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On Dedekind Criterion and Simple Extensions ofValuation RingsSudesh K. Khanduja a & Munish Kumar aa Department of Mathematics , Panjab University , Chandigarh, IndiaPublished online: 18 Feb 2010.

To cite this article: Sudesh K. Khanduja & Munish Kumar (2010) On Dedekind Criterion and Simple Extensions of ValuationRings, Communications in Algebra, 38:2, 684-696, DOI: 10.1080/00927870902829080

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Communications in Algebra®, 38: 684–696, 2010Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870902829080

ON DEDEKIND CRITERION AND SIMPLE EXTENSIONSOF VALUATION RINGS

Sudesh K. Khanduja and Munish KumarDepartment of Mathematics, Panjab University, Chandigarh, India

Let R be an integrally closed domain with quotient field K and S be the integralclosure of R in a finite extension L = K��� of K with � integral over R. Let f�x�

be the minimal polynomial of � over K and ��� be a maximal ideal of R. Kummerproved that if S = R���, then the number of maximal ideals of S which lie over �,together with their ramification indices and residual degrees can be determined fromthe irreducible factors of f�x� modulo ���. In this article, the authors give necessary andsufficient conditions to be satisfied by f�x� which ensure that S = R��� when R is thevaluation ring of a valued field �K� v� of arbitrary rank. The problem dealt with hereis analogous to the one considered by Dedekind in case R is the localization of ��� at arational prime p, which in fact gave rise to Dedekind Criterion (cf. [9]). The articlealso contains a criterion for the integral closure of any valuation ring R in a finiteextension of the quotient field of R to be generated over R by a single element, whichgeneralizes a result of Dedekind regarding the index of an algebraic number field.

Key Words: Non-Archimedean valued fields; Valued fields.

2000 Mathematics Subject Classification: 12J10; 12J25.

1. INTRODUCTION

Let R be an integrally closed domain with quotient field K and S be theintegral closure of R in a finite extension L = K��� of K with � integral over R.Let f�x� be the minimal polynomial of � over K and � be a maximal ideal of R.Let f �x� = g1�x�

e1 � � � gr �x�er be the factorization of the polynomial f �x� obtained by

reducing each coefficient of f�x� modulo �, into powers of distinct monic irreduciblepolynomials over R/�. A well-known theorem of Kummer asserts that if S = R���,then the ring S has exactly r maximal ideals ℘1� � � � � ℘r given by ℘i = �S + gi���Swhich lie over � and �S is the product of r pairwise comaximal ideals �1� � � � ��r

with �i = �S + gi���eiS, 1 ≤ i ≤ r (cf. [12, Chapter V, Theorem 34]). The following

question naturally arises:

For a given element � of S which generates the extension L/K, are there somenecessary and sufficient conditions to be satisfied by f�x� which ensure thatS=R���?

Received January 12, 2009. Communicated by A. Prestel.Address correspondence to Sudesh K. Khanduja, Department of Mathematics, Panjab

University, Chandigarh 160014, India; Fax: +91-172-2541132. E-mail: skhand@pu.ac.in

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ON DEDEKIND CRITERION AND SIMPLE EXTENSIONS 685

Indeed a similar problem was considered by Dedekind who proved thefollowing result known as Dedekind Criterion (see [2, Theorem 6.1.4], [9]).

Dedekind Criterion. Let K = ���� be an algebraic number field with f�x� as theminimal polynomial of the algebraic integer � over � and AK denote the ring ofalgebraic integers of K. Let p be a rational prime and f �x� = g1�x�

e1 � � � gr �x�er be the

factorization of the polynomial f �x� obtained by reducing coefficients of f�x� modulo p,as a product of powers of distinct irreducible polynomials over �/p�, with gi�x� monic.Let M�x� denote the polynomial 1

p�f�x�− g1�x�

e1 � � � gr�x�er � with coefficients from �.

Then p does not divide �AK ����� if and only if for each i, we have either ei = 1 orgi�x� does not divide �M�x�.

If ��p� denotes the localization of � at p� and S is the integral closureof ��p� in K, then using Lagrange’s Theorem and Cauchy’s Theorem for finitegroups, it can be easily seen that the condition p does not divide �AK ����� isequivalent to AK ⊆ ��p���� which is the same as requiring S = ��p����. Keepingthis in view, we extended Dedekind Criterion to discrete valuation rings in 2006(cf. [8]). In this article, we generalize this criterion to general valuation rings of Krullvaluations by reducing the problem to henselian valuation rings. In 2006, Ershov [3]also generalized Dedekind Criterion to arbitrary valuation rings. But our proof iscompletely different and more or less self-contained. Moreover, we also give somenecessary and sufficient conditions for the integral closure of a valuation ring R ina finite extension of its quotient field to be a simple extension R��� of R for some �.

Throughout the article, v is a valuation of arbitrary rank of a field K withvaluation ring Rv, value group Gv and residue field Rv/mv. For in the valuationring of a fixed prolongation w of v, will denote its w-residue, i.e., the image of under the canonical homomorphism from the valuation ring of w onto its residuefield. For g�x� belonging to Rv�x�, g�x� will denote the polynomial obtained byreplacing each coefficient of g�x� by its v-residue.

In this article, we prove the following theorem.

Theorem 1.1. Let v be a valuation of arbitrary rank of a field K and Rv�Gv� Rv/mv

be as above. Let L = K��� be a finite extension of K with � in the integral closure S ofRv in L. Let f�x� be the minimal polynomial of � over K and f �x� = g1�x�

e1 � � � gr �x�er

be the factorization of f �x� into a product of powers of distinct irreducible polynomialsover Rv/mv, with gi�x� monic. Then S = Rv��� if and only if for each i, we have eitherei = 1 or ei > 1, Gv has a smallest positive element v��� and gi�x� does not divide �M�x�,where M�x� = 1

��f�x�− g1�x�

e1 � � � gr�x�er �.

Before stating the next result of the article, we introduce some definitions andnotations.

Let �L�w�/�K� v� be a finite extension of valued fields. If Gv ⊆ Gw are thevalue groups of v� w respectively, then the index of ramification �Gw Gv� willbe denoted by e�w/v�. The degree of the extension Rw/mw over the residue fieldRv/mv of v, referred to as the residual degree of w/v, will be denoted by f�w/v�.If w1� w2� � � � � ws are all the prolongations of v to L, then a fundamental inequalityasserts that

∑si=1 e�wi/v�f�wi/v� ≤ �L K� (cf. [5, Theorem 3.3.4]). L is said to satisfy

the fundamental equality with respect to v if �L K� = ∑si=1 e�wi/v�f�wi/v�.

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686 KHANDUJA AND KUMAR

A finite extension L of �K� v� is said to be discretely ramified if Gv has asmallest positive element � and for each prolongation w of v to L with ramificationindex e, one has Gw = Gv + � �

e.

A finite extension L of �K� v� will be called almost separably ramified if for anyprolongation w of v to L with ramification index > 1, the residue field Rw/mw iseither a separable extension of Rv/mv or generated over it by a primitive elementwith a minimal polynomial g (g ∈ Rv�x� monic) such that max w�g�L��� is minimalpositive in Gw.

We say that a finite extension L of �K� v� has enough residues if for anyk distinct prolongations of v to L with same residual degree t, there are at least kdistinct monic irreducible polynomials of degree t over the residue field of �K� v�.

A ring extension S of a ring R is called simple if S = R��� for some � ∈ S.Now we are in a position to state the main result of the article.

Theorem 1.2. Let L be a finite extension of a valued field �K� v� with valuation ringRv, and let S be the integral closure of Rv in L. Then the extension S/Rv is simple ifand only if the following four conditions are satisfied:

(i) The residue field extensions with respect to all prolongations of v to L are simple;(ii) L satisfies the fundamental equality over �K� v�;(iii) L is either unramified or discretely and almost separably ramified over �K� v�; and(iv) L has enough residues over �K� v�.

It may be pointed out that in contrast to Theorem 1.1 we no longer assumethat L/K be simple in the above theorem; this rather occurs as a consequence of theequivalence.

By what has been said in the opening lines of the paragraph precedingTheorem 1.1, the following classical result of Dedekind [11, Theorem 4.34] is animmediate consequence of Theorem 1.2.

Corollary 1.3 (Dedekind). Let K be an algebraic number field with AK the ringof algebraic integers. Let i�K� denote the greatest common divisor of the indices�AK �����, where � runs over all generating elements in AK of K/�. A rational primep divides i�K� if and only if for some natural number t, the number of prime ideals ofAK lying over p with residual degree t, is strictly greater than the number of monicirreducible polynomials of degree t over the field with p elements.

