Sums of units in function fields II - The extension problem
aa r X i v : . [ m a t h . N T ] N ov Sums of units in function fields II - Theextension problem
Christopher Frei
Abstract
In 2007, Jarden and Narkiewicz raised the following question: Isit true that each algebraic number field has a finite extension L suchthat the ring of integers of L is generated by its units (as a ring)?In this article, we answer the analogous question in the function fieldcase.More precisely, it is shown that for every finite non-empty set S ofplaces of an algebraic function field F | K over a perfect field K , thereexists a finite extension F ′ | F , such that the integral closure of the ringof S -integers of F in F ′ is generated by its units (as a ring). In their paper [7], Jarden and Narkiewicz proved that, for every finitelygenerated integral domain R of characteristic 0 and every positive integer N ,there exists an element of R that can not be written as a sum of at most N units. This also follows from a result obtained by Hajdu [6], and applies inparticular to the case where R is the ring of integers of an algebraic numberfield. The author recently showed an analogous result for the case where R is a ring of S -integers of an algebraic function field of one variable over aperfect field [4].A related question is whether or not a ring R is generated by its units. Ifwe take R to be a ring of integers of an algebraic number or function field, Mathematics Subject Classification : Primary 11R58; Secondary 11R27.
Key words and phrases : function field, sums of units, generated by units . Frei The extension problem S -integers of quadratic function fields [4]. All of these results have in commonthat the unit group of the ring in question is of rank 1. The author is notaware of any general results for rings of integers whose unit groups havehigher rank.Among other problems, Jarden and Narkiewicz asked the following ques-tion, which was later called the extension problem. Problem 1. [7, Problem B] Is it true that each number field has a finiteextension L such that the ring of integers of L is generated by its units. This is of course true for finite abelian extensions of Q , since those arecontained in cyclotomic number fields by the Kronecker-Weber theorem, andthe ring of integers of a cyclotomic number field is generated by a root ofunity. The scope of this paper is an affirmative answer to the function fieldversion of Problem 1. Let us fix some basic notation before we state thetheorem.Regarding function fields, we use the notation from [9] and [10]. In partic-ular, an algebraic function field over a field K is a finitely generated extension F | K of transcendence degree 1. The algebraic closure of K in F is called the(full) constant field of F | K . An element t ∈ F is called a separating elementfor F | K , if the extension F | K ( t ) is finite and separable. Following [10], weregard the places P of F | K as the maximal ideals of discrete valuation rings O P of F containing K . In particular, the places correspond to (surjective)discrete valuations v P : F → Z ∪ {∞} of F over K . Let n be a positiveinteger. We say that a place P of F | K is a zero of an element f ∈ F of order n , if v P ( f ) = n >
0, and P is a pole of f of order n , if v P ( f ) = − n <
0. If S is a finite set of places of F | K then the ring O S of S -integers of F is theset of all elements of F that have no poles outside of S . Moreover, we write K × := K r { } . Theorem 2.
Let K be a perfect field, F | K an algebraic function field over K , and S = ∅ a finite set of places of F | K . Let O S be the ring of S -integersof F . Then there exists a finite extension F ′ | F such that the integral closureof O S in F ′ is generated by its units (as a ring). The basic idea to prove Theorem 2 is the following: First, choose a finite . Frei The extension problem { t, t , . . . , t n } of generators of O S over K . Then, for each 1 ≤ i ≤ n ,iteratively construct a finite extension F i | F such that(I.) t , t , . . . , t i are sums of units in the integral closure of O S in F i , and(II.) the integral closure of O S in F i is generated by units as a ring extensionof O S .Then the integral closure of O S in F n is generated by units and sums of unitsas an extension of K , thus it is generated by its units. Section 2 providesthe tools to construct the extension fields F i . In Section 3, everything is puttogether. The following lemma illustrates the idea explained at the end of the intro-duction.
Lemma 3.
Let K be a perfect field not of characteristic and a ∈ K × . Con-sider the extension of rational function fields K ( x ) | K ( t ) , where t = x + a /x .Then the integral closure of K [ t ] in K ( x ) is K [ x, x − ] , which is generated (asa ring) by its units.The only places of K ( t ) that are ramified in K ( x ) are the zeros of t − a and t + 2 a , both with ramification index .Proof. The minimal polynomial of x over K ( t ) is X − tX + a , whence K ( x ) = K ( t, y ), with y = t − a . (Here we used the assumption that K is not of characteristic 2.) One can verify the assertions about ramificationdirectly or use Proposition III.7.3 from [10].Obviously, x and x − are integral over K [ t ], and K [ t ] ⊆ K [ x, x − ]. Since K [ x, x − ], as a ring of fractions of the principal ideal domain K [ x ], is inte-grally closed, it is the integral closure of K [ t ] in K ( x ). Obviously, x , x − andall elements of K × are units in K [ x, x − ], and the lemma is proved.The main step in the construction of the extension fields F i is carried outin the following proposition, which is the most important component of ourproof of Theorem 2. . Frei The extension problem Proposition 4.
Let K be a perfect field not of characteristic , F | K analgebraic function field with full constant field K , t a separating element of F | K , and O the integral closure of K [ t ] in F . Assume that there is some a ∈ K × such that the zeros of t + 2 a and t − a in K ( t ) are unramified in F | K ( t ) .Let F ′ := F ( x ) , where x is a root of the polynomial f := X − tX + a ,and let O ′ be the integral closure of K [ t ] in F ′ . Then K is the full constantfield of F ′ | K , x is a unit in O ′ , t = x + a /x , and O ′ = O [ x ] .Proof. The roots of f ∈ O [ X ] in F ′ are x and a /x , whence x is a unit in O ′ . Obviously, t = x + a /x . If f is reducible over F then x, a /x ∈ O , andthe proposition holds trivially. Assume now that f is irreducible over F .The field F ′ is the compositum of F and K ( x ). Since the characteristicof K is not 2, the extension F ′ | F , and thus as well F ′ | K ( t ) is separable. ByLemma 3, the only places of K ( t ) that are ramified in K ( x ) are the zeros of t − a and t + 2 a , both with ramification index 2.Let P be a zero of t + 2 a or t − a in F ′ | K . By Abhyankar’s lemma (see,for example, Proposition III.8.9 from [10]), the ramification index of P over K ( t ) is 2. Here, we used the assumption that the zeros of t − a and t + 2 a in K ( t ) are unramified in the extension F | K ( t ). Therefore, the ramificationindex of P over F is 2.Again by Abhyankar’s lemma, every place Q of F ′ | K that is not a zeroof t + 2 a or t − a is unramified over F .Since there are ramified places in the extension F ′ | F , it is not a constantfield extension, so K is the full constant field of F ′ | K .We are left with the task of proving that O ′ = O [ x ]. Denote the differentof O ′ |O by D , and let δ ( x ) be the different of x , that is δ ( x ) = f ′ ( x ) = 2 x − t .It is well known that O ′ = O [ x ] if and only if D is the principal ideal of O ′ generated by δ ( x ) (see, for example, Theorem V.11.29 from [12]).Already knowing all ramification indices in the extension F ′ | F , we seethat the different D of O ′ |O is the product of all prime ideals of O ′ dividing( t + 2 a ) or ( t − a ) (use, for example, Theorem III.2.6 from [8] and theassumption K is not of characteristic 2).Since δ ( x ) = (2 x − t ) = t − a = ( t + 2 a )( t − a ), . Frei The extension problem O ′ generated by δ ( x ) satisfies( δ ( x )) = Y P | ( t ± a ) P = Y P | ( t ± a ) P = D .Here, P ranges over all prime ideals of O ′ dividing ( t + 2 a ) or ( t − a ). Aswe have already seen, the ramification index of each such P over the primeideal ( t + 2 a ) [or ( t − a )] of K [ t ] is 2. By unique ideal factorization, the idealof O ′ generated by δ ( x ) is D .For function fields of characteristic 2, we use a slightly modified form ofProposition 4. Proposition 5.
