3
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $0, J$.

$\endgroup$
2
  • $\begingroup$ I'm not sure what you mean this as a counterexample to? $\endgroup$
    – Tom Miller
    Apr 7, 2019 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. $\endgroup$
    – user6
    Apr 7, 2019 at 16:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .