Ytukyg

Let $(R,M)$ be a Noetherian local ring. Show that

i) If there exists a non-maximal prime ideal of $R$, then $M^{n+1} subset M^{n}$ for every $n in mathbb{N}$

ii) If $I$ is a proper ideal of $R$ and $sqrt{I} neq M$ then $I +M^{n+1} subset I +M^{n}$ for every $n in mathbb{N}$

I think $M^{n+1} subset M^{n}$ is obvious without condition about the existence of non-maximal prime ideal, doesn't it? I have no idea for this problem. Can anyone help me?

The $subset$ subset symbol is definitely meant to indicate strict inclusion here.

For the first part, since trivially $I^n subseteq I^m$ for any ideal $I$ and $n geq m$, the problem is just to prove that $M^{n+1} not= M^{n}$ given the assumption that the ring is not $0$-dimensional.

We will show contrapositively that $M^{n+1} = M^n$ implies that the ring is $0$-dimensional.

Recall that Nakayama's lemma, in one of its incarnations, states the following:

Nakayama's lemma: Let $I$ be an ideal contained in the Jacobson radical of $R$ and $M$ be a finitely generated ideal. If $IM = M$ then $M = 0$.

In a local ring, the Jacobson radical is the unique maximal ideal, which contains our non-maximal ideal $M$. Moreover $M$ is f.g. by the Noetherian assumption. So if we have $MM^n = M^n$ then Nakayama says that $M^n = 0$.

If the maximal ideal is nilpotent, then every nonunit is nilpotent, and this implies that the ring has a unique prime ideal (good elementary exercise).

Hence the ring is $0$-dimensional.

For the second part, note that $(R/I, M/I)$ is a local Noetherian ring, too. Again we show the contrapositive: if $I + M^{n+1} = I + M^n$ then $sqrt{I} = M$. By the first part, if $I + M^{n+1} = I + M^n$ then $R/I$ is $0$-dimensional. So $M$ is the only prime ideal of $R$ containing $I$. But the radical of an ideal is the intersection of the primes containing it, so $sqrt{I} = M$.