2. SOME PRELIMINARY RESULTS AND REFORMULATIONOF THEOREM 1.1

We shall denote by vx the Gaussian valuation of the field K�x� of rationalfunctions in an indeterminate x which extends v and is defined on K�x� by

vx(∑

i

aixi) = min

i v�ai��� ai ∈ K�

The following lemma is already known. For the sake of completeness, this is provedhere (cf. [7, Lemma 2.1]).

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ON DEDEKIND CRITERION AND SIMPLE EXTENSIONS 687

Lemma 2.A. Let v be valuation of a field K with valuation ring Rv and g�x� belongingto Rv�x� be a monic polynomial of degree m such that g�x� is irreducible over Rv/mv.Let w be a prolongation of v to a finite extension L of K. If h�x� = ∑m−1

i=0 aixi belonging

to K�x� is a polynomial of degree less than m and � in the valuation ring of w is suchthat � is a root of g�x�, then

w�h���� = mini

w�ai�i� = min

iv�ai� = vx�h�x��� (1)

Proof. Clearly, the lemma needs to be proved when m > 1, which implies thatw��� = 0. Suppose to the contrary that (1) does not hold, then the triangle inequalitywould imply

w�h���� > mini

w�ai�i� = v�aj�� �say�

which gives∑m−1

i=0 �ai/aj��i = 0; this is impossible as the minimal polynomial of �

over Rv/mv is g�x� having degree m. Hence the lemma is proved.

We shall denote by �Kh� vh� the henselization of �K� v� and by Rvh its valuationring. The unique prolongation of vh to the algebraic closure of Kh will again bedenoted by vh. Let �� f�x� be as in Theorem 1.1 and pd ≥ 1 be the degree ofinseparability of K���/K. Note that for any K-conjugate �i of �, the degree ofinseparability of the extension Kh��i�/K

h is also pd, because Kh/K is separable.Indeed if g�x� belonging to K�x� is such that g�xp

d� = f�x� and g�x� splits into

irreducible factors over Kh as G1�x� � � � Gs�x�, then each Gi�xpd� is irreducible over

Kh. Consequently, f�x� factors over Kh as a product of distinct monic irreduciblepolynomials. In view of this, the following theorem to be used in the sequel, is animmediate consequence of a well known result (cf. [4, 17.17]). Its proof is omitted.

Theorem 2.B. Let v be a valuation of arbitrary rank of a field K and L = K��� be afinite extension of K with � in the integral closure S of Rv in L. Let f�x� = ∏s

i=1 Fi�x�be the factorization of the minimal polynomial f�x� of � over K into a product ofdistinct monic irreducible polynomials over the henselization Kh of �K� v�. Then thereare exactly s prolongations of v to L. Let �i be a root of Fi�x�; the valuations w1� � � � � ws

of L defined by

wi

(∑j

aj�j)= vh

(∑j

aj�ji

)� aj ∈ K� (2)

are the prolongations of v to L.

Notations. In what follows �� f�x�� S are as in Theorem 2.B, f�x� = ∏si=1 Fi�x� will

denote the factorization of f�x� into distinct monic irreducible polynomials overKh. f �x� = g1�x�

e1 � � � gr �x�er will stand for the factorization of f �x� into powers of

distinct irreducible polynomials over Rv/mv with each gi�x� monic. Whenever r = s,then it will be assumed that �Fi�x� = gi�x�

ei , 1 ≤ i ≤ r. We shall denote by whi the

unique prolongation of the valuation vh of Kh to Kh��i�, where �i is a fixed rootof Fi�x�. This notation is justified because wh

i is a henselization of the valuation wi

defined by (2).

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688 KHANDUJA AND KUMAR

With the above notations, we prove the following lemma.

Lemma 2.1. Suppose that S = Rv���, then the following hold:

(a) r = s;(b) Gv has a smallest positive element in case ei > 1 for some i, 1 ≤ i ≤ r.

Proof. Since Fi�x� is irreducible over Kh, in view of Hensel’s Lemma (cf. [4, 16.7]),

there exists a positive integer di and a monic polynomial �i�x� ∈ Rv�x� with �i�x�irreducible over Rv/mv, such that �Fi�x� = �i�x�

di . So assertion (a) of the lemma isproved once it is shown that �Fi�x� and �Fj�x� are relatively prime polynomials fori �= j. Let w1� � � � � ws be as in Theorem 2.B. It is known that the maximal idealsof S are ℘1� � � � � ℘s, where ℘i = mwi

∩ S (cf. [10, Theorem 5.8.4]). Fix any i, 1≤ i ≤s. By Chinese Remainder Theorem, there exists an element � ∈ S such that �≡ 0�mod ℘i� and � ≡ 1 �mod ℘j�, j �= i. In view of the hypothesis S = Rv���, thereexists H�x�∈Rv�x� such that � = H���. Then

wi�H���� > 0� wj�H���− 1� > 0� j �= i� (3)

Keeping in mind (2), we can rewrite (3) as vh�H��i�� > 0 and vh�H��j�− 1� > 0, i.e.,

�H��i� = 0� �H��j� = 1� j �= i� (4)

As each �l�x� is irreducible over Rv/mv and has �l as a root, it follows from (4) that�i�x� divides �H�x� and �j�x� does not divide �H�x� when j �= i. Therefore, �i�x� �=�j�x�, which proves that �Fi�x� and �Fj�x� are relatively prime and hence assertion (a).

(b) Denote g1�x�e1 � � � gr�x�

er by G�x� and write f�x� = G�x�+ �M�x�, � ∈ mv,M�x� ∈ Rv�x�, �M�x� �= 0. In view of assertion (a), it may be assumed that �Fi�x� =gi�x�

ei , 1 ≤ i ≤ r. Suppose that some ei > 1, say e1 > 1. Claim is that v��� is thesmallest positive element of Gv. It will be first shown that for each positive � in Gv,we have

e1 − 1e1

v��� < �� (5)

Assume to the contrary that there exists a positive element � in Gv such that

e1� ≤ �e1 − 1�v���� (6)

Choose a ∈ mv with v�a� = �. Using (6), it will be shown that

= g1���e1−1g2���

e2 � � � gr���er

a∈ S� (7)

Since G�x�/g1�x� is a monic polynomial of degree less than deg f�x�, doesnot belong to Rv��� and thus (7) will contradict the hypothesis S = Rv��� therebyproving (5). In view of the fact S = ⋂r

i=1 Rwi(see [5, Corollary 3.1.4]), it is to be

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ON DEDEKIND CRITERION AND SIMPLE EXTENSIONS 689

verified that wi�� ≥ 0� 1 ≤ i ≤ r. Note that for i ≥ 2, g1�x� does not divide �Fi�x�;therefore, wi�g1���� = 0 and consequently,

wi�� = wi�G����− v�a�� i ≥ 2� (8)

Keeping in mind the fact 0 = f��� = G���+ �M��� and (6), we see that

wi�G���� = wi��M���� ≥ wi��� = v��� ≥ � = v�a�� (9)

It follows from (8) and (9) that wi�� ≥ 0 for i ≥ 2. Arguing as above, it can beeasily seen that

w1�� = �e1 − 1�w1�g1����− v�a� (10)

and e1w1�g1���� = w1�G���� = w1��M���� ≥ v���; consequently, w1�g1���� ≥ v���

e1;

this inequality together with (6) gives

w1�g1���� ≥v���

e1≥ �

e1 − 1� (11)

It follows from (10) and (11) that w1�� ≥ 0, which proves (7) and hence (5).

We now deduce from (5) that v��� is the smallest positive element of Gv.Suppose to the contrary that there exists an element � in Gv such that 0 < � < v����set �′ = � if � ≤ 1

2v��� and �′ = v���− � if � > 12v���. Since e1 > 1, we have 0 <

�′ ≤ 12v��� ≤ e1−1

e1v���. This contradicts (5) and hence assertion (b) of the lemma is

proved.

Lemma 2.2. If r = s and Rwhi= Rvh��i� for 1 ≤ i ≤ r, then S = Rv���.

Proof. Let n denote the degree of f�x�. Suppose to the contrary that there existsan element

∑n−1i=0 ai�

i = H��� of S such that H��� does not belong to Rv���. Thenmin0≤i≤n−1 v�ai��, say v�a�, is negative. Define T�x� belonging to Rv�x� by T�x� =H�x�

a. Since degree of �T�x� is less than n, there exists i such that gi�x�

ei does not divide�T�x�, say g1�x�

e1 does not divide �T�x�. By division algorithm, write

H�x� = q�x�F1�x�+ r�x�� deg r�x� < degF1�x�� (12)

putting x = �1 in (12), we obtain H��1� = r��1�. Recall that H��� belongs toS = ⋂r

i=1 Rwi; in particular, w1�H���� = vh�H��1�� ≥ 0. So r��1� = H��1� ∈ Rwh

1. By

hypothesis, Rwh1= Rvh��1�; therefore, r�x� ∈ Rvh�x�; keeping in mind that v�a� < 0,

it now follows from (12) that �F1�x� = g1�x�e1 divides �T�x� which is not so. This

contradiction proves that S = Rv���. �

Lemma 2.3. If S = Rv���, then Rwhi= Rvh��i� for 1 ≤ i ≤ s.

Proof. By Lemma 2.1, r = s. Recall that �i is a root of Fi�x� whose degreewill be denoted by ni. If ei = 1, then the equality Rwh

i= Rvh��i� holds in view

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690 KHANDUJA AND KUMAR

of Theorem 2.B and Lemma 2.A. Suppose that ei > 1 for some i, say e1 > 1.Then by Lemma 2.1, Gv has a smallest positive element, say v���. Suppose to thecontrary that Rwh

i�= Rvh��i� for some i. So there exists an element = ∑ni−1

j=0 �j�ji

in Rwhi, �j ∈ Kh, such that does not belong to Rvh��i�. Choose � ∈ K with

v��� = minj vh��j�� < 0. Then

whi �/�� ≥ −v��� ≥ v���� (13)

Choose aj ∈ K, 0 ≤ j < ni, with vh��j�− aj� > 0 and hence vh�

�j�− aj� ≥ v���. Set

A�x� = ∑ni−1j=0 ajx

j . Using triangle law, it is clear that whi �

�− A��i�� ≥ v���. So in

view of (13), we conclude that

whi �A��i�� ≥ v���� i.e.,

A���

�∈ Rwi

�

It is known that Rwiis a localization of S at a maximal ideal of S (see

[10, Theorem 5.8.4]). Therefore,

A���

�= B���

C���(14)

for some B�x�� C�x� ∈ Rv�x� of degree less than n and C��� not belonging to mwi.

Since degA�x� < ni and �A�x� is a non-zero polynomial by choice, gi�x�ei does not

divide �A�x�. As wi�C���� = 0, gi�x� does not divide �C�x�. Consequently,

gi�x�ei does not divide �A�x��C�x�� (15)

By division algorithm, write

A�x�C�x� = f�x�q�x�+ r�x�� (16)

It follows on using (14) and (16) that r��� = A���C��� = �B��� and hence r�x� =�B�x�. In particular, vx�r�x�� > 0, which in view of (16) shows that f �x� divides�A�x��B�x�. This contradicts (15) and proves the lemma.

The following theorem follows immediately from Lemmas 2.1–2.3.

Theorem 2.4. Let the notations be as in the paragraph preceding Lemma 2.1. ThenS = Rv��� if and only if r = s and Rwh

i= Rvh��i� for 1 ≤ i ≤ r.

3. REDUCTION OF THEOREM 1.1 TO HENSELIAN CASE

In this section, assuming that Theorem 1.1 holds for henselian base fields, weprove it in general. Retaining the notations of the second section, we first prove twosimple lemmas.

Lemma 3.1. If r < s and Gv has a smallest positive element v���, then there exists isuch that ei > 1 and gi�x� divides �M�x�, where M�x� is as in Theorem 1.1.

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ON DEDEKIND CRITERION AND SIMPLE EXTENSIONS 691

Proof. Recall that f�x� = ∏si=1 Fi�x� is the factorization of f�x� into distinct monic

irreducible polynomials over Kh. Since r < s, there exists an index k such that

gk�x� divides �Ft�x� and gk�x� divides �Fu�x� for some t �= u� (17)

So ek > 1. For each i, there exists �i�x� ∈ g1�x�� � � � � gr�x�� and Hi�x� ∈ Rvh�x� suchthat

Fi�x� = �i�x�di + �Hi�x�� di ≥ 1� (18)

On multiplying, we see that there exists �1�x� ∈ Rvh�x� such that

f�x� = F1�x� � � � Fs�x� =s∏

i=1

�i�x�di + �

s∑i=1

( s∏j=1�j �=i

�j�x�dj

)Hi�x�+ �2�1�x�

=r∏

i=1

gi�x�ei + �

s∑i=1

( s∏j=1�j �=i

�j�x�dj

)Hi�x�+ �2�1�x�

and hence M�x� = ∑si=1

(∏sj=1�j �=i �j�x�

dj)Hi�x�+ ��1�x� is divisible modulo � by

gk�x� in view of (17) and (18).

Lemma 3.2. Suppose that r = s and v��� is the smallest positive element of Gv. IfFi�x� = gi�x�

ei + �Mi�x��Mi�x� ∈ Rvh�x�, then for any i, gi�x� divides �Mi�x� if and onlyif gi�x� divides �M�x�, where M�x� = 1

��f�x�− g1�x�

e1 � � � gr�x�er �.

Proof. On multiplying Fi�x�, we see that there exist ��x� ∈ Rvh�x� such that

M�x� =r∑

i=1

( r∏j=1�j �=i

gj�x�ej

)Mi�x�+ ���x��

The lemma follows immediately from the above equation.For proving Theorem 1.1, suppose first that S = Rv��� and ek > 1 for some k.

Then by Lemma 2.1, r = s and Gv has a smallest positive element say v���. It is to beshown that gk�x� does not divide �M�x� when ek > 1. In view of the assumption thatTheorem 1.1 holds for the extension Kh��k�/K

h and ek > 1, gk�x� does not divide�Mk�x� and hence by Lemma 3.2, gk�x� does not divide �M�x�.

For the converse, observe that if ei = 1 for all i, then by virtue of Lemma 3.1,r = s and by the henselian case of Theorem 1.1, Rwh

i= Rvh��i�, 1 ≤ i ≤ r, which gives

S = Rv��� on applying Theorem 2.4.It only remains to be shown that S = Rv��� when Gv has a smallest positive

element v��� and gi�x� does not divide �M�x� for those i for which ei > 1. Note thatthe above hypothesis in view of Lemma 3.1 implies that r = s. Applying Lemma 3.2,we see that gi�x� does not divide �Mi�x� for those i for which ei > 1. So by thehenselian case of Theorem 1.1, we have Rwh

i= Rvh��i�, 1 ≤ i ≤ r and consequently

by Theorem 2.4, S = Rv���.

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4. PROOF OF THEOREM 1.1 IN THE HENSELIAN CASE

In this section, we prove the following theorem.

Theorem 4.1. Let v be a henselian valuation of arbitrary rank of a field K withvaluation ring Rv, residue field Rv/mv and value group Gv. Let L = K��� be a finiteextension of K with � belonging to the valuation ring Rw of the unique prolongation wof v to L. Let f�x� be the minimal polynomial of � over K and g�x� ∈ Rv�x� be a monicpolynomial with g�x� irreducible over Rv/mv and f �x� = g�x�e. Let f�x� = g�x�e +fe−1�x�g�x�

e−1 + · · · + f0�x� be the g�x�-expansion of f�x�, with deg fi�x� < deg g�x�for each i. Then the following statements are equivalent:

(i) Rw = Rv���;(ii) Either e = 1 or vx�f0�x�� is the least positive element of Gv;(iii) If e > 1, then Gv has a smallest positive element v��� and g�x� does not divide

�M�x�, where M�x� = f�x�−g�x�e

�.

Proof. (i) ⇒ (ii). Assume that Rw = Rv��� and e > 1. As shown in the lastparagraph of the proof of Lemma 2.1, for proving (ii) it is enough to prove that

e− 1e

vx�f0�x�� < � (19)

for every positive � in Gv. (19) will be established by first showing that

�e− 1�w�g���� < � (20)

for every positive � in Gv and then proving that

w�g���� = vx�f0�x��

e� (21)

Suppose to the contrary that (20) does not hold. So there exists a ∈ K with v�a� > 0such that w�g���e−1� ≥ v�a�. Then g���e−1

abelongs to Rw, but

g���e−1

adoes not belong to

Rv���, which is a contradiction to the assumption Rw = Rv���. Thus (20) is proved.Putting x = � in the g�x�-expansion of f�x�, we obtain

g���e = −�fe−1���g���e−1 + · · · + f0����� (22)

Note that w�g���ifi���� �= w�g���jfj���� when 0 ≤ i < j ≤ e− 1 and fi���fj��� �= 0,for otherwise by virtue of Lemma 2.A, we would have

�j − i�w�g���� = w�fi���/fj���� = vx�fi�x�/fj�x�� = � �say�

for some indices 0 ≤ i < j ≤ e− 1; this is impossible in view of (20). The aboveargument also shows that w�g���e� �= w�fi���g���

i� for 1 ≤ i ≤ e− 1. It now followsfrom (22) on applying the strong triangle law that

w�g���e� = min0≤i≤e−1

w�g���ifi���� = w�f0�����

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ON DEDEKIND CRITERION AND SIMPLE EXTENSIONS 693

The above equality together with Lemma 2.A proves (21) and completes the proofof (i) ⇒ (ii).

(ii) ⇒ (i) Set � = vx�f0�x��

e. By hypothesis

�e− 1�� < � (23)

for every positive � in Gv. Using (23) and arguing exactly as for the proof of (21),we see that

w�g���� = �� (24)

To prove that Rw = Rv���, let G��� be an element of Rw, where G�x� ∈ K�x�,degG�x� < deg f�x�. We have to show thatG�x� ∈ Rv�x�. LetG�x� = ∑e−1

i=0 Gi�x�g�x�i

be the g�x�-expansion ofG�x�. Using the triangle inequality, (24) and Lemma 2.A, wesee that

w�G���� ≥ mini w�Gi����+ i�� = min

i vx�Gi�x��+ i���

in fact the above inequality cannot be strict, for otherwise vx�Gi�x��+ i� =vx�Gj�x��+ j� for some i < j ≤ e− 1, which would imply that

0 < �j − i�� = vx�Gi�x��− vx�Gj�x�� ∈ Gv

contrary to (23). So we conclude that

vx�Gi�x��+ i� ≥ 0� 0 ≤ i ≤ e− 1� (25)

It follows from (25) that Gi�x� ∈ Rv�x� for each i, for if there exist j suchthat vx�Gj�x�� = −v�b� < 0, b ∈ K, then (25) would imply that j� ≥ v�b� > 0,contradicting (23). This proves that G�x� ∈ Rv�x� as desired.

To prove the equivalence of (ii) and (iii), write f0�x� =∑

cixi, vx�f0�x�� = v�c�,

c ∈ K. Suppose that e > 1 and let v��� denote the least positive element of Gv.Clearly, �M�x�, when divided by g�x�, leaves the remainder

∑�ci/��x

i, which is anonzero polynomial if and only if v�c� = v���. So g�x� does not divide �M�x� if andonly if vx�f0�x�� = v���. This completes the proof of the theorem. �

5. PROOF OF THEOREM 1.2

Assume first that S/Rv is simple, say S = Rv���. Then L = K��� is simple overK, and we are in the setting of Theorem 1.1. Let f�x�, gi�x�, Fi�x� and wi beas in the paragraph preceding Lemma 2.1. In view of the fundamental inequality[5, Theorem 3.3.4] and the fact that the wi-residue of � is a root of gi�x� and hencef�wi/v� ≥ deg gi�x�, conditions (i) and (ii) of the theorem are immediately provedonce we show that for 1 ≤ i ≤ r,

e�wi/v� ≥ ei� (26)

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694 KHANDUJA AND KUMAR

For simplicity of notation, we verify (26) for i = 1. Clearly, it needs to be verifiedonly when e1 > 1. Then Gv has a smallest positive element, say v��� and g1�x� ��M�x� by Theorem 1.1, where

f�x� = g1�x�e1 � � � gr�x�

er + �M�x�� (27)

So w1�M���� = 0. Also w1�gi���� = 0 for i ≥ 2. Substituting x = � in (27), it followsthat

w1�g1���� =v���

e1∈ Gw1

(28)

which proves (26) and hence conditions (i) and (ii). Since the gi�x� are distinctpolynomials, this also proves condition (iv).

Keeping in mind Theorem 1.1 and (28), it is clear that L is either unramifiedor discretely ramified over �K� v�. For condition (iii), it remains to check that Lis almost separably ramified over �K� v�. It suffices to check this in the henseliancase, i.e., in the setting of Theorem 4.1 when e > 1. Assume that the residue fieldextension is not separable, i.e., the polynomial g is not separable.

Let L′ denote the normal hull of L/K and w′ the unique prolongation of wto L′. Let KT denote the inertia field of w′ over K. Then as is well known, theinertia field of w′ over L is the composite field KT �L (see [4, 19.10]). If wT denotesthe restriction of w′ to KT �L, then by ramification theory the value groups of wand wT are the same (cf. [4, 19.12]). Keeping in mind Theorem 4.1(ii), (21) and theequality e�w/v� = e, we see that w�g���� is the minimum positive element of Gw. Forcondition (iii) it only remains to verify that for any � in L, with w�g���� > 0, wehave

w�g���� ≤ w�g����� (29)

If w�g���� > 0, then there exist a K-conjugate �1 of � in KT �L such that w′�� − �1� >0. Denote � − �1 as � and write the Taylor expansion

g��� = g��1�+ �g′��1�+�2

2! g′′��1�+ � � � (30)

Since g�x� is not a separable polynomial and �K� v� is henselian,

w′�g′��1�� = w′�g′���� > 0�

Keeping in mind that w�g���� is the minimum positive element of w′�KT �L�, wesee that w′�g′��1�� ≥ w�g���� and w′��� ≥ w�g����. It now follows from (30) and thestrong triangle law that w�g���� = w�g���� proving (29) and hence condition (iii).

For the converse, assume conditions (i)–(iv). Let w1� � � � � ws be all theprolongations of v to L. By (i) and (iv), we can choose g1� � � � � gs belonging to Rv�x�monic such that g1� � � � � gs are distinct minimal polynomials of primitive elements�1� � � � � �s (�i ∈ Rwi

) for the residue field extensions, where in the ramified case wechoose �i such that wi�gi��i�� is minimal positive in Gwi

: by condition (iii), this

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ON DEDEKIND CRITERION AND SIMPLE EXTENSIONS 695

is possible, if the residue extension (i.e., gi) is not separable, and in the separablecase, if wi�gi��i�� is not minimal positive, replace �i by �i + �i, where wi��i� isminimal positive in Gwi

then using Taylor’s expansion, one has wi�gi��i + �i�� =wi��i�.

By the weak approximation theorem [5, Theorem 3.2.7], we find � ∈ S suchthat �− �i ∈ mwi

for each i, and in the ramified case, wi��− �i� > wi��i� with �i asabove.

Then v has s distinct prolongations to K��� with the same residue fieldand value group as the extensions wi. By condition (ii), this implies thatL=K���. Moreover, the minimal polynomial f�x� of � over K satisfies f �x� =g1�x�

e1 � � � gs�x�es and by the choice of the gi, r = s. So, by Theorems 2.4 and 4.1,

S = Rv���.The following examples are applications of Theorems 1.1 and 1.2.

Example 5.1. Let v be a valuation of arbitrary rank of a field K whose valuegroup has a smallest positive element say �0. Let g�x� ∈ Rv�x� be a monic polynomialsuch that g�x� is an irreducible and separable1 polynomial over Rv/mv. Let e beany positive integer and A0�x�� � � � � Ae�x� be polynomials over Rv each of degree lessthan deg g�x� such that

vx�Ai�x�� > 0� 0 ≤ i ≤ e− 1� vx�A0�x�� = �0�

Then by Generalized Schönemann Irreducibility Criterion (cf.[6, Theorem 1.1]), the polynomial

f�x� = g�x�e + Ae−1�x�g�x�e−1 + · · · + A0�x��

is irreducible over the henselization Kh of �K� v�. Let � be a root of f�x�. So v hasa unique prolongation, say w to K��� and by Theorem 1.1, Rw = Rv���.

Example 5.2. Let K = Fp�x���t�� with the t-adic valuation v, let L1 =K�p√x + p

√t�,

and let L2 = K� p√x� p

√t�. Then L1 and L2 have same residue field and value group

extension (namely, Fp�p√x� and 1

p�), satisfy conditions (i), (ii), and (iv), and are

discretely ramified, but L1 is almost separably ramified whereas L2 is not. So, byTheorem 1.2, the integral closure S1 of Rv in L1 is simple: S1 = Rv�

p√x + p

√t�, whereas

the integral closure of Rv in L2 is not.

ACKNOWLEDGMENTS

The authors are highly thankful to Dr. Peter Roquette, Emeritus ProfessorUniversität Heidelberg for several helpful suggestions and to the referee of thearticle. In fact it was the referee who suggested Theorem 1.2 along with Example 5.2.The financial support by National Board for Higher Mathematics, Mumbai isgratefully acknowledged.

1As shown in [1], the condition g�x� is a separable polynomial is not necessary.

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