1
$\begingroup$

If $(R,m)$ be an Artinian ring then we know that $m^n=0$ for some integer $n$. Now if $(R,m)$ be a ring such that $m^n=0$, is this Artinian? If no, please give me an example. thanks

  • 1
    For future question like this, you could attempt a search at the DaRT [like this](http://ringtheory.herokuapp.com/search/results/?has=1&has=55&has=105&lacks=3) And as always, if you find a gap and have a suggestion to plug the gap, please use the submission form. – 2017-02-06
  • 1
    Incidentally, it's a little awkward to write a title whose answer is "yes" and then write a question in the body whose answer is "no." Watson's answer is addresses the body, while his answer to the title would be "yes" – 2017-02-06
  • 0
    @rschwieb excuse me. I do not know english well. – 2017-02-06

1 Answers 1

4

No. Take $R=k[X_1,X_2,\dots]/(X_1,X_2,\dots)^2$. It is not noetherian (so not artinian), but it is local with nilpotent maximal ideal $m=(x_1,x_2,\dots)$ (here $x_i$ denotes the equivalence class of $X_i$ in $R$).

  • 1
    I think this is isomorphic to [this ring at DaRT](http://ringtheory.herokuapp.com/rings/ring/62/) which, incidentally, I added at your suggestion :) – 2017-02-06
  • 1
    This reminds me of another lingering task: link the construction tags to pages describing the construction (in this case, the trivial extension.) – 2017-02-06
  • 0
    @S Ali Mousavi : notice that $m$ is actually the nilradical of $R$. More generally, the nilradical of $A/I$ is $\sqrt I / I$. – 2017-02-06
  • 0
    @Watson thank you very much. Only, $k$ can be any field? – 2017-02-06
  • 0
    @SAliMousavi : yes $k$ is any field here. – 2017-02-06
  • 0
    My comment above is proved [here](http://math.stackexchange.com/questions/1622209). Moreover, it can be checked _directly_ that $R$ is not artinian. By saying "it is not noetherian", I wanted to point out that it is not for dimensional reasons that it fails to be artinian – indeed it _has_ dimension 0 (being also local, it has actually only _one_ prime ideal, namely $m$ itself). – 2017-02-07
  • 1
    @Watson Turns out that I had already implemented Keyword searches but they were hidden in the navbar! Soon I will put it in the body of the main page. After that, I think I'll add mouseover descriptions for keywords to make them a little more informative. – 2017-02-15