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.


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 $R/I^n$.

  • 1
    $\begingroup$ I need to prove that $R/I^n$ has only one maximal right ideal (that's the definition on the non-commutative case). You proved that there is a unique two sided ideal. I don't know how to prove that $I^n /subseteq J$ implies $I$ or $I^{n-1}$ contained in $J$, with $J$ maximal right ideal. How do I do that? $\endgroup$ – Makuta Apr 29 '15 at 19:54
  • $\begingroup$ you are right! There is something missing in my proof... $\endgroup$ – Simone Apr 29 '15 at 20:07

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.


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.