Let K be a perfect field of characteristic , F | K an algebraicfunction field with full constant field K , t a separating element of F | K , and O the integral closure of K [ t ] in F . Assume that there is some a ∈ K suchthat the zero of t + a in K ( t ) is unramified in F | K ( t ) .Let F ′ := F ( x ) , where x is a root of the polynomial f := X +( t + a ) X +1 ,and let O ′ be the integral closure of K [ t ] in F ′ . Then K is the full constantfield of F ′ | K , x is a unit in O ′ , t = x + 1 /x + a , and O ′ = O [ x ] .Proof. Again, x is a unit in O ′ , since x and 1 /x are the roots of the monicpolynomial f ∈ O [ X ]. Clearly, t = x + 1 /x + a . The proposition holds againtrivially if f is reducible over F . Assume from now on that f is irreducibleover F .Putting y := x/ ( t + a ), we get F ′ = F ( x ) = F ( y ) and y + y = 1 / ( t + a ) .We use Proposition III.7.8 from [10] to prove that the only places of F | K that are ramified in F ′ are the zeros of t + a . Indeed, for each such zero P ,we have v P (cid:0) / ( t + a ) − (1 / ( t + a ) − / ( t + a )) (cid:1) = v P (1 / ( t + a )) = − P is unramified over K ( t ). For each place Q of F | K that is not a zeroof t + a , we have v Q (1 / ( t + a ) ) ≥ F | K thatare ramified in F ′ are exactly the zeros of t + a , and that the respective . Frei The extension problem K isthe full constant field of F | K and that the different D of O ′ |O is of the form D = Y P | ( t + a ) P .Here, P ranges over all prime ideals of O ′ dividing ( t + a ). On the otherhand, the different of x is δ ( x ) = f ′ ( x ) = t + a , and the ideal of O ′ generatedby t + a is given by ( t + a ) = Y P | ( t + a ) P = D .Note that the ramification index of every prime ideal P of O ′ over the primeideal ( t + a ) of K [ t ] is 2, since ( t + a ) is unramified in F and the ramificationindex of P over F is 2.Therefore, D = ( δ ( x )), which suffices to prove that O ′ = O [ x ].The following lemma shows a way to enlarge O , while still maintainingthe property that O ′ = O [ x ] from the previous propositions. The resultsare probably not new, but the author is not aware of an adequate reference.Recall that, for any place P of an algebraic function field, O P denotes thediscrete valuation ring with maximal ideal P . Lemma 6.
Let F | K be an algebraic function field with perfect constant field K , F ′ | F a finite separable extension, and x ∈ F ′ with F ′ = F ( x ) . Let S ⊆ T be sets of places of F | K , and assume that x is integral over O S . Then wehave:(a) If O P [ x ] is integrally closed for all P / ∈ S then O S [ x ] is integrally closedas well.(b) If O S [ x ] is integrally closed then O T [ x ] is integrally closed as well.(c) If x is algebraic over K then O T [ x ] is integrally closed.Proof. Denote the integral closure of O S in F ′ by O ′ . Clearly, O S [ x ] ⊆ O ′ .To prove (a) , we need to show that O S [ x ] = O ′ . Let S ′ be the set of placesof F ′ | K lying over places in S . We have O ′ = \ P ′ / ∈ S ′ O P ′ = \ P / ∈ S \ P ′ | P O P ′ = \ P / ∈ S ( O P [ x ]). . Frei The extension problem P ′ denotes places of F ′ | K and P denotes places of F | K . The thirdequality follows from the assumption that O P [ x ] is integrally closed and thefact that x is integral over O P , for all P / ∈ S . Therefore, it is sufficient toshow that \ P / ∈ S ( O P [ x ]) = \ P / ∈ S O P ! [ x ].Clearly, the right-hand side of the above equality is included in the left-handside. Now let f be an arbitrary element of T P / ∈ S ( O P [ x ]). Denote the degree[ F ′ : F ] by n . Then, for each P / ∈ S , there is some polynomial g P ∈ O P [ X ]of degree smaller than n , with f = g P ( x ). Since { , x, . . . , x n − } is a basis of F ′ | F , all g P are equal and thus elements of (cid:0)T P / ∈ S O P (cid:1) [ X ]. This shows theother inclusion.To prove (b) , notice that, for all P / ∈ S , O P is the localization of O S atthe unique prime ideal P of O S corresponding to the place P . Therefore, O P [ x ] can be seen as ring of fractions of O S [ x ] with denominators in themultiplicative set O S r P . Assume that O S [ x ] is integrally closed. By theabove argument, O P [ x ] is integrally closed for all P / ∈ S , in particular for all P / ∈ T , so (b) follows from (a) .The special case of (b) with S = ∅ is exactly (c) .As an immediate consequence of Lemma 6 (c) and the primitive elementtheorem, we get that finite constant field extensions have property (II.) fromthe overview presented at the end of Section 1 (see also the third paragraphof Remark 6.1.7 from [5] for a more general formulation): Corollary 7.
Let F | K be an algebraic function field with perfect constantfield K , S a set of places of F | K , and K ′ | K a finite extension. Then theintegral closure of O S in K ′ F is K ′ O S . To use Propositions 4 and 5, we need to ensure that we can always findan a as required. This is accomplished by the following lemma. Lemma 8.
Let F | K be an algebraic function field with perfect constant field K , and t ∈ F r K . Then there is a finite extension K | K and an element a ∈ K × , such that the zeros of t − a and t + a in K ( t ) are unramified in theextension K F | K ( t ) .If F is separable over K ( t ) then K F is separable over K ( t ) . . Frei The extension problem Proof.
The first part of the lemma clearly holds if K is infinite, since thereare only finitely many ramified places in F | K ( t ), so we can put K := K .In the general case, consider the algebraic closure K of K in some alge-braically closed field Φ ⊇ F and the constant field extension KF | K of F | K .Since K is infinite, we find some a ∈ K , such that the zeros of t − a and t + a in K ( t ) are unramified in KF . Put K := K ( a ). Then the zeros of t − a and t + a in K ( t ) are unramified in K F , as desired. Indeed, let P ′ be a placeof KF | K lying over the zero P of, say, t + a in K ( t ). Put P ′ := P ′ ∩ K F and P := P ′ ∩ K ( t ). We know that P ′ | P is unramified. From P ′ | P | P andthe fact that constant field extensions are unramified, it follows that P ′ | P isunramified. Now P ′ | P ′ | P implies that P ′ | P is unramified.The assertion regarding separability holds because if F is separable over K ( t ) then K F is generated over K ( t ) by separable elements. For convenience, let us state the theorem again.
Theorem 2.
Let K be a perfect field, F | K an algebraic function field over K , and S = ∅ a finite set of places of F | K . Let O S be the ring of S -integersof F . Then there exists a finite extension F ′ | F such that the integral closureof O S in F ′ is generated by its units (as a ring). It is enough to prove Theorem 2 under the assumption that K is the fullconstant field of F | K , since then the general case follows as well.Denote the characteristic of K by p ≥
0, and assume first that p = 2. Wefind a separating element t of F | K such that O S is the integral closure of K [ t ] in F . To this end, choose places Q ∈ S and R , R ′ / ∈ S of F | K . By thestrong approximation theorem, we can find an element t ∈ F that satisfiesthe conditions v R ( t ) = 1, v R ′ ( t ) = X P ∈ S r { Q } deg P , v P ( t ) = −
1, for all P ∈ S r { Q } , and v P ( t ) ≥
0, for all places
P / ∈ S ∪ { R, R ′ } . . Frei The extension problem t has degree 0, it follows that v Q ( t ) <
0. There-fore, the poles of t are exactly the elements of S . Moreover, t is not a p -thpower, since p does not divide v R ( t ) = 1. It follows that F is separable over K ( t ) (see, for example, Proposition III.9.2 (d) from [10]) and the integralclosure of K [ t ] in F is exactly O S .Choose some non-constant elements t , . . . , t n of O S , such that O S = K [ t, t , . . . , t n ] (for example, let { t , . . . , t n } be an integral basis of O S over K [ t ] and omit a possible constant).Lemma 8 permits us to find a finite extension K | K and some a ∈ K × ,such that the zeros of t − a and t + 2 a in K ( t ) are unramified in K F . ByCorollary 7, the integral closure of O S in K F is K O S = K [ t, t , . . . , t n ].Proposition 4 yields a finite extension F | K of K F | K , such that t is asum of units in the integral closure O of O S in F , and O = K O S [ x ] = K [ t, t , . . . , t n , x ], for some unit x of O . Moreover, K is the full constantfield of F | K .We inductively construct finite extensions F | K , . . . , F n | K n of F | K with the following properties. If O i denotes the integral closure of O S in F i then we have, for i ∈ { , . . . , n } : • O i = K i [ t, s , . . . , s i , t i +1 , . . . t n , x , x , . . . , x i ], where x , . . . , x i areunits of O i , and for all 1 ≤ j ≤ i there is some m with s p m j = t j . • t , s , . . . , s i are sums of units of O i . • K i is the full constant field of F i | K i .For i = 0, the function field F | K has all desired properties. Let i ∈{ , . . . , n } and assume that we have constructed F i − | K i − . The figure onpage 10 shows the relations between the rings and fields constructed in thefollowing paragraphs.Take the maximal non-negative integer m such that t i is a p m -th power in F i − (the maximum exists since t i is not constant), and let s i be the p m -th rootof t i . (If p = 0, simply put s i := t i .) Then s i ∈ O i − , since s i has the samepoles as t i . Therefore, O i − = K i − [ t, s , . . . , s i , t i +1 , . . . , t n , x , . . . , x i − ].Since s i is not a p -th power in F i − , it is a separating element of F i − | K i − (again, we used Proposition III.9.2 (d) from [10]). By Lemma 8, there is somefinite extension K i | K i − and some a ∈ K × i such that the zeros of s i − a and s i + 2 a in K i ( s i ) are unramified in K i F i − , and K i F i − is separable over K i ( s i ). . Frei The extension problem F i Proposition 4 F i O [ x i ] ❖❖❖❖❖ ⊆ O i ♥♥♥♥♥♥♥ K i F i − K i F i − O ❖❖❖❖❖❖ ⊆ K i O i − ♥♥♥♥♥ K i ( s i ) F i − K i [ s i ] ◆◆◆◆ O i − ♦♦♦♦♦♦ Figure 1: The rings and fields occurring in the induction step.Denote the integral closure of K i [ s i ] in K i F i − by O . By Proposition 4,there is a finite extension F i | K i of K i F i − | K i , such that the integral closure of O in F i is O [ x i ], for some unit x i , and s i is a sum of units in O [ x i ]. Moreover, K i is the full constant field of F i | K i .By our convention, O i is the integral closure of O S in F i , and thus aswell the integral closure of O i − in F i . By Corollary 7, the integral closure of O i − in K i F i − is K i O i − . Since s i ∈ O i − , we have O ⊆ K i O i − . Let U bethe set of poles of s i in K i F i − , and V ⊇ U the set of poles of t in K i F i − .Then O = O U and K i O i − = O V . Since O U [ x i ] = O [ x i ] is integrally closed,Lemma 6 (b) implies that O V [ x i ] = K i O i − [ x i ] is integrally closed as well.Therefore, K i O i − [ x i ] is O i , the integral closure of O i − in F i . We concludethat O i = K i [ t, s , . . . , s i , t i +1 , . . . , t n , x , . . . , x i ],as desired. The elements x , . . . , x i − are units in O i , because they are unitsin O i − ⊆ O i . Moreover, x i is a unit in O i , since it is a unit in O [ x i ] ⊆ O i .Therefore, t , s , . . . , s i are sums of units of O i , and the induction is complete.Now put F ′ | K ′ := F n | K n , and Theorem 2 is proved whenever the char-acteristic of K is not 2. In characteristic 2, the proof is exactly the sameas above, except that we always write a instead of 2 a and use Proposition 5instead of Proposition 4. Acknowledgements
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Glas. Mat. , 43(63)(2):293–307, 2008.Christopher FreiTechnische Universit¨at GrazInstitut f¨ur Analysis und Computational Number TheorySteyrergasse 30, 8010 Graz, AustriaE-mail: [email protected], 43(63)(2):293–307, 2008.Christopher FreiTechnische Universit¨at GrazInstitut f¨ur Analysis und Computational Number TheorySteyrergasse 30, 8010 Graz, AustriaE-mail: [email protected]