The following question seems natural to ask in view of this question and its comments/answers:

Let $R \subseteq S$ be commutative Noetherian rings, let $q$ be a maximal ideal of $S$, $p$ a maximal of $R$ and $p=R \cap q$.

The localization of $R$ at $p$, $R_p$, is a local ring with unique maximal ideal $pR_p$ and the localization of $S$ at $q$, $S_q$, is a local ring with unique maximal ideal $qS_q$.

Since $p=R \cap q$, we can view $R_p$ as a subring of $S_q$.

Let $pS \subseteq I \subseteq q$ be an ideal of $S$ (not necessarily prime), and then $I_q \subseteq S_q$ is an ideal of $S_q$ (where $I_q:=IS_q$).

Assume that $(pR_p)S_q = I_q = qS_q$, where $(pR_p)S_q$ is the ideal of $S_q$ generated by $pR_p$ (this makes sense, since $pR_p \subset R_p \subseteq S_q$).

Question: Is it true that $I=q$?

If not, would it help to further assume one or more of the following conditions:

(i) $R \subseteq S$ is flat. (ii) $R \subseteq S$ is integral. (iii) $\dim(R)=\dim(S) < \infty$ ($\dim$ is Krull dimension). (iv) $R$ and $S$ are regular rings.

Remark: The name 'sandwich' comes from the assumption that $(pR_p)S_q = I_q = qS_q$; call it 'the sandwich equation'.

A non-counterexample: Without condition (iii), the following is not a counterexample: $R=k[x]$, $S=k[x,y]$, $p=(x)$, $q=(x,y)$, $I=(x(x-1),y)$. Indeed, the sandwich equation is not satisfied: only the right equality is satisfied, but not the left equality.

Any hints and comments are welcome! Thank you.


1 Answer 1


No. Let $R$ is the ring of integers of a number field $K$, $p$ be a nonzero prime ideal, $S$ be the ring of the integers of $K(e^{2i\pi/r})$ where $r$ is a prime number coprime to $r$ with $r-1 > [K:\mathbb{Q}]$.

Let $q$ be any prime ideal of $S$ lying over $p$: then $K(e^{2i\pi/r})/K$ is unramified at $p$ thus $(pR_p)S_q=qS_q$. So if $I$ is any ideal of $S$ with $pS\subset I \subset q$, $I_q=qS_q$ too.

If the statement holds, then it follows (by characterization of prime ideals from Dedekind domains, else we can take eg $I=qq’$, where $q’$ is another prime ideal of $S$ lying over $p$) that $pS=q$, thus that $p$ is inert in $K(e^{2i\pi}/r)$, and this certainly isn’t systematic – the final part gives an elementary example.

Now, assume $K=\mathbb{Q}$: even in this setting where (i),(ii),(iii),(iv) all hold, $p$ is inert in $K(e^{2i\pi/r})$ iff $S/pS$ is an integral domain, iff $\mathbb{Z}[e^{2i\pi/r}]/(p)$ is a domain, iff $\mathbb{Z}[T]/(\Phi_r,p)$ is a domain, iff $\Phi_r$ is irreducible in $\mathbb{F}_p[T]$.

Now fix $r > 3$ and choose $p$ congruent to $1$ mod $r$, then $\Phi_r|X^r-1|X^{p-1}-1$ which is split with simple roots in $\mathbb{F}_p[T]$, so that $\Phi_r$ cannot be irreducible and thus $p$ isn’t inert and we can find a “problematic” ideal $I$.

  • $\begingroup$ Thank you very much; very nice! Please, which of conditions (i)--(iv) is satisfied by your counterexample and which is not? $\endgroup$
    – user237522
    Apr 7, 2021 at 22:28
  • $\begingroup$ All the conditions are satisfied – it’s a ring of cyclotomic integers over $\mathbb{Z}$! $\endgroup$
    – Aphelli
    Apr 7, 2021 at 22:43
  • $\begingroup$ Thank you very much, interesting. Please, could you find a counterexample in polynomial rings? $\endgroup$
    – user237522
    Apr 7, 2021 at 22:55
  • $\begingroup$ The following is not a counterexample, since the sandwich equation is not satisfied: $R=k[x^2,y^2]$, $S=k[x,y]$, $p=(x^2,y^2)$, $q=(x,y)$, $I=(x^2,y)$ or $I=(x,y^2)$. $\endgroup$
    – user237522
    Apr 7, 2021 at 22:56
  • $\begingroup$ If I am not wrong, $k[x^2,y^2] \subset k[x,y]$ is not flat. $\endgroup$
    – user237522
    Apr 7, 2021 at 22:59

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