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If $P$ is a prime ideal of a commutative ring $R$, such that $P^2 \subset P^r \subset P$, does it follow that $1 \leq r \leq 2$ or $P^2=P$ ?

More generally, does $P^a \subset P^b \subset P^c$ imply $c \leq b \leq a$ or $P^a=P^c$? (The condition $P^a=P^c$ implies $P^a=P^r=P^c$ for any $c \leq r \leq a$ ; but I'm not sure it implies $P=P^2=P^3=\cdots$). Are there conditions under which this is true, e.g. Noetherian?

In a noetherian domain $R$, $P^2 = P$ can't hold provided that $P \neq (0)$ (an interesting counter-example is here for non-noetherian rings), because of Nakayama's lemma: if $r>s$ and $P^r = P^s = P^s P^{r-s}$ then there is $i \in I = P^{r-s}$ such that $x=ix$ for all $x \in M=P^s$, which is finitely generated as $R$-module. Since $R$ is a domain and $P \neq (0)$ (so $M \neq 0$), we get $i=1 \in P^{r-s} \subset P$ (since $r-s \geq 1$), so that $P=R$, contradiction.

In Dedekind domains, my initial question should have a positive answer. Moreover, any prime power between $P^a$ and $P^c$ must be a power of $P$ itself, just by taking the radicals.

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    If it's any help, it suffices to only consider maximal ideals by localizing. – 2017-01-23
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    If $P$ is an ideal, then $P^k$ is an ideal for every $k$ so that $P^{k+1}\subseteq P^k$ – 2017-01-23
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    @Ofir : yes, I forgot to mention that the converse of my "claim" is obviously true. – 2017-01-23
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    Not necessarily: take $(R,m)$ a local notherian ring which is not a field and $m^2=0$. – 2017-01-23
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    @user26857 : thank you for this comment! By taking $R = k[x]/(x^2) \times k, m = (x)/(x^2) \times k$, we get $(0) \neq P^2 = P^3 = \cdots \subsetneq P$. Would you have sufficient conditions under which my claim becomes true (aside from Dedekind domain)? For instance, $k[x,y]$ has the property in the yellow box but has dimension 2. – 2017-01-24
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    Suficient conditions I don't know, but Noetherian domains also work since $\bigcap_{n\ge1}P^n=(0)$. – 2017-01-24
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    @user26857 : this is a very good sufficient condition for me! Thank you very much! – 2017-01-24

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