$R/I^n$ is a local ring 
Let $I$ be a two sided ideal of a ring $R$ such that $I$ is maximal as a right ideal. I need to show that $R/I^n$ is a local ring, for every $n \geqslant 1$.

For $n=1$ I was able to show that the quotient $R/I$ is a division ring and so it is a local ring (because the non-invertible elements form a group).
For $n>1$ I tried to use induction, but got stuck. Am I on the right track? Do you have any suggestions? Thanks. 
 A: Let me treat the non-commutative version. In case $n=1$, your ring is not necessarily a division ring, but it is always a simple ring, so it is a square matrix ring over a division ring (you do not need this for what follows). 
Let $n\geq 1$, then the two-sided ideals in $R/I^n$ are exactly the two-sided ideals of $R$ containing $I^n$. Now, since a maximal ideal is prime, then if $I^n\subseteq J$, for $J$ another maximal ideal, then either $I$ or $I^{n-1}$ is contained in $J$. If $I\subseteq J$, then they are equal by maximality, if $I^{n-1}\subseteq J$, then repeat this game, until you get, finally, that $I=J$. 
Thus, $I/I^n$ is the unique maximal ideal of $R/I^n$. 
A: You need to show that for each $n$, $R/I^n$ has a unique maximal ideal (just the definition). The ideals in $R/I^n$ are in bijective, inclusion-preserving correspondence to the ideals of $R$ which contain $I^n$ (the bijection is induced by the quotient map). Therefore, the image of $I$ in $R/I^n$ is a unique maximal ideal, thus $R/I^n$ must be local.
That is at least how it works for a commutative ring, where we don't have to care about left and right-ideals. I am not sure if this changes when we assume $I$ to only be maximal as a right-ideal.
