A Dedekind Criterion over Valued Fields

Abstract

Let $(K,\nu)$ be an arbitrary-rank valued field, let $R_{\nu}$ be the valuation ring of $(K,\nu)$, and let $K(\alpha)/K$ be a separable finite field extension generated over $K$ by a root of a monic irreducible polynomial $f\in R_{\nu}[X]$. We give some necessary and sufficient conditions for $R_{\nu}[\alpha]$ to be integrally closed. We further characterize the integral closedness of $R_{\nu}[\alpha]$ which is based on information about the valuations on $K(\alpha)$ extending $\nu$. Our results enhance and generalize some existing results as well as provide applications and examples.

1. Introduction

Given a valued field $(K,\nu)$, we denote by $\overline{K}$ an algebraic closure of $K$; by $R_{\nu}$, the valuation ring of $\nu$; by $M_{\nu}$, the maximal ideal of $R_{\nu}$; by $k_{\nu}=R_{\nu}/M_{\nu}$, the residue field of $\nu$; and by $\Gamma_{\nu}$, the (totally ordered abelian) value group of $\nu$. We denote the set of elements $g\in\Gamma_{\nu}$ such that $g>0$ by $\Gamma_{\nu}^{+}$ and a minimum element of $\Gamma_{\nu}^{+}$, if any, by $\min(\Gamma_{\nu}^{+})$. We also denote by $\nu^{G}$ the Gaussian extension of $\nu$ to the field $K(X)$ of rational functions; i.e., given $f(X)=\sum\nolimits_{i=0}^{m}a_{i}X^{i}\in K[X]$, we put $\nu^{G}(f)=\min\{\nu(a_{0}),\dots,\nu(a_{m})\}$ and extend to $K(X)$ as $\nu^{G}(f/g)=\nu^{G}(f)-\nu^{G}(g)$ for $f,g\in K[X]$ and $g\neq 0$.

Let $(K,\nu)$ be a valued field of arbitrary rank, let $f\in R_{\nu}[X]$ be a monic irreducible separable polynomial, let $\alpha\in\overline{K}$ be a root of $f$, let $L=K(\alpha)$ be the simple field extension over $K$ generated by $\alpha$, and let $S$ be the integral closure of $R_{\nu}$ in $L$. Assume that $\overline{f}=\prod\nolimits_{i=0}^{s}\overline{\phi_{i}}^{\,l_{i}}$ is the monic irreducible factorization of $\overline{f}$ over $k_{\nu}$, and $\phi_{i}\in R_{\nu}[X]$ is a monic lifting of $\overline{\phi_{i}}$ for $i=1,\dots,s$. For the sake of brevity, we will refer to these notations and assumptions as Assump’s.

Under Assump’s, if $R_{\nu}$ is a discrete valuation ring and $M_{\nu}$ does not divide the index ideal $[S:R_{\nu}[\alpha]]$, then the well-known theorem of Dedekind (see [1, Proposition 8.3] for instance) gives the factorization of the ideal $M_{\nu}S$; namely, $M_{\nu}S=\prod\nolimits_{i=1}^{s}{\mathfrak{p}}_{i}^{l_{i}}$, where ${{\mathfrak{p}}_{i}=M_{\nu}S+\phi_{i}(\alpha)S}$ with residue degree $\operatorname{deg}(\phi_{i})$. Dedekind in [2] gave a criterion for the divisibility of ${[S:R[\alpha]]}$ by $M_{\nu}$ that was also extended in [3]. Considering an arbitrary valuation $\nu$ in general, Ershov in [4] introduced a nice generalized version of Dedekind’s Criterion. Namely, he showed that if we write $f$ in the form

$$ f=\prod\limits_{i=1}^{s}\phi_{i}^{l_{i}}+\pi T $$

for some $\pi\in M_{\nu}$ and ${T\in(R_{\nu}-M_{\nu})[X]}$; then $R_{\nu}[\alpha]$ is integrally closed (i.e. $R_{\nu}[\alpha]=S$) if and only if either $l_{i}=1$ for all $i=1,\dots,s$ or, else, $\nu(\pi)=\min(\Gamma_{\nu}^{+})$ and $\overline{\phi_{i}}$ does not divide $\overline{T}$ for all those $i=1,\dots,s$ with $l_{i}\geq 2$. Khanduja and Kumar gave a different elegant proof of Ershov’s result in [5, Theorem 1.1].

Assuming Assump’s, the following Theorem 2.5 gives a new characterization of the integral closedness of $R_{\nu}[\alpha]$, where we utilize the Euclidean division of $f$ by $\phi_{i}$ for all $i=1,\dots,s$, $l_{i}\geq 2$, with a motivation to enhance its application as compared to [5, Theorem 1.1]. Theorem 2.5 further improves [5, Theorem 4.1] as it does not require $K$ to be Henselian. Using our techniques, moreover, we give a simpler proof of some significant result proved in [6, Theorem 1.3] which gives a complete characterization of the integral closedness of $R_{\nu}[\alpha]$ which is based on the valuations of $L$ extending $\nu$ and their values at $\phi_{i}(\alpha)$. We also compute the ramification indices and residue degrees of all valuations of $L$ extending $\nu$ (Corollary 2.10). Some further applications and examples are given in Section 3.

2. The Main Results

Keeping the notations of Assump’s, denote by $(K^{h},\nu^{h})$ a Henselization of $(K,\nu)$ and by $\overline{\nu^{h}}$ the unique extension of $\nu^{h}$ to the algebraic closure $\overline{K^{h}}$ of $K^{h}$.

We begin this section with the following important well-known result which we present without proof (see, for instance, [7, 17.17]). The result asserts a one-to-one correspondence between the valuations on $L$ extending $\nu$ and the irreducible factors of $f$ over $K^{h}$.

Lemma 2.1

Keep the notation and assumptions of Assump’s. Let $f=\prod\nolimits_{j=1}^{t}f_{j}$ be the factorization of $f$ into a product of distinct monic irreducible polynomials over $K^{h}$. Then there are exactly $t$ extensions $\omega_{1},\dots,\omega_{t}$ of $\nu$ to $L$. Moreover, if $\alpha_{j}$ is a root of $f_{j}$ in $\overline{K^{h}}$ for $j\in\{1,\dots,t\}$, then the valuation $\omega_{j}$ corresponding to $f_{j}$ is precisely the valuation on $L$ satisfying $\omega_{j}(h(\alpha))=\overline{\nu^{h}}(h(\alpha_{j}))$ for all $h\in K[X]$.

The following result is a generalization of [8, Lemma 2.1] to arbitrary-rank valuations.

Lemma 2.2

Keep the notation and assumptions of Assump’s and Lemma $2.1$.

(i) For every $i=1,\dots,s$, there is some $j=1,\dots,t$ such that $\omega_{j}(\phi_{i}(\alpha))>0$.

(ii) $\omega_{j}(p(\alpha))\geq\nu^{G}(p(X))$ for every $j=1,\dots,t$ and every nonzero $p\in R_{\nu}[X]$.

