How to show the length of $A$-module $A/\mathfrak p$ is $\infty$?

Question

Suppose $A$ is a noetherian commutative ring with unity, $\mathfrak p$ is a minimal prime ideal of $A$ which is not maximal, how to show the length of $A$-module $A/\mathfrak p$ is $\infty$?

Answer 1

In fact, if there is an injection of $ A$-homomorphism:$A/q\rightarrow A/p$,where $q$ is another prime ideal, then you can show that $q=p$. In other words, $Ass(R/p)=\{p\}$.

Answer 2

Since $\mathfrak{p}$ is assumed prime, $A/\mathfrak{p}$ is an integral domain. So, you might as well assume that $A$ is a Noetherian domain to begin with, and show that $A$ has infinite length as an $A$-module. Let $\mathfrak{m}$ be a maximal ideal of $A.$ Take the localization $A_{\mathfrak{m}}.$ You now have a local Noetherian domain, again call it $A$, and call its maximal ideal $\mathfrak{m},$ as well. In this setup, $0$ is the minimal prime ideal, and is not maximal. Consider the ideals $\mathfrak{m}^n$ for positive integers $n.$ They form a decreasing chain $$\mathfrak{m} \supset \mathfrak{m}^2 \supset \ldots.$$ Suppose this chain stabilizes at some point, that is, $\mathfrak{m}^n = \mathfrak{m}^{n+1}$ for some $n.$ Since $\mathfrak{m}$ is the unique maximal ideal of $A$, it is the Jacobson radical of $A$. Since $A$ is Noetherian, all ideals are finitely generated, in particular $\mathfrak{m}^n.$ Since $\mathfrak{m}^n = \mathfrak{m}^{n+1} = \mathfrak{m} \cdot \mathfrak{m}^n,$ Nakayama's lemma implies that $\mathfrak{m}^n = 0.$ But $A$ is a domain, so $0$ is a prime ideal. But then $\mathfrak{m} = 0,$ contradicting the assumption that $\mathfrak{m} \neq 0.$ So, the chain does not stabilize, giving you an infinite chain of distinct submodules of $A,$ so that $A$ has infinite length.

Now $A$ is local, but the chain of ideals above corresponds to an infinite decreasing chain of ideals in the original noetherian domain (and also such a chain in $A/\mathfrak{p}$). The argument I have given is essentially that for Proposition 8.6 in Atiyah and Macdonald (First edition). You can find all the facts you need about localization, and the correspondence between prime ideals in a ring and the primes of a localization of that ring in Chapter 3 of A-M. The chapters may have changed in various editions over the years. My book, like me, is getting on in years

You do not need the machinery of localization and Nakayama, however. The ring $R/\mathfrak{p}$ is an integral domain that is not a field. It will have a nonzero maximal ideal $\mathfrak{m}.$ Choose any nonzero $a \in \mathfrak{m}$ and consider the chain of ideals $$(a) \supset (a^2) \supset \ldots.$$ If $(a^n) = (a^{n+1})$ for some $n$, then $a^n=ra^{n+1}$ for some nonzero $r$ in the domain. But then $a^n(1-ra)=0.$ As the ring is a domain and $a^n \neq 0,$ we have $ra=1$ so that $a$ is a unit, which is impossible, since $a$ lies in a proper ideal. This argument is more in line with my original suggestion, and is much simpler.