Let $A$ be a noetherian ring, and $\mathfrak q$ a $\mathfrak p$-primary ideal in $A$, then every chain of primary ideals from $\mathfrak q$ to $\mathfrak p$ has a finit length, the set of the length of all these chains is bounded, and every maximal chain has same length.

I've have already proofed that every chain form $\mathfrak q$ to $\mathfrak p$ is finite, and the set of the length of all these chains is bounded, so I just have to prove the last part.

I'm trying to prove by contradiction, I'm assuming that there is two maximal chains $\mathfrak q=\mathfrak q_0\subsetneq\dots\subsetneq \mathfrak q_n=\mathfrak p $, and $\mathfrak q=\mathfrak q'_1\subsetneq\dots\subsetneq\mathfrak q'_{m}$, and that $m<n$,so we have that exist a $\mathfrak q_i\neq \mathfrak q'_j$ for all $j\in \{0,\dots,m\}$, and using maximality , it never satisfies this relation $\mathfrak q'_j\subsetneq\mathfrak q_i\subsetneq \mathfrak q'_{j+1}$.

So what I want to do is to find a way to merge this two chains, and get that the shorter is not maximal, I think that it follows from some intersection of this $\mathfrak q_i$ with some of the $\mathfrak q'_j$, but I can not see how to get this done.

Is this proof going correctly? Any tips on how to proceed from that?

Thanks in advance.

  • $\begingroup$ I don't think you are on the right way. Usually this theorem is proven by reducing to the local case first (localizing at $p$), and then to a primary ring by considering $R/q$. $\endgroup$ – user26857 May 25 '17 at 22:13
  • 2
    $\begingroup$ (p. 45) people.brandeis.edu/~igusa/Math205bS10/Math205b_S10_82.pdf $\endgroup$ – Hamed May 29 '17 at 22:54

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