(iii) For every $j=1,\dots,t$, there exists a unique $i=1,\dots,s$ such that $\omega_{j}(\phi_{i}(\alpha))>0$. Moreover, $\omega_{j}(\phi_{k}(\alpha))=0$ for all $k\neq i$, $k=1,\dots,s$.

(iv) Equality holds in (ii) if and only if $\overline{\phi_{i}}$ does not divide $\overline{(p/a)}$ for the unique index $i$ associated to $\omega_{j}$ in (iii), where $a$ is any coefficient of $p$ of a minimum $\nu$-valuation.

Proof

(i): Since $k_{\nu^{h}}=k_{\nu}$; therefore, $\prod\nolimits_{i=1}^{s}\overline{\phi_{i}}^{l_{i}}=\prod\nolimits_{j=1}^{t}\overline{f_{j}}$. So, for a fixed $i=1,\dots,s$, there is some $j=1,\dots,t$ such that $\overline{\phi_{i}}$ divides $\overline{f_{j}}$. Since $f_{j}$ is irreducible, it follows from Hensel’s Lemma that $\overline{f_{j}}=\overline{\phi_{i}}^{u_{i}}$ for some $1\leq u_{i}\leq l_{i}$. Let $\alpha_{j}\in\overline{K^{h}}$ be a root of $f_{j}$. As $f_{j}(\alpha_{j})=0$, we have $\overline{\phi_{i}(\alpha_{j})}^{u_{j}}=\overline{f_{j}(\alpha_{j})}=\overline{0}$ modulo $M_{\overline{\nu^{h}}}$. Thus, $\phi_{i}(\alpha_{j})^{u_{i}}\in M_{\overline{\nu^{h}}}$ and so $\phi_{i}(\alpha_{j})\in M_{\overline{\nu^{h}}}$. Now, by Lemma 2.1, $\omega_{j}(\phi_{i}(\alpha))=\overline{\nu^{h}}(\phi_{i}(\alpha_{j}))>0$ as desired.

(ii): Set $p_{1}=p/a$, where $a$ is a coefficient of $p$ of the least $\nu$-valuation. As $\nu^{G}(p_{1})=0$, $p_{1}\in R_{\nu}[X]$. Since $S=\bigcap\nolimits_{j=1}^{t}R_{\omega_{j}}$ (see [9, Corollary 3.1.4]), it follows that, for every $j=1,\dots,t$, we have $p_{1}(\alpha)\in R_{\nu}[\alpha]\subseteq S\subseteq R_{\omega_{j}}$ and

$$ \omega_{j}(p(\alpha))=\omega_{j}(a)+\omega_{j}(p_{1}(\alpha))=\nu(a)+\omega_{j}(p_{1}(\alpha))=\nu^{G}(p(X))+\omega_{j}(p_{1}(\alpha))\geq\nu^{G}(p(X)) $$

as claimed.

(iii): Fix a $j=1,\dots,t$. Since $\prod\nolimits_{i=1}^{s}\phi_{i}(\alpha)^{l_{i}}\equiv f(\alpha)\equiv 0\;(\operatorname{mod}M_{\omega_{j}})$; therefore, $\omega_{j}(\prod\nolimits_{i=1}^{s}\phi_{i}(\alpha)^{l_{i}})>0$. Thus, $\omega_{j}(\phi_{i}(\alpha))>0$ (and so $\phi_{i}(\alpha)\in M_{\omega_{j}}$) for some $i=1,\dots,s$. For $k=1,\dots,s$ with $k\neq i$, as $\overline{\phi_{i}}$ and $\overline{\phi_{k}}$ are coprime modulo $M_{\nu}$, we let $s_{k},t_{k}\in R_{\nu}[X]$ be such that $\overline{s_{k}}\overline{\phi_{i}}+\overline{t_{k}}\overline{\phi}_{k}\equiv 1\;(\operatorname{mod}M_{\nu})$. Then $s_{k}(\alpha)\phi_{i}(\alpha)+t_{k}(\alpha)\phi_{k}(\alpha)=1+h(\alpha)$ for some $h\in M_{\nu}[X]$. As $\nu^{G}(h)>0$, it follows from (ii) that $\omega_{j}(h(\alpha))>0$ and so $h(\alpha)\in M_{\omega_{j}}$. Since $\phi_{i}(\alpha)\in M_{\omega_{j}}$ and $s_{k}(\alpha)\in R_{\nu}[\alpha]\subseteq S\subseteq R_{\omega_{j}}$; therefore, $s_{k}(\alpha)\phi_{i}(\alpha))\in M_{\omega_{j}}$. Thus, ${t_{k}(\alpha)\phi_{k}(\alpha)\in R_{\omega_{j}}-M_{\omega_{j}}}$. Hence, $\omega_{j}(t_{k}(\alpha)\phi_{k}(\alpha))=0$ and so $\omega_{j}(\phi_{k}(\alpha))=0$, yielding the uniqueness of $i$ such that $\omega_{j}(\phi_{i}(\alpha))>0$.

(iv): Define the map $\psi_{j}:k_{\nu}[X]\to R_{\omega_{j}}/M_{\omega_{j}}$ by $\overline{p}(X)\mapsto p(\alpha)+M_{\omega_{j}}$. Since $M_{\nu}\subseteq M_{\omega_{j}}$, $\psi_{j}$ is a well-defined ring homomorphism. As $\omega_{j}(p(\alpha))=\nu^{G}(p(X))+\omega_{j}(p_{1}(\alpha))$ (see (ii)), it follows that $\omega_{j}(p(\alpha))=\nu^{G}(p(X))$ if and only if $\omega_{j}(p_{1}(\alpha))=0$, if and only if $p_{1}(\alpha)\in R_{\omega_{j}}-M_{\omega_{j}}$, if and only if $\overline{p_{1}}(X)\not\in\operatorname{ker}\psi_{j}$. By (iii), let $\phi_{i}$ be such that $\omega_{j}(\phi_{i}(\alpha))>0$. Then $\phi_{i}(\alpha)\in M_{\omega_{j}}$ and so $\overline{\phi_{i}}\in\operatorname{ker}\psi_{j}$. Since $\operatorname{ker}\psi_{j}$ is a principal ideal of $k_{\nu}[X]$ and $\overline{\phi_{i}}$ is irreducible over $k_{\nu}$, $\operatorname{ker}\psi_{j}$ is generated by $\overline{\phi_{i}}$. It follows that $\omega_{j}(p(\alpha))=\nu^{G}(p)$ if and only if $\overline{\phi_{i}}$ does not divide $\overline{p_{1}}$.  ☐

Keeping the notation of Assump’s, in what follows we let $q_{i},r_{i}\in R_{\nu}[X]$ be the quotient and the remainder upon the Euclidean division of $f$ by $\phi_{i}$ for $i=1,\dots,s$.

In [5, Lemma 2.1(b)], it was shown that $\Gamma_{\nu}^{+}$ contains a smallest element in case $R_{\nu}[\alpha]$ is integrally closed and $l_{i}\geq 2$ for some $i=1,\dots,s$. Below, we prove this fact differently with something more.

Lemma 2.3

Keep the notation and assumptions of Lemma $2.2$. If $R_{\nu}[\alpha]$ is integrally closed and $I=\{i\mid l_{i}\geq 2,i=1,\dots,s\}$ is not empty, then $\Gamma_{\nu}^{+}$ has a minimum element with $\min(\Gamma_{\nu}^{+})=\nu^{G}(r_{i})$ for every $i\in I$.

Proof

For $i\in I$, let $q_{i}^{*},r_{i}^{*}\in R_{\nu}[X]$ be the quotient and remainder upon the Euclidean division of $q_{i}$ by $\phi_{i}$. Since $\overline{\phi_{i}}$ divides both $\overline{f}$ and $\overline{q_{i}}\overline{\phi_{i}}$; therefore, $\overline{\phi_{i}}$ divides $\overline{r_{i}}$. But, as $\phi_{i}$ is monic, $\operatorname{deg}(\overline{\phi_{i}})=\operatorname{deg}(\phi_{i})>\operatorname{deg}(r_{i})\geq\operatorname{deg}(\overline{r_{i}})$. This implies that $\overline{r_{i}}$ is zero and so $\nu^{G}(r_{i})>0$. Thus, $\nu^{G}(r_{i})\in\Gamma_{\nu}^{+}$. Now as $\overline{f}=\overline{q_{i}}\overline{\phi_{i}}$ and $\overline{\phi_{i}}^{2}$ divides $\overline{f}$, we see that $\overline{\phi_{i}}$ must divide $\overline{q_{i}}$. Applying a similar argument to the expression $\overline{q_{i}}=\overline{q_{i}^{*}}\overline{\phi_{i}}+\overline{r_{i}^{*}}$, we get that $\overline{r_{i}^{*}}$ is zero. Thus, $\nu^{G}(r_{i}^{*})>0$ and so $\nu^{G}(r_{i}^{*})\in\Gamma_{\nu}^{+}$. To the contrary, suppose that $\tau_{i}\in\Gamma_{\nu}^{+}$ is such that $\tau_{i}<\nu^{G}(r_{i})$, and set $\delta_{i}=\min\{\tau_{i},\nu^{G}(r_{i})-\tau_{i},\nu^{G}(r_{i}^{*})\}$. As $\delta_{i}\in\Gamma_{\nu}^{+}$, let $d_{i}\in R_{\nu}$ be such that $\nu(d_{i})=\delta_{i}$ and set $\theta_{i}=q_{i}(\alpha)/d_{i}$. Let $\omega$ be a valuation of $L$ extending $\nu$. We show that $\theta_{i}\in R_{\omega}$ and, since $\omega$ is arbitrary, it would follow that $\theta_{i}\in S$ [9, Corollary 3.1.4]. As $f(\alpha)=0$; therefore, $\theta_{i}=-r_{i}(\alpha)/(d_{i}\phi_{i}(\alpha))$. By Lemma 2.2, let $j\in\{1,\dots,s\}$ be the unique index such that $\omega(\phi_{j}(\alpha))>0$ and $\omega(\phi_{k}(\alpha))=0$ for all $k\in\{1,\dots,s\}-\{j\}$. If $i\neq j$, then $\omega(\phi_{i}(\alpha))=0$ and

$$ \omega(\theta_{i})=\omega(r_{i}(\alpha))-\omega(d_{i})=\omega(r_{i}(\alpha))-\nu(d_{i})\geq\nu^{G}(r_{i})-\delta_{i}>\delta_{i}-\delta_{i}=0, $$

and so $\theta_{i}\in R_{\omega}$ in this case. Assume, on the other hand, that $i=j$. If $\omega(\phi_{i}(\alpha))>\delta_{i}$, then as $q_{i}^{*}$ is monic and $\omega(q_{i}^{*}(\alpha))\geq\nu^{G}(q_{i}^{*})=0$ (Lemma 2.2), we have

$$ \omega(q_{i}(\alpha))\geq\min\{\omega(q_{i}^{*}(\alpha)\phi_{i}(\alpha)),\omega(r_{i}^{*}(\alpha))\}\geq\min\{\omega(\phi_{i}(\alpha)),\nu^{G}(r_{i}^{*})\}\geq\delta_{i}. $$

So, $\omega(\theta_{i})=\omega(q_{i}(\alpha))-\omega(d_{i})\geq\delta_{i}-\delta_{i}=0$, which implies that $\theta_{i}\in R_{\omega}$ in this case too. If, on the other hand, $\omega(\phi_{i}(\alpha))\leq\delta_{i}$; then

$$ \omega(\theta_{i})=\omega(r_{i}(\alpha))-\omega(d_{i})-\omega(\phi_{i}(\alpha))\geq\nu^{G}(r_{i})-\delta_{i}-\delta_{i}\geq\nu^{G}(r_{i})-\tau_{i}-\delta_{i}\geq\delta_{i}-\delta_{i}=0. $$

So $\theta_{i}\in R_{\omega}$ in this case as well. It follows now from the above argument that $\theta_{i}\in S$. But, as $q_{i}$ is monic and $1/d_{i}\not\in R_{\nu}$, it is clear that $\theta_{i}\not\in R_{\nu}[\alpha]$, contradicting the assumption that $R_{\nu}[\alpha]$ is integrally closed. Hence, $\nu^{G}(r_{i})$ is the minimum element of $\Gamma_{\nu}^{+}$ as claimed.  ☐

Lemma 2.4

Keep the notation and assumptions of Lemma $2.2$. If $\min(\Gamma_{\nu}^{+})=\sigma$, then $\omega(\phi_{i}(\alpha))=\sigma/l_{i}$ for all $i\in\{1,\dots,s\}$ with $\nu^{G}(r_{i})=\sigma$ and for every valuation $\omega$ of $L$ extending $\nu$ such that $\omega(\phi_{i}(\alpha))>0$.

Proof

Let $i\in\{1,\dots,s\}$ and let $\omega$ be a valuation of $L$ extending $\nu$ such that $\omega(\phi_{i}(\alpha))>0$. Write $f$ in the form $f=m_{i}\phi_{i}^{l_{i}}+n_{i}\phi_{i}+r_{i}$, with $m_{i},n_{i}\in R_{\nu}[X]$ and $\nu^{G}(m_{i})=0$, while $\overline{\phi_{i}}$ does not divide $\overline{m_{i}}$, $\nu^{G}(n_{i})>0$, and $\operatorname{deg}(r_{i})<\operatorname{deg}(\phi_{i})$. Notice that if $l_{i}=1$ then $m_{i}=q_{i}$ and $n_{i}=0$. By Lemma 2.2, $\omega(n_{i}(\alpha))\geq\nu^{G}(n_{i})\geq\sigma$, $\omega(m_{i}(\alpha))=\nu^{G}(m_{i})=0$, and $\omega(r_{i}(\alpha))=\nu^{G}(r_{i})=\sigma$ as $\overline{\phi_{i}}$ divides neither $\overline{m_{i}}$ nor $\overline{r_{i}}$. We then have

$$ l_{i}\omega(\phi_{i}(\alpha))=\omega(m_{i}(\alpha)\phi_{i}^{l_{i}}(\alpha))=\omega(n_{i}(\alpha)\phi_{i}(\alpha)+r_{i}(\alpha))=\omega(r_{i}(\alpha))=\nu^{G}(r_{i})=\sigma $$

as claimed.  ☐

Now we get to our first main result which computationally enhances [5, Theorem 1.1] as well as improves [5, Theorem 4.1] in the sense that $K$ is not assumed to be Henselian.

