# Height of minimal primes

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

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!

• 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. Commented Jan 20, 2021 at 8:45
• @user26857 is showing the height is at most $n$ a consequence of applying Krull's theorem in reverse? Commented Jan 20, 2021 at 10:08
• No. It is a consequence of the definition of height. Commented Jan 20, 2021 at 10:11
• @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? Commented Jan 20, 2021 at 21:32