# Prime $\frak p\subset q\implies\exists$ prime $\frak p_1,q_1:p\subseteq p_1\subset q_1\subseteq q$ with no prime ideals between $\frak p_1,q_1$

This statement is Exercise 1.6 in this book as well as Exercise 3 in this note:

Statement. Let $$R$$ be a commutative ring with prime ideals $$\mathfrak{p}$$ and $$\mathfrak{q}$$. If $$\mathfrak{p}\subsetneq\mathfrak{q}$$, then there are prime ideals $$\mathfrak{p}_1$$ and $$\mathfrak{q}_1$$ in $$R$$ such that $$\mathfrak{p}\subseteq\mathfrak{p}_1\subsetneq\mathfrak{q}_1\subseteq\mathfrak{q}$$ and there are no other prime ideals between $$\mathfrak{p}_1$$ and $$\mathfrak{q}_1$$.

Here are my attempts. Consider the quotient ring $$R/\mathfrak{p}$$, which is an integral domain now.

Besides, we have $$\mathfrak{p}/\mathfrak{p}=\langle 0\rangle$$ and $$\mathfrak{q}/\mathfrak{p}\ne\langle 0\rangle$$, both of which are prime ideals in $$R/\mathfrak{p}$$ as well.

Then how can we find such prime ideals $$\mathfrak{p}_1$$ and $$\mathfrak{q}_1$$? I was thinking of minimal prime ideals over $$\mathfrak{p}$$, but since $$\mathfrak{p}$$ itself is prime, it make no sense to consider it now. Any help will be appreciated.

Update. The proof of this result should be as follows: Consider the set $$S=\{(\mathfrak{p}',\mathfrak{q}')\in\operatorname{Spec}(R)\times\operatorname{Spec}(R)\mid\mathfrak{p}\subseteq\mathfrak{p}'\subsetneq\mathfrak{q}'\subseteq\mathfrak{q}\},$$ which is certainly nonempty as $$(\mathfrak{p},\mathfrak{q})$$ is contained in it. Next, define a partial order on $$S$$ by $$(\mathfrak{p}_1,\mathfrak{q}_1)\le(\mathfrak{p}_2,\mathfrak{q}_2)\iff\mathfrak{p}_1\subseteq\mathfrak{p}_2~\text{and}~\mathfrak{q}_2\subseteq\mathfrak{q}_1.$$ One can easily verify that $$\le$$ defines a partial order on $$S$$. Consider an ascending chain $$(\mathfrak{p}_i,\mathfrak{q}_i)_{i\in\Gamma}$$ in $$S$$. Define $$\mathfrak{p}_0:=\bigcup_{i\in\Gamma}{\mathfrak{p}_i}\quad\text{and}\quad\mathfrak{q}_0:=\bigcap_{i\in\Gamma}{\mathfrak{q}_i}.$$ Again, both $$\mathfrak{p}_0$$ and $$\mathfrak{q}_0$$ are clearly prime ideals in $$R$$, but there is a question:

Question. Is $$\mathfrak{p}_0\subsetneq\mathfrak{q}_0$$?

If it is, then $$(\mathfrak{p}_0,\mathfrak{q}_0)\in S$$ as well. Finally, we simply let $$(\mathfrak{p}_1,\mathfrak{q}_1)$$ be the maximal element in $$S$$ ensured by Zorn's lemma.

I initially did not pay much attention to such question, but thought it would hold automatically. However, after rethinking it for a long time, I still do not get that. Any help will be appreciated :)

## 1 Answer

Take any element $$x\in \mathfrak{q} \setminus \mathfrak{p}$$. Choose a prime ideal $$\mathfrak{p}_0$$ maximal among those which contain $$\mathfrak{p}$$, are contained in $$\mathfrak{q}$$ and do not contain $$x$$. Symmetrically, choose $$\mathfrak{q}_0$$ minimal among those primes which contain $$\mathfrak{p}_0 + xR$$ and are contained in $$\mathfrak{q}$$. Then, any prime between $$\mathfrak{p}_0$$ and $$\mathfrak{q}_0$$ would be equal to one of these, depending on whether it contains $$x$$ or not.