Theorem 2.5

Keep the notation and assumptions of Lemma $2.2$.

(i) If $l_{i}=1$ for all $i=1,\dots,s$, then $R_{\nu}[\alpha]$ is integrally closed.

(ii) If $I=\{i\mid l_{i}\geq 2,i=1,\dots,s\}$ is not empty, then $R_{\nu}[\alpha]$ is integrally closed if and only if $\nu^{G}(r_{i})=\min(\Gamma_{\nu}^{+})$ for every $i\in I$.

Proof

(i): Assume that $l_{i}=1$ for all $i=1,\dots,s$. An arbitrary element of $S$ is of the form $\theta=h(\alpha)/b$ for some $b\in R_{\nu}$ and $h\in R_{\nu}[X]$, with $\nu^{G}(h)=0$ and ${\operatorname{deg}(h)<\operatorname{deg}(f)}$. Since $f$ is monic, $\operatorname{deg}(\overline{h})\leq\operatorname{deg}(h)<\operatorname{deg}(f)=\operatorname{deg}(\overline{f})$. As $l_{i}=1$ for all $i=1,\dots,s$, there is some $i=1,\dots,s$ such that $\overline{\phi_{i}}$ does not divide $\overline{h}$. For such a fixed $i$, let $\omega$ be a valuation of $L$ extending $\nu$ such that $\omega(\phi_{i}(\alpha))>0$, which exists by Lemma 2.2. Hence, $\omega(h(\alpha))=\nu^{G}(h)=0$. If $\nu(b)>0$, then ${\omega(\theta)=\omega(h(\alpha))-\omega(b)=0-\nu(b)<0}$. Thus $\theta\not\in S$, which is a contradiction. Hence, $\nu(b)=0$, which implies that $\theta\in R_{\nu}[\alpha]$. This shows that $S=R_{\nu}[\alpha]$ and so $R_{\nu}[\alpha]$ is integrally closed.

(ii): Assume that $I\neq\varnothing$. If $R_{\nu}[\alpha]$ is integrally closed, then it follows from Lemma 2.3 that $\nu^{G}(r_{i})$ is the minimum element of $\Gamma_{\nu}^{+}$ for every $i\in I$, as claimed.

Conversely, put $\min(\Gamma_{\nu}^{+})=\sigma$ and let $\pi\in R_{\nu}$ be such that $\nu(\pi)=\sigma$. Assume that $\nu^{G}(r_{i})=\sigma$ for every $i\in I$. We aim at proving that $R_{\nu}[\alpha]$ is integrally closed. By an appropriate choice of a lifting of $\overline{\phi_{i}}$, we begin by showing that we can also assume that $\nu^{G}(r_{i})=\sigma$ for $i\not\in I$. Let $i\not\in I$, and assume that $\nu^{G}(r_{i})>\sigma$. If $\delta\in\Gamma_{\nu}^{+}$ with $\sigma<\delta<2\sigma$, then $\delta-\sigma\in\Gamma_{\nu}^{+}$ with $\delta-\sigma<2\sigma-\sigma=\sigma$ contradicting the minimality of $\sigma$. So there is no element of $\Gamma_{\nu}^{+}$ lying strictly between $\sigma$ and $2\sigma$. So, $\nu^{G}(r_{i})\geq 2\sigma$. Let $q_{i}^{*},r_{i}^{*}\in R_{\nu}[X]$ be the quotient and remainder upon the Euclidean division of $q_{i}$ by $\phi_{i}$. Put $\phi_{i}^{**}=\phi_{i}+\pi$, $q_{i}^{**}=q_{i}-\pi q_{i}^{*}$, and $r_{i}^{**}=r_{i}-\pi r_{i}^{*}+\pi^{2}q_{i}^{*}$. Then

$$ q_{i}^{**}\phi_{i}^{**}+r_{i}^{**}=(q_{i}-\pi q_{i}^{*})(\phi_{i}+\pi)+r_{i}-\pi r_{i}^{*}+\pi^{2}q_{i}^{*}=q_{i}\phi_{i}+r_{i}=f. $$

It can be easily checked that $q_{i}^{**}$ and $r_{i}^{**}$ are the quotient and remainder upon the Euclidean division of $f$ by $\phi_{i}^{**}$ (if $\operatorname{deg}(r_{i}^{**})\geq\operatorname{deg}(\phi_{i}^{**})$; then we replace $r_{i}^{**}$ by the remainder upon the Euclidean division of $r_{i}^{**}$ with $\phi_{i}^{**}$ and replace $q_{i}^{**}$ with $q_{i}^{**}+Q_{i}$, where $Q_{i}$ is the quotient upon the Euclidean division of $r_{i}^{**}$ by $\phi_{i}^{**}$). Since $l_{i}=1$, $\overline{r_{i}^{*}}$ is nonzero, and so $\nu^{G}(\pi r_{i}^{*})=\nu(\pi)=\sigma$. As $\nu^{G}(r_{i})\geq 2\sigma$ and $\nu^{G}(\pi^{2}q_{i}^{*})\geq\nu(\pi^{2})=2\sigma$, it follows that $\nu^{G}(r_{i}^{**})=\nu^{G}(\pi r_{i}^{*})=\sigma$. So, replacing $\phi_{i}$ by $\phi_{i}+\pi$, we can assume that $\nu^{G}(r_{i})=\sigma$. We thus assume in the remainder of the proof that $\nu^{G}(r_{i})=\sigma$ for all $i=1,\dots,s$. We finally get to proving that $R_{\nu}[\alpha]$ is integrally closed. Assume to the contrary that there exists some $\theta\in S-R_{\nu}[\alpha]$. Then $\theta$ can be written as $\theta=g(\alpha)/b$ for some $b\in R_{\nu}$ and $g\in R_{\nu}[X]$ with $\nu(b)\geq\sigma$, $\nu^{G}(g)=0$, and $\operatorname{deg}(g)<\operatorname{deg}(f)$. Given $i=1,\dots,s$, let $m_{i}\geq 0$ be the highest power of $\overline{\phi_{i}}$ dividing $\overline{g}$. Since $\operatorname{deg}(g)<\operatorname{deg}(f)$, there must exist some $i=1,\dots,s$ such that $m_{i}\leq l_{i}-1$. For such an $i$, apply the Euclidean division of $g$ by $\phi_{i}^{m_{i}}$ to get $g=S_{i}\phi_{i}^{m_{i}}+T_{i}$, where $S_{i},T_{i}\in R_{\nu}[X]$, while $\overline{\phi_{i}}$ does not divide $\overline{S_{i}}$, and $\nu^{G}(T_{i})\geq\sigma$. By Lemma 2.2, let $\omega$ be a valuation of $L$ extending $\nu$ such that $\omega(\phi_{i}(\alpha))>0$. Since $\overline{\phi_{i}}$ does not divide $\overline{S_{i}}$ and $S_{i}$ is monic, it follows from Lemma 2.2 that $\omega(S_{i}(\alpha))=\nu^{G}(S_{i})=0$. Using Lemma 2.4, we then have $\omega(S_{i}(\alpha)\phi_{i}(\alpha)^{m_{i}})=m_{i}\omega(\phi_{i}(\alpha))=m_{i}\sigma/l_{i}$. Since $\omega(T_{i}(\alpha))\geq\nu^{G}(T_{i})\geq\sigma$ (by Lemma 2.2), it follows that

