$ is in $J'$. Since $a(r-b)=x \in I$ it follows that $r-b$ is in $J$ and hence in $J'$. Therefore $r = r-b + b$ is in $J'$. Now if $I+I'=I$, then $J'=J$ and if $I+I'=R$, then $J'=R$.