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 1


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.


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