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Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial of degree $n$. Let $\theta$ be a root of $f(X)$. Let $A = \mathbb{Z}[\theta]$. Let $p$ be a prime number. Suppose the discriminant of $f(X)$ is a power of $p$. Suppose $pA = \alpha^n A$, where $\alpha \in A$.

My question: Is $A$ integrally closed?

Motivation Let $p$ be an odd prime number. Let $\theta$ be a $p$-th primitive root of unity. Let $\alpha = 1 - \theta$. Then it is well known that the discriminant of the minimal polynomial of $\theta$ is a power of $p$ and $pA = \alpha^n A$. It is also well known that $A$ is integrally closed.

This is a related question.

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I think the technique of localization is what you want to use for this, and I think that this readily gives a positive answer to your question. –  Lubin Jul 23 '12 at 1:59
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1 Answer

We denote by $|S|$ the number of elemets of a finite set $S$.

Lemma 1 Let $A$ be a local doman which is not a field. Let $\mathfrak{m}$ be its maximal ideal. Suppose $A$ satisfies the following conditions.

(1) $\mathfrak{m}$ is principal, i.e. there exists $t \in A$ such that $\mathfrak{m} = tA$.

(2) $0 = \bigcap_n \mathfrak{m}^n$

Then $A$ is a discrete valuation ring.

Proof: Since $A$ is not a field, $t \neq 0$. Let $x$ be a non-zero element of $\mathfrak{m}$. By (2), there exists largest integer $n \geq 1$ such that $x \in \mathfrak{m}^n$. We denote $n$ = ord($x$). Since $x \in t^nA$, there exists $u \in A$ such that $x = t^n u$. Clearly $u$ is invertible in $A$. Hence $xA = t^nA$.

Let $I$ be a non-zero ideal of $A$ such that $I \subset \mathfrak{m}$. Let $n$ = min{ord($x$);$x \in I - {0}$}. There exists $y$ such that $n$ = ord($y$). If $x \in I$ - {$0$}, ord($x$) $\geq n$. Hence $x \in t^nA$. Since $t^nA = yA$, $x \in yA$. Hence $I = yA$. Therefore $A$ is a principal ideal domain. Hence $A$ is a discrete valuation ring. QED

Lemma 2 Let $A$ be a Noetherian local doman which is not a field. Let $\mathfrak{m}$ be its maximal ideal. Suppose $\mathfrak{m}$ is principal. Then $A$ is a discrete valuation ring.

Proof: There exists $t \in A$ such that $\mathfrak{m} = tA$. Since $A$ is not a field, $t \neq 0$. Let $I = \bigcap_n \mathfrak{m}^n$. By Lemma 1, it suffices to prove that $I = 0$. Let $x \in I$. There exists $y_n \in A$ such that $x = t^n y_n$ for each integer $n \geq 1$. Since $x = t^n y_n = t^{n+1} y_{n+1}$, $y_n = ty_{n+1}$. Hence $y_nA \subset y_{n+1}A$. Since $A$ is Noetherian, $y_kA = y_{k+1}A$ for some integer $k$. Hence $y_{k+1} = y_k u$. Since $y_k = ty_{k+1} = tu y_k$. Hence $(1 - tu)y_k = 0$. Since $1 - tu$ is invertible, $y_k = 0$. Hence $x = 0$. Therefore $I = 0$. QED

Lemma 3 Let $A$ be an integral domain. Let $K$ be the field of fractions of A. Let $\Omega$ be the set of maximal ideals of $A$. Then $A = \bigcap_{P\in\Omega} A_P$, where we regard each $A_P$ as a subring of $K$.

Proof: Let $x \in \bigcap_{P\in\Omega} A_P$. Let $I$ = {$a \in A$; $ax \in A$}. Clearly $I$ is an ideal. Suppose $I \neq A$. There exists a maximal ideal $P$ such that $I \subset P$. Since $x \in A_P$, there exists $s \in A - P$ such that $sx \in A$. Since $s \in I$, $s \in P$. This is a contradiction. Hence $I = A$. Therefore $1 \in I$. Hence $x \in A$. Hence $\bigcap_{P\in\Omega} A_P \subset A$.

The other inclusion is clear. QED

Lemma 4 Let $K$ be an algebraic number field. Let $n = [K : \mathbb{Q}]$. Let $B$ be the ring of algebraic integers in $K$. Let $p$ be a prime number. Suppose $pB = \alpha^nB$. Then $\alpha B$ is a prime ideal of $B$ and $|B/\alpha B| = p$.

Proof: Let $P$ be a prime ideal of $B$ lying over $p\mathbb{Z}$. Since $\alpha^n \in P$, $\alpha \in P$. Hence $pB \subset P^n$. Hence $pB = P^n$. Hence $\alpha^nB = P^n$. Hence $\alpha B = P$. Cearly $|B/\alpha B| = p$. QED

Definition Let $K$ be an algebraic number field. Let $n = [K : \mathbb{Q}]$. Let $\sigma_1,\dots,\sigma_n$ be distinct homomorphisms $K \rightarrow \mathbb{C}$. Let $\omega_1,\dots,\omega_n$ be elements of $K$. We denote det($\sigma_i(\omega_j)$) by $\Delta(\omega_1,\dots,\omega_n)$.

Lemma 5 Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial of degree $n$. Let $\theta$ be a root of $f(X)$. Let $A = \mathbb{Z}[\theta]$. Let $p$ be a prime number. Suppose $pA = \alpha^n A$, where $\alpha \in A$. Then $\alpha A$ is a prime ideal of $A$ and $|A/\alpha A| = p$.

Proof: Let $K$ be the field of fractions of $A$. $1, \theta,\dots,\theta^{n-1}$ is a basis of $A$ as a free $\mathbb{Z}$-module. Hence $\alpha,\alpha\theta,\dots,\alpha\theta^{n-1}$ is a basis of $\alpha A$ as a free $\mathbb{Z}$-module. Hence $|A/\alpha A| = |\Delta(\alpha,\alpha\theta,\dots,\alpha\theta^{n-1})/\Delta(1, \theta,\dots,\theta^{n-1})| = |N_{K/\mathbb{Q}}(\alpha)|$. Since $|N_{K/\mathbb{Q}}(\alpha)|$ is $p$ by Lemma 4, $A/\alpha A$ is a field. Hence $\alpha A$ is a prime ideal of $A$. QED

Proposition Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial of degree $n$. Let $\theta$ be a root of $f(X)$. Let $A = \mathbb{Z}[\theta]$. Let $p$ be a prime number. Suppose the discriminant of $f(X)$ is a power of $p$. Suppose $pA = \alpha^n A$, where $\alpha \in A$. Then $A$ is integrally closed.

Proof: Let $\Omega$ be the set of non-zero prime ideals of $A$. By this, $\Omega$ is the set of maximal ideals of $A$. By Lemma 3, $A = \bigcap_{P\in\Omega} A_P$. Hence it suffices to prove that $A_P$ is integrally closed for each $P \in \Omega$.

Let $P \in \Omega$. Suppose $P$ does not divide $p$. Since the discriminant $d$ of $f(X)$ is a power of $p$, $P$ does not divide $d$. Hence $A_P$ is integrally closed by my answers to this question.

Suppose $P$ divides $p$. Since $P$ divides $\alpha$, By Lemma 5, $P = \alpha A$. Hence $A_P$ is a discrete valuation ring by Lemma 2. Hence $A_P$ is integrally closed. QED

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