# In a commutative local Noetherian ring $R$ with maximal ideal $J$, if $J$ is not nilpotent then $R$ is an integral domain.

We've just proved this result:

Let $$R$$ be a commutative, local, Noetherian ring. Suppose that $$J$$ (the maximal ideal) is principal. Then every nonzero ideal of $$R$$ is a power of $$J$$.

And now we want to prove this, which is supposedly a corollary:

Let $$R$$ be a commutative, local, Noetherian ring. If $$J$$ is not nilpotent then $$R$$ is an integral domain and $$0$$ and $$J$$ are the only prime ideals of $$R$$.

Obviously we can't immediately apply the previous result because $$J$$ is not necessarily principal. Since $$R$$ is Noetherian, one can write $$J = x_1 R + \dots + x_n R$$ for some $$x_i \in R$$. Then we may take the quotient $$R / (x_2 R + \dots + x_n R)$$. Then $$J / (x_2 R + \dots + x_n R)$$ will be a unique maximal ideal in this ring. Moreover, this ideal is principal, being generated by $$\overline{x_1}$$ (the overline denoting equivalence class).

However, I don't see how we can use the fact that $$J$$ is not nilpotent now.

There is also a second part to this corollary which states that, if $$J$$ is nilpotent (with the other conditions being the same), then $$R$$ is Artinian and $$J$$ is the only prime ideal of $$R$$. This part comes with a hint, saying that, assuming $$J^s = 0$$, the only ideals in $$R$$ are $$R, J, J^2, \dots, J^s$$. But this would only follow from what we proved before if $$J$$ is principal, surely? Is it possible that the statement of the corollary misses that condition out?

Here's a counterexample. Consider $$R= \mathbb C[X,Y,Z]_{\langle X,Y,Z \rangle}$$. This local Noetherian ring has dimension $$3$$ and hence cannot have only $$2$$ prime ideals. But if the maximal ideal is principal $$J=(j)$$ then of course any element can be written as $$\mu j^n$$ for some $$\mu \in R^* \ , n\in \mathbb N$$. Thus if $$J$$ and hence $$j$$ is not nilpotent, the ring is an integral domain with only prime ideals $$0, J$$.