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 $0, J$.

  • $\begingroup$ I'm not sure what you mean this as a counterexample to? $\endgroup$ – Tom Miller Apr 7 at 15:46
  • 2
    $\begingroup$ Clearly the maximal ideal is not nilpotent. You need the assumption that it is principal which is equivalent to saying the ring has Krull dimension 1. $\endgroup$ – Ignorant Mathematician Apr 7 at 16:24

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.