Let $R$ be a Noetherian ring and an integral domain. Let $\mathfrak p \subset \mathfrak q$ be prime ideals with $\mathrm{ht}(\mathfrak q) = \mathrm{ht}(\mathfrak p) + n$. I am trying to show that there exist $x_1,\dots,x_n \in R$ such that $\mathfrak q$ is a minimal prime of $\mathfrak p + (x_1,\dots,x_n)$.

I know that $\mathrm{ht}(x_1,\dots,x_n) \leq n$ and by Krull's theorem, any minimal prime of $(x_1,\dots,x_n)$ must have height $\leq n$. The problem is I don't know much about the ideal $\mathfrak p + (x_1,\dots,x_n)$, it would be helpful I think to know its height but I don't believe there is a general result on the height of the sum of two ideals. Any hints on approaching this?


1 Answer 1


Consider the quotient ring $R/P$. $Q/P$ is a prime in $R/P$ of height $n$, hence there exists $\bar x_1, \dots, \bar x_n \in R/P$ such that $Q/P$ is a minimal prime over the ideal $(\bar x_1, \dots, \bar x_n)$. This is equivalent to saying that $Q$ is a minimal prime over $P+(x_1, \dots, x_n)$.

Let me know if this helps!

  • $\begingroup$ The claim that the height of $q/p$ equals $n$ is wrong. One can prove that the height of $q/p$ is at most $n$, and this is enough. $\endgroup$
    – user26857
    Commented Jan 20, 2021 at 8:45
  • $\begingroup$ @user26857 is showing the height is at most $n$ a consequence of applying Krull's theorem in reverse? $\endgroup$ Commented Jan 20, 2021 at 10:08
  • $\begingroup$ No. It is a consequence of the definition of height. $\endgroup$
    – user26857
    Commented Jan 20, 2021 at 10:11
  • $\begingroup$ @user26857 why is the height possibly less than $n$? Is it because when you modulo $\mathfrak{p}$ that the chain of $n+1$ terms is further reduced as some ideals may become equivalent? $\endgroup$ Commented Jan 20, 2021 at 21:32

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