$$ \omega(g(\alpha))=\min\{\omega(S_{i}(\alpha)\phi_{i}(\alpha)^{m_{i}}),\omega(T_{i}(\alpha))\}=\min\{m_{i}\sigma/l_{i},\sigma\}=m_{i}\sigma/l_{i}<\sigma. $$

Thus, $\omega(\theta)=\omega(g(\alpha))-\omega(b)=\omega(g(\alpha))-\nu(b)<\sigma-\sigma=0$. Hence, $\theta\not\in R_{\omega}$ and so $\theta\not\in S$. This contradiction leads to the conclusion that $S=R_{\nu}[\alpha]$, as desired.  ☐

The following corollary is immediate.

Corollary 2.6

Keep the assumptions of Theorem $2.5$. If $\Gamma_{\nu}^{+}$ does not have a minimum element, then $R_{\nu}[\alpha]$ is integrally closed if and only if $l_{i}=1$ for all $i=1,\dots,s$.

The following corollary shows, in particular, that Theorem 2.5 is a new version of the generalized Dedekind criterion which computationally improves [4, Theorem 1] and [5, Theorem 1.1] in the case of separable extensions.

Corollary 2.7

Keep the assumptions of Theorem $2.5$. If $\Gamma_{\nu}^{+}$ has a minimum element $\sigma$ and $I=\{i\mid l_{i}\geq 2$, $i=1,\dots,s\}$ is not empty, then $R_{\nu}[\alpha]$ is integrally closed if and only if $\overline{\phi_{i}}$ does not divide $\overline{M}$ for every $i\in I$, where $M=\frac{f-\prod\nolimits_{i=1}^{s}\phi_{i}^{l_{i}}}{\pi}$ for any $\pi\in R_{\nu}$ with $\nu(\pi)=\sigma$.

Proof

Let $i\in I$. Since $\overline{r_{i}}=\overline{f}-\overline{q_{i}}\;\overline{\phi_{i}}$ and $\overline{\phi_{i}}$ divides $\overline{f}$; therefore, $\overline{r_{i}}$ is divisible by $\overline{\phi_{i}}$. But as $\operatorname{deg}(\overline{r_{i}})\leq\operatorname{deg}(r_{i})<\operatorname{deg}(\phi_{i})=\operatorname{deg}(\overline{\phi_{i}})$, $\overline{r_{i}}$ must be zero. Thus,

$$ \overline{q_{i}}=\overline{\phi_{i}^{l_{i}-1}}\prod\limits_{j=1,j\neq i}^{s}\overline{\phi_{j}^{l_{j}}}. $$

Let $H_{i}\in R_{\nu}[X]$ be such that $q_{i}=\phi_{i}^{l_{i}-1}\prod\nolimits_{j=1,j\neq i}^{s}\phi_{j}^{l_{j}}+\pi H_{i}$ with $\pi\in R_{\nu}$ such that $\nu(\pi)=\sigma$. Then

$$ f=\bigg{(}\phi_{i}^{l_{i}-1}\prod\limits_{j=1,j\neq i}^{s}\phi_{j}^{l_{j}}+\pi H_{i}\bigg{)}\phi_{i}+r_{i}. $$

Put

$$ M=\frac{f-\prod\limits_{j=1}^{s}\phi_{j}^{l_{j}}}{\pi}\in R_{\nu}[X]. $$

Then

$$ M=\frac{\Big{(}\phi_{i}^{l_{i}-1}\prod\limits_{j=1,j\neq i}^{s}\phi_{j}^{l_{j}}+\pi H_{i}\Big{)}\phi_{i}+r_{i}-\prod\limits_{j=1}^{s}\phi_{j}^{l_{j}}}{\pi}=H_{i}\phi_{i}+\frac{r_{i}}{\pi}. $$

Since $M,H_{i}\phi_{i}\in R_{\nu}[X]$, we must have $\frac{r_{i}}{\pi}\in R_{\nu}[X]$ and so $\nu^{G}\bigl{(}\frac{r_{i}}{\pi}\bigr{)}\geq 0$. Clearly, $\overline{\phi_{i}}$ divides $\overline{M}$ if and only if $\overline{\phi}$ divides $\overline{(\frac{r_{i}}{\pi})}$. As $\operatorname{deg}(\overline{(\frac{r_{i}}{\pi})})\leq\operatorname{deg}(\overline{r_{i}})<\operatorname{deg}(\overline{\phi_{i}})$ (see above), we conclude that $\overline{\phi_{i}}$ divides $\overline{M}$ if and only if $\overline{(\frac{r_{i}}{\pi})}$ is zero; i.e., $\nu^{G}(r_{i})>\sigma$. Contrapositively, $\overline{\phi_{i}}$ does not divide $\overline{M}$ if and only if $\nu^{G}(r_{i})=\sigma$.  ☐

Our second main result, Theorem 2.9 below, gives a characterization of the integral closedness of $R_{\nu}[\alpha]$ which is based on characterization of the extensions of $\nu$ to $L$ (see also [6, Theorem 1.3], where the proof of our result is simpler and selfcontained).

In 1850, Eisenstein introduced his infamous criterion for testing irreducibility of polynomials over valued fields in [10]. In 2008, Brown gave a simple proof of the most general version of Eisenstein–Schönemann irreducibility criterion in [11]. Namely, if $p\in{𝕑}$ is prime and $f\in{𝕑}[x]$ is such that $f=\phi^{n}+a_{n-1}\phi^{n-1}+\dots+a_{0}$ for some monic polynomial $\phi\in{𝕑}[x]$ whose reduction modulo $p$ is irreducible and $a_{i}\in{𝕑}[x]$ with $\deg(a_{i})<\deg(\phi)$ for $i=0,\dots,n-1$, then $f$ is irreducible over ${𝕈}$ if $\operatorname{gcd}(\nu_{p}^{G}(a_{0}),n)=1$ and $n\nu_{p}^{G}(a_{i})\geq(n-i)\nu_{p}^{G}(a_{0})>0$ for every $i$ where $\nu_{p}$ is the $p$-adic valuation. In preparation for Theorem 2.9, we introduce the following definition and prove some lemma that partially generalizes the Eisenstein–Schönemann irreducibility criterion.

