# In a Noetherian local ring every ideal is a power of the Jacobson radical

I want to prove that for $$R$$ a commutative Noetherian local ring, if the Jacobson radical $$J$$ is nilpotent then $$R$$ is Artinian.

My question comes from these lecture notes where the comment in the bracket confuses me. I want to use the fact that in a commutative Noetherian ring, $$\bigcap_{k=1}^{\infty} J^k=0$$ and show that $$R,J,J^2,\dots, J^s=0$$ are all the ideals in $$R$$. But how can I show that every ideal is a power of the Jacobson radical?

• Is this true? If $R=\mathbb C[X,Y]/(X^2,Y^2)$ the maximal ideal is $\mathfrak m=(x,y)$ and the ideal $(x)$ is not a power of $\mathfrak m$. – user26857 Apr 30 '19 at 15:41
• This claim comes from a lecture note but now I think it is not true by your example. – anon Apr 30 '19 at 15:50
• For a proof see here. – user26857 Apr 30 '19 at 16:20

But how can I show that every ideal is a power of the Jacobson radical?

You can't, because that would imply the ideals of the ring are linearly ordered, which of course may not be the case.

Here is the DaRT search showing at least two commutative, local, Noetherian, non-uniserial rings.

To make progress on the problem, I suggest taking the chain of powers of the radical and working out that there must be a finite composition series. That would imply the ring is Artinian.

Another way to do it, if you know a commutative zero dimensional Noetherian ring is Artinian, is to note that $$J$$ is a nilpotent maximal ideal, and hence can be the only prime ideal in the entire ring. Consequently, the ring is zero dimensional.

• Could you suggest how to show that it is Artinian? – anon Apr 30 '19 at 15:41
• @Nicky I went ahead and just moved the suggestion into the solution. – rschwieb Apr 30 '19 at 15:51