Let $R$ be a Noetherian ring and $I$ be an ideal of $R$. Suppose $P$ is an associated prime of $I$ and $Q\subseteq P$. Then is $Q$ an associated prime, too?
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1One obvious way in that this can fail is that if $P$ is an associated prime of $I$, then $P$ contains $I$. Surely $Q$ can be contained in $P$ but not contain $I$, e.g. if $I$ is a non-zero ideal in an integral domain $R$, and $P$ is an associated prime of $I$, then $(0)$ is a prime ideal contained in $P$ which is not an associated prime of $I$. – walkar Nov 01 '23 at 19:29
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For $I = 0$, this is true iff the associated primes have height at most $1$. The if direction is just the fact that minimal primes are associated.
Conversely, suppose that $R$ is Noetherian and has the property that prime ideals contained in associated primes are associated (in other words, the set of associated primes is closed under generalizations). If $P$ were an associated prime of height $n > 1$, then $R_P$ would be a Noetherian ring in which every prime is associated. Thus $R_P$ has finitely many primes, and is a local ring of dimension $n > 1$. This is absurd because a Noetherian ring of dimension $d$ necessarily has infinitely many primes of height $k$ for each $0 < k < d$. See here.
Can you come up with an example of a Noetherian ring with an associated prime of height $2$?
Badam Baplan
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$k[x,y]/(x^2,xy)$ is a $1$-dimensional ring, so that's not going to work. But you're on the right track. – Badam Baplan Nov 03 '23 at 05:55