If $R$ is commutative, local notherian ring with $J$ nilpotent then $R$ is Artinian. I am attempting to show that if $R$ is commutative, local notherian ring with $J$ nilpotent then $R$ is Artinian.
Clearly as $\exists k$ such that $J^k=0$ then if there is no other ideal other than $R$ and $J^i$ then we are done but how do I show this?
That is how do I show that if $I$ is an ideal of $R$ then it is equal to one of $R,J,J^2,...,J^k=0$?
Thanks for any help
 A: The conjecture implicit in the line of proof you suggest is not true except for the trivial case when  $n = 1$. For $n > 1$, let $K$ be a field and let $R = K[x, y]/I$ where $I$ is the ideal generated by all monomials $x^iy^j$ where $i + j = n$. Then as a quotient of the polynomial ring $K[x, y]$, $R$ is Noetherian. Any homomorphism from $R$ to a field must annihilate $x$ and $y$, hence as $K$ is a field, if $J$ is the ideal generated by $x$ and $y$, then $R$ is local with maximal ideal $J$ and $J^n = 0$. But $R$ has lots of ideals other than the powers of $J$, e.g., the ideal generated by $x - y$.
A: I'm assuming here that $J=\operatorname{rad}R = \mathfrak m$, where $\mathfrak m$ is the (unique) maximal ideal of $R$. It's certainly not the case that every ideal in $R$ is of the form $\mathfrak m^i$ for some $i$. For example, we could have $R=k[x,y]/(x^n,y^m)$ for some field $k$. In that example $(x,y^2)$ is not of the form $\mathfrak m^i$. To answer your question: the simplest way to show that $R$ is an artinian $R$-module is to use the fact that in an exact sequence 
$$
  0 \to E \to F \to G \to 0
$$ 
of $R$-modules, $F$ is artinian if and only if $E$ and $G$ are. 
Start with the exact sequence 
$$
  0 \to \mathfrak m \to R \to k \to 0
$$
where $k=R/\mathfrak m$ is the residue field of $R$. Clearly $k$ is an artinian $R$-module, so we only have to show that $\mathfrak m$ is artinian. Consider the exact sequence 
$$
  0 \to \mathfrak m^2 \to \mathfrak m \to \mathfrak m/\mathfrak m^2 \to 0
$$
The module $\mathfrak m/\mathfrak m^2$ is actually a $k$-vector space with at most the number of generators of $\mathfrak m$ as an ideal. In other words, $\mathfrak m/\mathfrak m^2$ is a finite-dimensional $k$-vector space, hence artinian over $R$. So we are reduced to showing that $\mathfrak m^2$ is artinian. It's clear that we can iterate this to arrive at an exact sequence 
$$
  0 \to \mathfrak m^{i+1} \to \mathfrak m^i \to \mathfrak m^i/\mathfrak m^{i+1} \to 0
$$
where $\mathfrak m^{i+1}=0$. The first and last terms of the sequence are artinian, so $\mathfrak m^i$, and hence $\mathfrak m^j$ for all $j$, are artinian. Our original exact sequence shows that $R$ itself is artinian. 
