Ring $R’$ with $R\subset R’\subset T$ where $R$, $T$ integrally closed and obtained by a purely inseparable extension

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Let $R$ be an integrally closed domain with field of fractions $K$. Let $M/K$ be a purely inseparable extension and let $T$ be $R$'s integral closure in $M$. Consider a subring $R\subset R'\subset T$ and put $L$ its field of fractions and $S$ the integral closure of $R$ in $L$. We have $R\subset R'\subset S\subset T$ and purely inseparable extensions $K\subset L\subset M$.

Question. is $R'$ automatically integrally closed, i.e. is $R'=S$ ?

$
\newcommand{\P}{\mathfrak{P}}
\newcommand{\p}{\mathfrak{p}}
\newcommand{\Spec}{\mathrm{Spec}}
\newcommand{\N}{\mathbb{N}}
$I don't remember exactly how the question came about but I know it came up when reading the proof of the theorem below from Gerald J. Janusz's Algebraic Number Fields. For more details see the final sentence of the post.

Theorem 6.1 Let $R$ be a Dedekind ring with field of fractions $K$ and let $L$ be a finite field extension of $K$. Then the integral closure of $R$ in $L$ is a Dedekind domain.

The proof is in two steps. Considering the intermediate field $K\subset E\subset L$ of all separable elements of $L$ over $K$ one reduces the question to two special cases: separable extension and purely inseparable extension. In the purely inseparable case the author proves something that can be generalized as follows:

Lemma. Let $R$ be integrally closed with field of fractions $K$ of characteristic $p$ and consider a purely inseparable extension $L/K$. Let $S$ be the integral closure of $R$ in $L$ and consider a subring $R\subset R'\subset S$. Then the inclusion $i:R\to R'$ induces a bijection on spectra, i.e. $i^*:\Spec(R')\to\Spec(R),\P\mapsto\P\cap R$ is a bijection.

Proof. One first notices that
$$S=\{x\in L\mid \exists n\in\N,~x^{p^n}\in R\}$$
Indeed, if $x\in L$ satisfies $x^{p^n}\in R$ then $x$ is integral over $R$ so that $x\in S$ by definition of $S$. Conversely, if $x\in L$ then, $L/K$ being purely inseparable, $x^{p^n}\in K$ for some $n$. If furthermore $x\in S$, then $x$ is integral over $R$ and so is $x^{p^n}\in K$. Since $R$ is integrally closed we have $x^{p^n}\in R$, and so both inclusions are proven.

  1. $\mathbf{i^*}$ is one to one. First of all, if $\P$ is nonzero prime ideal of $R'$ then so is $\p=i^*(\P)$. Indeed, if $x\in\P$ is nonzero then for some $n$ we have
    $$x^{p^n}\in R\setminus 0$$
    and thus $x^{p^n}$ is a nonzero element of $\p$. Furthermore, we can reconstruct $\P$ from $\p$. Indeed, one shows
    $$\P=\{x\in R'\mid \exists n\in \N,~x^{p^n}\in\p\}$$
    Indeed, if $x\in R'$ satisfies $x^{p^n}\in\p\subset\P$, then, $\P$ being prime, $x\in\P$. And conversely, if $x\in\P$ then for some $n$, $x^{p^n}\in R$ so that $x^{p^n}\in\P\cap R=\p$.
  2. $\mathbf{i^*}$ is onto. Let $\p$ be a prime of $R$ and put
    $$\p'=\{x\in R'\mid\exists n\in\N,~x^{p^n}\in\p\}$$
    $\p'$ is a prime ideal of $R'$. Indeed, if $x,y\in\p'$ and $r\in R'$ then there exist exponents $a,b,c\geq 0$ with $x^{p^a},y^{p^a}\in\p$ and $r^{p^c}\in R$. Thus if $n=\max\{a,b,c\}$ we have $(x+ry)^{p^n}\in \p$ so that $\p'$ is an ideal of $R'$. And if $xy\in\p'$ with $x,y\in R'$ and $y\notin\p'$ then for some $n\in\N$, $x^{p^n}\cdot y^{p^n}=(xy)^{p^n}\in\p$. With $\p$ prime and $y\notin\p'$ we have $x^{p^n}\in\p$ i.e. $x\in\p'$, and so $\p'$ is prime. Now one easily sees
    $$\p=R\cap\p'=i^*(\p')$$
    Indeed, $\p\subset\p'$ thus $\p\subset R\cap\p'$, and if $x\in R\cap\p'$ then $x^{p^n}\in\p$ with $x\in R$ and with $\p$ being a prime ideal of $R$, $x\in\p$. Therefore $i^*$ is onto.$\quad\blacksquare$

I was surprised that there would be a bijection on spectra without further hypothesis. The author of the book doesn't state the result above, rather he proves and uses it only for the subring $R'=S$ and I wondered whether this generalization is indeed a generalization or whether all subrings $R'$ are automatically integrally closed (whence my question, I think.)

Best Answer

  • If $R'/R$ is integral and $Frac(R')/Frac(R)$ is purely inseparable then it is immediate that $\mathfrak{p}\to \mathfrak{p}\cap R$ is a bijection $Spec(R')\to Spec(R)$.

  • $\Bbb{F}_p[x^p,x^{p+1}]$ is not integrally closed in its fraction field and it is between the PID $\Bbb{F}_p[x]$ and $\Bbb{F}_p[x^p]$.

  • The problem of non-separable extensions $L/K$ is that $O_L$ doesn't have to be a finitely generated $O_K$-module (can't use the trace map which is identically $0$) and the $n=\sum_i f_i e_i$ stuffs may fail.

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