# A sandwich theorem for local rings

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.

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 $$I$$.

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