Atiyah-Macdonald Ex8.6 Is there anybody can give a proof? I can prove "finite" only, but I cannot prove "bounded".
Here is the exercise:
Let A be a Noetherian ring and Q a P-primary ideal in A. Consider chains of primary ideals from Q to P. Show that all such chains are of finite bounded length, and that all maximal chains have the same length.
 A: EDIT Dear User, this is just to tell you that I have trimmed a bit the Google list. It contains all the solutions to the AM exercises I have been able to find on the web, but I'm sure there are many others. If you know some of these others, please add them to this community wiki answer by editing it. TIDE
This is not an answer, and this is a community wiki. 
I don't know if it's a good idea, but if you're interested you can help me collect here links to online solutions to Atiyah-MacDonald exercises. 
Here is my Google list of such links. 
It's very messy, and some links are broken. 
If it turn out to be a bad idea, I can delete this "answer", and give the link in a comment. 
Here is a MathOveflow Errata for Atiyah-MacDonald. 
A: Let $A$ a noetherian ring, let $P$ be a prime ideal of $A$ and $Q$ a $P$-primary ideal. Consider the ring $B = A_P / Q A_P$.
Prove:
1) Every ideal of $B$ is $\bar{P}$-primary.
2) There is a 1:1 correspondence between the set of $P$-primary ideals of $A$ that contain $Q$ and the set of ideals of $B$. This correspondence preserves containments.
3) $B$ is an artinian ring, so $B$ is a $B$-module of finite length.
4) Every chain of ideals of $B$ has lenght $ \leq l_B(B)$. Every maximal chain of ideals of $B$ has lenght $= l_B(B)$. ($l_B(B)$ is the length of $B$ as a $B$-module.)
