Let $K$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$. Assume $R$ and $R'$ be two discrete valuation rings of $K$ such that $R \subset R'$.

Lemma: Is $R =R'$ or $R \cong R'$ ?

Since $R$, $R'$ are discrete valuation ring,then both has unique maximal ideals $m$ and $m'$ respectively such that $m \subset m'$.

If we show that $m' \subset m$, then $m=m'$ will imply $R=R'$.

Am I right here?

Now, let $0 \neq x \in m'$, then either $x \in R$ or $x ^{-1} \in R$. If $x^{-1} \in R$, then $x^{-1} \in R'$, which implies $x$ is invertible in $R'$ and hence $x \notin m'$, a contradiction. Thus $m' \subset m$ and we already have $m \subset m'$ as $R \subset R'$. Thus $m=m'$, which forces $R=R'$.


My main question:

Consider the Cyclotomic extension $\mathbb{Q}_p(\zeta_p)$ of $\mathbb{Q}_p$. We know that both $\mathbb{Z}_p[\zeta_p]$ and $\mathbb{Z}_p[\pi], \ \pi=\sqrt[(p-1)]{-p}$ are discrete valuation rings of $\mathbb{Q}_p(\zeta_p)$.

I want to show $\mathbb{Z}_p[\zeta_p]=\mathbb{Z}_p[\pi]$.

Clearly $\mathbb{Z}_p[\zeta_p]\subset \mathbb{Z}_p[\pi]$ because $\zeta_p \in \mathbb{Z}_p[\pi]$ as $\pi=\sqrt[(p-1)]{-p}$.

Thus by above Lemma, we have $\mathbb{Z}_p[\zeta_p]=\mathbb{Z}_p[\pi]$.

Am I correct ?

Thanks for help

  • $\begingroup$ @TorstenSchoeneberg, because a discrete valuation ring contains $unique$ maximal ideal. If $R' \neq R$, then we get a contradiction. Second, I have used the field $K$, when I say $0 \neq x \in m'$ ,then either $x \in R$ or $x^{-1} \in R$ and right here I am using $K$ as a fraction field of $R$. $\endgroup$
    – MAS
    Dec 8, 2020 at 19:35
  • $\begingroup$ @TorstenSchoeneberg, here I thought $\zeta_p \in \mathbb{Z}_p[\pi=\sqrt[(p-1)]{-p}]$ looking at $\mathbb{Z}_p[\pi=\sqrt[(p-1)]{-p}]=\mathbb{Z}_p[\zeta_p, \sqrt[(p-1)]{p}]$. May be I am wrong?? $\endgroup$
    – MAS
    Dec 8, 2020 at 19:37
  • $\begingroup$ As for the specific $p$-adic situation, in general $\mathbb Z_p[\sqrt[p-1]{-p}]$ does not contain an element $\sqrt[p-1]{p}$, and if I'm not mistaken, that was pointed out already here: math.stackexchange.com/q/3800738/96384. $\endgroup$ Dec 8, 2020 at 21:37
  • $\begingroup$ @TorstenSchoeneberg, I think yes, you are right. I am wrong $\endgroup$
    – MAS
    Dec 9, 2020 at 5:15
  • $\begingroup$ @TorstenSchoeneberg, ok. Thanks $\endgroup$
    – MAS
    Dec 11, 2020 at 2:07

1 Answer 1


Lemma 1: Let $R$ be a discrete valuation ring (with maximal ideal $m =(\pi)$), and let $(R', m')$ be any local ring such that $R \subseteq R'$. Then either $m \subseteq m'$ or $R'$ contains the field of fractions of $R$. In particular, if $R, R'$ are valuation rings of the same fraction field $K$, then $m \subseteq m'$ (unless $R' = K$ which you tacitly exclude.)

Proof: If $\pi \notin m'$, then $\pi \in (R')^\times$ and hence $R'$ contains $R[\pi^{-1}] = Frac(R)$.

Note that if we drop the condition that the valuation of $R$ is discrete, the result is not necessarily true even for two valuation rings of the same fraction field $K$:

(Counter-)Example 1: Let $K$ be the field of Laurent series in two variables $x,y$ as described in this answer. Let $R$ be the valuation ring for the (rank two) valuation described there (called $A$ in that answer). Note that $R$ consists of those series where the coefficients of $x^r y^s$ are zero if either $r$ is negative, or $r = 0$ and $s$ is negative.

Now consider the following bigger subring $R'$ of $K$: It consists of those series where the coefficients of $x^r y^s$ are zero if $r$ is negative. Note that $R \subsetneq R' \subsetneq K$: e.g. $y^{-1} \in R' \setminus R$. In fact $R'$ is the valuation ring to the rank one, discrete valuation $v'$ which just measures the degree of $x$ in $K$ viewed as $E((x))$, where $E:=\mathbb C((y))$ is the ground field and residue field, all its elements having trivial valuation $v'(e)=0$.

Now $y \in m$ but $y \notin m'$.

ADDED: Also look through the "similar but different", but maybe psychologically easier, example here.

Lemma 2: If $R$ is a valuation ring of the field $K$, and $R'$ is any proper subring of $K$ containing $R$, then $R'$ is local, and its maximal ideal $m'$ is contained in the maximal ideal $m$ of $R$.

Proof: Exactly as outlined in your question: Let $x$ be any non-zero non-unit in $R'$. Since $x$ is no unit in $R'$, we cannot have $x^{-1} \in R \subseteq R'$. So by $R$ being a valuation ring, we have $x \in R$, and again by assumption cannot have $x \in R^\times$, so $x\in m$. This means that all non-units of $R$ must be in $m$. But then the sum of two non-units in $R'$ is a non-unit, and it's clear that a non-unit multplied by any element of $R'$ is still a non-unit, i.e. the non-units in $R'$ form an ideal, which then must be the unique maximal ideal of $R'$, and we showed all its elements must bei in $m$.

Note that (Counter-)Example 1 also gives an example of this where the inclusion $m' \subset m$ is proper.

Lemma 3: In the situation of lemma 2, assume $m=m'$. Then $R=R'$.

Proof: It suffices to show $R'^\times \subseteq R$. But if some unit $x$ of $R'$ were not in $R$, then by $R$ being a valuation ring we would have $x^{-1} \in R$. Now obviously $x^{-1} \notin m = m'$ (else it couldn't be a unit in $R'$), so $x^{-1} \in R^\times$, but then $x \in R^\times$ all along.

Corollary ("Yes" to your question): If $R \subseteq R'$ are two proper valuation rings of the field $K$, and $R$ is a discrete valuation ring, then $R=R'$.

Proof: Combine lemmas 1,2,3.

However, all of this is not the main problem in your main question. The difficult part in showing that


is to show any inclusion $\pi \in \mathbb{Q}_p(\zeta_p)$ or $\zeta_p \in \mathbb{Q}_p[\pi]$ to begin with. That exact problem has been solved on this site both with Hensel's Lemma and Krasner's Lemma, and without anything of that price category, you will not be able to get any argument started here. Once one has one of those inclusions, one could continue with the general machinery above, but one might as well conclude with basic dimension/rank arguments and/or properties of extensions of complete DVRs, because when dealing with $p$-adics one has to use those all the time anyway.

  • $\begingroup$ Thanks a lot. Nice answer. Regarding,the main question, Indeed I followed Professor Lubin's technique in order to it. $\endgroup$
    – MAS
    Dec 11, 2020 at 2:17

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