Definition

We say that a monic polynomial $g\in R_{\nu}[X]$ is $\nu$-Eisenstein–Schönemann if there exists a monic polynomial $\psi\in R_{\nu}[X]$ such that $\overline{\psi}$ is irreducible, $\overline{g}$ is a positive power of $\overline{\psi}$, and $\nu^{G}(r)=\min(\Gamma_{\nu}^{+})$, where $r\in R_{\nu}[X]$ is the remainder upon the Euclidean division of $g$ by $\psi$. In particular, if $\psi(x)=x$, then $g$ is said to be $\nu$-Eisenstein.

Lemma 2.8

Keep the assumptions of Theorem $2.5$. If $g\in R_{\nu}[X]$ is monic and $\nu$-Eisenstein–Schönemann, then $g$ is irreducible over $K$.

Proof

Let $\psi\in R_{\nu}[X]$ be monic such that $\overline{\psi}$ is irreducible, $\overline{g}=\overline{\psi}^{l}$, and $\nu^{G}(r)=\min(\Gamma_{\nu}^{+})=\sigma$, where $r\in R_{\nu}[X]$ is the remainder upon the Euclidean division of $g$ by $\psi$. Suppose to the contrary that $g=h_{1}h_{2}$ for some nonconstant and monic $h_{1},h_{2}\in R_{\nu}[X]$. Then $\overline{h_{1}}=\overline{\psi}^{l_{1}}$ and $\overline{h_{2}}=\overline{\psi}^{l_{2}}$ for some positive $l_{1}$ and $l_{2}$ with $l_{1}+l_{2}=l$. Assume that the Euclidean division of each of $g$, $h_{1}$, and $h_{2}$ by $\psi$ yields

$$ g=q\psi+r,\quad h_{1}=q_{1}\psi+r_{1},\quad h_{2}=Q_{2}\psi+r_{2}. $$

It is clear that $r$ is the remainder upon the Euclidean division of the product $r_{1}r_{2}$ by $\psi$. Since both $\overline{h_{1}}$ and $\overline{h_{2}}$ are positive powers of $\overline{\psi}$, both of $\overline{r_{1}}$ and $\overline{r_{2}}$ must be zero. So, $\nu^{G}(r_{1})\geq\sigma$ and $\nu^{G}(r_{2})\geq\sigma$. Thus, $\nu^{G}(r)\geq 2\sigma>\sigma$ (as $\sigma>0$), which is a contradiction. Hence, $g$ is irreducible over $R_{\nu}$ and, consequently, irreducible over $K$ (by Gauss’s Lemma as $R_{\nu}$ is integrally closed).  ☐

Theorem 2.9

Keep the assumptions of Theorem $2.5$. The following are equivalent:

(i) $R_{\nu}[\alpha]$ is integrally closed.

(ii) $\nu$ has exactly $s$ distinct extensions $\omega_{1},\dots,\omega_{s}$ to $L$, and if $I=\{i\mid l_{i}\geq 2,i=1,\dots,s\}$ is not empty; then $l_{i}\omega_{i}(\phi_{i}(\alpha))$ is the minimum element of $\Gamma_{\nu}^{+}$ for every $i\in I$, where $\omega_{i}$ is a valuation satisfying $\omega_{i}(\phi_{i}(\alpha))>0$ which exists by Lemma $2.2$.

Proof

Assume that $R_{\nu}[\alpha]$ is integrally closed. Since $k_{\nu}=k_{\nu^{h}}$ and $\overline{f}=\prod\nolimits_{i=1}^{s}\overline{\phi_{i}^{l_{i}}}$, Hensel’s Lemma yields a factorization $f=\prod\nolimits_{i=1}^{s}f_{i}$ over $K^{h}$ such that $\overline{f_{i}}=\overline{\phi_{i}^{l_{i}}}$ for $i=1,\dots,s$. In order for us to invoke Lemma 2.1, we need to show that the factors $f_{1},\dots,f_{s}$ are all irreducible over $K^{h}$. If $i\in\{1,\dots,s\}-I$, then $f_{i}$ is immediately irreducible over $K^{h}$ since $\overline{f_{i}}=\overline{\phi_{i}}$ is irreducible. If $i\in I$, then we set to show that $f_{i}$ is $\nu^{h}$-Eisenstein–Schönemann and thus irreducible by Lemma 2.8. Since $R_{\nu}[\alpha]$ is integrally closed and $l_{i}\geq 2$, it follows from Lemma 2.3 that $\Gamma^{+}$ has a minimum element $\sigma$ and $\nu^{G}(r_{i})=\sigma$. Notice that as $\Gamma_{\nu}=\Gamma_{\nu^{h}}$; therefore, $\sigma$ is the minimum element of $\Gamma_{\nu^{h}}^{+}$ as well. Let $q_{i}^{*},r_{i}^{*}\in R_{\nu^{h}}[X]$ be, respectively, the quotient and remainder upon the Euclidean division of $f_{i}$ by $\phi_{i}$. Letting $G_{i}=\prod\nolimits_{j=1,j\neq i}^{s}f_{j}$, we write $f=f_{i}G_{i}=q_{i}^{*}\phi_{i}G_{i}+r_{i}^{*}G_{i}$. Using the Euclidean division again to divide $r_{i}^{*}G_{i}$ by $\phi_{i}$, let $r_{i}^{*}G_{i}=q_{i}^{**}\phi_{i}+r_{i}^{**}$, with $q_{i}^{**},r_{i}^{**}\in R_{\nu^{h}}[X]$. Then

$$ f=q_{i}^{*}\phi_{i}G_{i}+q_{i}^{**}\phi_{i}+r_{i}^{**}=(q_{i}^{*}G_{i}+q_{i}^{**})\phi_{i}+r_{i}^{**}. $$

Owing to the uniqueness of the remainder, $r_{i}=r_{i}^{**}$. Thus, ${\nu^{h}}^{G}(r_{i}^{**})={\nu^{h}}^{G}(r_{i})=\nu^{G}(r_{i})=\sigma$. If ${\nu^{h}}^{G}(r_{i}^{*})>\sigma$, then ${\nu^{h}}^{G}(r_{i}^{*}G_{i})>\sigma$ and so ${\nu^{h}}^{G}(r_{i}^{**})>\sigma$; a contradiction. Thus, ${\nu^{h}}^{G}(r_{i}^{*})=\sigma$ and we conclude that $f_{i}$ is $\nu^{h}$-Eisenstein–Schönemann as desired. It follows now by Lemma 2.1 that there are exactly $s$ valuations $\omega_{1},\dots,\omega_{s}$ of $L$ extending $\nu$; and by Lemma 2.4$l_{i}\omega_{i}(\phi_{i}(\alpha))=\sigma$ for the valuation $\omega_{i}$ of $L$ extending $\nu$ with $\omega_{i}(\phi_{i}(\alpha))>0$.

Conversely, assume that there are exactly $s$ valuations $\omega_{1},\dots,\omega_{s}$ of $L$ extending $\nu$, and if $I=\{i\mid l_{i}\geq 2,i=1,\dots,s\}$ is not empty, then $l_{i}\omega_{i}(\phi_{i}(\alpha))$ is the minimum element of $\Gamma_{\nu}^{+}$ for every $i\in I$ and every $\omega_{i}$ satisfying $\omega_{i}(\phi_{i}(\alpha))>0$. If $I=\varnothing$, then $R_{\nu}[\alpha]$ is integrally closed by Theorem 2.5. Assume that $I\neq\varnothing$. Following Theorem 2.5, in order to show that $R_{\nu}[\alpha]$ is integrally closed, it suffices to prove that $\nu^{G}(r_{i})=\sigma$ for every $i\in I$, where $\sigma=\min(\Gamma_{\nu}^{+})$. Let $\omega_{i}$ be the valuation of $L$ extending $\nu$ such that $\omega_{i}(\phi_{i}(\alpha))>0$ (by Lemma 2.2). Then, by assumption, $l_{i}\omega_{i}(\phi_{i}(\alpha))=\sigma$. Write $f$ in the form $f=m_{i}\phi_{i}^{l_{i}}+n_{i}\phi_{i}+r_{i}$ for $m_{i},n_{i}\in R_{\nu}[X]$ with $\nu^{G}(m_{i})=0$. Thus $\overline{\phi_{i}}$ does not divide $\overline{m_{i}}$, $\nu^{G}(n_{i})>0$, and $\operatorname{deg}(r_{i})<\operatorname{deg}(\phi_{i})$. Since $f(\alpha)=0$, we have $r_{i}=-m_{i}\phi_{i}^{l_{i}}-n_{i}\phi_{i}$. We can see (using Lemma 2.2(ii)) that

$$ \omega_{i}(n_{i}(\alpha)\phi_{i}(\alpha))=\omega_{i}(n_{i}(\alpha))+\omega_{i}(\phi_{i}(\alpha))>\omega_{i}(n_{i}(\alpha))\geq\nu^{G}(n_{i})\geq\sigma, $$

and (where $\omega_{i}(m_{i}(\alpha))=\nu^{G}(m_{i})=0$ by Lemma 2.2(iv))

$$ \omega_{i}(m_{i}(\alpha)\phi_{i}(\alpha)^{l_{i}})=\omega_{i}(\phi_{i}(\alpha)^{l_{i}})=l_{i}\omega_{i}(\phi_{i}(\alpha))=\sigma. $$

So,

$$ \omega_{i}(r_{i}(\alpha))=\omega_{i}\left(-m_{i}(\alpha)\phi_{i}(\alpha)^{l_{i}}-n_{i}(\alpha)\phi_{i}(\alpha)\right)=\sigma. $$

Since $\operatorname{deg}(r_{i})<\operatorname{deg}(\phi_{i})$, $\overline{\phi_{i}}$ does not divide $\overline{r_{i}}$. So, by Lemma 2.2(iv), $\nu^{G}(r_{i})=\omega_{i}(r_{i}(\alpha))=\sigma$ and the proof is complete.  ☐

With the notation of Theorem 2.9, given a valuation $\omega_{i}$ of $L$ extending $\nu$, we denote the ramification index $[\Gamma_{\omega_{i}}:\Gamma_{\nu}]$ by $e(\omega_{i}/\nu)$ and the residue degree $[k_{\omega_{i}}:k_{\nu}]$ by $f(\omega_{i}/\nu)$. The following fundamental inequality is well known (see [9, Theorem 3.3.4] for instance):

$$ \sum\limits_{i=1}^{s}e(\omega_{i}/\nu)f(\omega_{i}/\nu)\leq[L:K]. $$

When $R_{\nu}[\alpha]$ is integrally closed, we calculate in the next corollary the ramification indices $e(\omega_{i}/\nu)$ and residue degrees $f(\omega_{i}/\nu)$ and show consequently that the above inequality is indeed an equality.

Corollary 2.10

Keep the notation and assumptions of Theorem $2.9$. If $R_{\nu}[\alpha]$ is integrally closed, then $e(\omega_{i}/\nu)=l_{i}$ and $f(\omega_{i}/\nu)=\operatorname{deg}(\phi_{i})$ for all $i=1,\dots,s$ and, furthermore, $\sum\nolimits_{i=1}^{s}e(\omega_{i}/\nu)f(\omega_{i}/\nu)=[L:K]$.

Proof

We show first that $e(\omega_{i}/\nu)\geq l_{i}$ and $f(\omega_{i}/\nu)\geq\operatorname{deg}(\phi_{i})$ for every $i=1,\dots,s$. If $l_{i}=1$ for some $i=1,\dots,s$, then clearly $e(\omega_{i}/\nu)\geq l_{i}$. Since $\overline{f_{i}}=\overline{\phi_{i}}$, it follows that, for any root $\alpha_{i}$ of $f_{i}$, $\overline{\phi_{i}}$ is the minimal polynomial of $\overline{\alpha_{i}}$ over $k_{\nu}$ and so

$$ \operatorname{deg}(\phi_{i})=\operatorname{deg}(\overline{\phi_{i}})=[k_{\nu}(\overline{\alpha_{i}}):k_{\nu}]\leq[k_{\omega_{i}}:k_{\nu}]=f(\omega_{i}/\nu). $$

If $l_{i}\geq 2$ for some $i=1,\dots,s$, then it follows from Theorem 2.9(ii) that $\omega_{i}(\phi_{i}(\alpha))=\sigma/l_{i}$, where $\sigma=\min(\Gamma^{+}_{\nu})$. So, $\Gamma_{\nu}\subseteq\Gamma[\sigma/l_{i}]\subseteq\Gamma_{\omega_{i}}$ and

$$ l_{i}=[\Gamma_{\nu}[\sigma/l_{i}]:\Gamma_{\nu}]\leq[\Gamma_{\omega_{i}}:\Gamma_{\nu}]=e(\omega_{i}/\nu). $$

Also, for a root $\alpha_{i}$ of $f_{i}$, we have $\overline{\phi_{i}}(\overline{\alpha_{i}})^{l_{i}}=\overline{f_{i}}(\overline{\alpha_{i}})=0$ implying that $\overline{\phi_{i}}(\overline{\alpha_{i}})=0$ in $k_{\omega_{i}}$. Since $\overline{\phi_{i}}$ is monic and irreducible over $k_{\nu}$; therefore,

$$ \operatorname{deg}(\phi_{i})=\operatorname{deg}(\overline{\phi_{i}})=[k_{\nu}(\overline{\alpha_{i}}):k_{\nu}]\leq[k_{\omega_{i}}:k_{\nu}]=f(\omega_{i}/\nu). $$

Now, the above argument yields

$$ \sum\limits_{i=1}^{s}e(\omega_{i}/\nu)f(\omega_{i}/\nu)\geq\sum\limits_{i=1}^{s}l_{i}\operatorname{deg}(\phi_{i})=\sum\limits_{i=1}^{s}l_{i}\operatorname{deg}(\overline{\phi_{i}})=\operatorname{deg}(\overline{f})=\operatorname{deg}(f)=[L:K]. $$

Thus, by this and the fundamental inequality, we get the claimed equality

$$ \sum\limits_{i=1}^{s}e(\omega_{i}/\nu)f(\omega_{i}/\nu)=[L:K]. $$

Furthermore, since $l_{i}\leq e(\omega_{i}/\nu)$ and $\operatorname{deg}(\phi_{i})\leq f(\omega_{i}/\nu)$ for all $i=1,\dots,s$ with

$$ \sum\limits_{i=1}^{s}l_{i}\operatorname{deg}(\phi_{i})=\sum\limits_{i=1}^{s}e(\omega_{i}/\nu)f(\omega_{i}/\nu), $$

we conclude that $l_{i}=e(\omega_{i}/\nu)$ and $\operatorname{deg}(\phi_{i})=f(\omega_{i}/\nu)$ for every $i=1,\dots,s$.  ☐

3. Applications and Examples

Corollary 3.1

Keep the assumptions of Theorem $2.5$ with $f(X)=X^{n}-a\in R_{\nu}[X]$ irreducible of degree $n\geq 2$ and $a\in M_{\nu}$.

1. If $\Gamma_{\nu}^{+}$ has no minimum element, then $R_{\nu}[\alpha]$ is not integrally closed.

2. If $\min(\Gamma_{\nu}^{+})=\sigma$, then $R_{\nu}[\alpha]$ is integrally closed if and only if $\nu(a)=\sigma$.

Proof

This is a direct application of Theorem 2.5.  ☐

Corollary 3.2

Keep the assumptions of Theorem $2.5$. Let $\min(\Gamma_{\nu}^{+})=\sigma$ and let $g\in R_{\nu}[X]$ be monic. If $g$ is $\nu$-Eisenstein and $L=K(\theta)$ for some root $\theta$ of $g$, then $R_{\nu}[\theta]$ is integrally closed.

Proof

By Lemma 2.8, $g$ is irreducible over $K$. Now, the remaining part is straightforward from Theorem 2.5.  ☐

Corollary 3.3

Let $f(X)=X^{n}-a\in R_{\nu}[X]$, $\min(\Gamma^{+})=\sigma$, $\nu(a)=m\sigma$ for some $m\in{𝕅}$. Let $L=K(\theta)$ for a root $\theta$ of $f(X)$. If $m$ and $n$ are coprime; then $f$ is irreducible over $R$ and $R[\theta^{v}/\pi^{u}]$ is the integral closure of $R$ in $L$, where $\pi\in R_{\nu}$ is such that $\nu(\pi)=\sigma$, and $u,v\in{𝕑}$ are the unique integers such that $mv-nu=1$ and $0\leq v<n$.

Proof

Let $A=a^{v}/\pi^{nu}$. Then $\nu(A)=(mv-nu)\sigma=\sigma$. By Lemma 2.8, $g(X)=X^{n}-A$ is irreducible over $R_{\nu}$. Furthermore, $\theta^{v}/\pi^{u}$ is a root of $g$. So $[K(\theta^{v}/\pi^{u}):K]=n$. Therefore, $K(\theta)=K(\theta^{v}/\pi^{u})$ and $f$ is irreducible over $K$. By Corollary 3.2, $R_{\nu}[\theta^{v}/\pi^{u}]$ is integrally closed.  ☐

Example 1

Let $\geq$ be the lexicographic order on ${𝕑}^{2}$; i.e., $(a,b)\geq(c,d)$ if and only if ($a<c$) or ($a=c$ and $b\leq d$). Then $({𝕑}^{2},\geq)$ is a totally ordered abelian group. Let $F$ be a field and $K=F(X,Y)$, the field of rational functions over $F$ in indeterminates $X$ and $Y$. Define the valuation $\nu:K\to{𝕑}^{2}\cup\{\infty\}$ by $0\neq\sum\nolimits_{i,j}a_{i,j}X^{i}Y^{j}\mapsto\min\{(i,j)\mid a_{i,j}\neq 0\}$ for $\sum\nolimits_{i,j}a_{i,j}X^{i}Y^{j}\in F[X,Y]$, $0\mapsto\infty$, and $\nu^{G}(f/g)=\nu^{G}(f)-\nu^{G}(g)$ for $f,g\in F[X,Y]$ with $g\neq 0$. Then, obviously, $\nu$ is a discrete valuation on $K$ of rank 2 whose value group is $\Gamma_{\nu}=({𝕑}^{2},\geq)$. Let $f(Z)=Z^{3}+aZ+b\in R_{\nu}[Z]$ be irreducible and $L=K(\alpha)$ for some root $\alpha$ of $f$. Assume that $\nu(a)>(0,0)$ and $\nu(b)>(0,0)$. Then $\overline{f}(Z)=Z^{3}$. Let $r$ be the remainder upon the Euclidean division of $f$ by $Z$. Noting that $\min(\Gamma_{\nu}^{+})=(0,1)$, it follows from Theorem 2.5 that $R_{\nu}[\alpha]$ is integrally closed if and only if $\nu^{G}(r)=(0,1)$. In particular, if $f(Z)=Z^{3}+Y$, then $R_{\nu}[\alpha]$ is integrally closed; while if $f(Z)=Z^{3}+YZ+X$, then $R_{\nu}[\alpha]$ is not integrally closed.

Example 2

Let $(F,\nu)$ be a valued field and let $K=F(X)$ be the field of rational functions over $F$ in an indeterminate $X$. Given some positive irrational real $\lambda$, define the valuation $\omega:K\to{𝕉}\cup\{\infty\}$ as follows: $\omega(0)=\infty$, for $0\neq f(X)=\sum\nolimits_{i=0}^{n}a_{i}X^{i}\in F[X]$, set ${\omega(f)=\min\{\nu(a_{i})+i\lambda,i\}}$, and for $f,g\in F[X]$ with $g\neq 0$, $\omega(f/g)=\omega(f)-\omega(g)$ (see [9, Theorem 2.2.1]). Let $f(Z)=Z^{3}+aZ+b\in R_{\omega}[Z]$ be irreducible and $L=K(\alpha)$ for some root $\alpha$ of $f$. If $(F,\nu)$ is the trivial valued field, then $\Gamma_{\omega}=\lambda{𝕑}$. So, in this case, if $\nu(a)>0$ and $\nu(b)>0$, then $\overline{f}(Z)=Z^{3}$. Hence, by Theorem 2.5, $R_{\omega}[\alpha]$ is integrally closed if and only if $\nu(b)=\lambda$. In particular, if $f(Z)=Z^{3}+X$, then $R_{\omega}[\alpha]$ is integrally closed. If $F={𝕈}$ and $\nu$ is the $p$-adic valuation on ${𝕈}$ for some prime integer $p$, then $\Gamma_{\omega}={𝕑}+\lambda{𝕑}$, which is dense in ${𝕉}$ and, thus, $\inf(\Gamma_{\omega}^{+})=0$. So, according to Theorem 2.5, $R_{\omega}[\alpha]$ is integrally closed if and only if $\overline{f}$ is square-free. In particular, if $\nu(a)>0$ and $\nu(b)>0$; then $\overline{f}(Z)=Z^{3}$ and so $R_{\omega}[\alpha]$ is not integrally closed.