Monomial discrete valuations in k[[X]]

arXiv:math/0307238v1 [math.AC] 17 Jul 2003 Monomial discrete valuations in k[[X]] Francisco J. Herrera Govantes Miguel ´ Angel Olalla Acosta Jos´e Luis Vicente C´ordoba July, 2003 Abstract [2, 3] prove that all the rank one discrete valuations of k((X1, X2)), centered in the ring k[[X1, X2]], come from the usual order function, i.e. there exists a finite number of transformations such that we obtain a new field k((Y1, Y2)) where the lifting of vis a monomial valuation given by v(Y1) = v(Y2) = 1. In this work we generalize this result to the rank mdiscrete valuation of K=k((X)), centered in R=k[[X]]. We prove that, if the dimension of vis n−m, the maximum since [1], then there exists an inmediate extension Lof Kwhere the valuation is monomial. Therefore we compute explicitly the residue field of the valuation. 1 Preliminaries. Remark 1.1.In this remark we remember some definitions and results given by I. Kaplansky in [4]. Let Kbe a valued field, let vbe the valuation.Let Lbe an extension of Kand v′a valuation that extends v. We say that the extension K⊆Lis immediate if the values group and the residue field of vand v′are the same. We say that a valued field Kis maximal if it doesn’t admit proper immediate extensions. A well ordered set {ai} ⊂ K, without last element in K, is called pseudo-convergent if v(aj−ai)< v(ak−aj) for all i < j < k. Easily we have that, if {ai}is pseudo-convergent, then v(aj−ai) = v(ai+1 −ai),∀i < j. So we can use the abbreviation ωifor v(aj−ai), with j > i. Let us note that {ωi}is an increasing set of elements of Γ. An element a∈Kis called limit of the pseudo-convergent set {ai}if v(a−ai) = ωifor all i. [4] prove the following results: •If K⊂Lis an immediate extension, then every element of L\Kis a limit of any pseudo-convergent subset of Kwithout limit in K. •A valued field Kis maximal if and only if there exists a limit for all its pseudoconvergent subsets. 1 In [5], Krull shown the existence of, at least, one maximal immediate extension for all valued field. Therefore, if b Kis the completion of Kby the valuation vand bvis the unique valuation of b Kthat extends v, then K⊂b Kis an immediate extension and b Kis a maximal valued field. Kaplansky shown the unicity of the maximal immediate extension of a valued field K that satisfy a certain condition, that he calls hypothesis A. In the case of zero characteristic this hypothesis A is empty. Finally, [4] gives an useful structure theorem for all maximal valued field. If ∆ is a field and Γ an ordered abelian group, the set of all formal series Xaitαicon ai∈∆, αi∈Γ y {αi}well ordered, is a field with the usual sum and times. We shall denote this field by ∆(tΓ). In ∆(tΓ) we can consider the valuation νtgiven by νt X i≥1 aitαi =α1with a16= 0. Krull shown in [5] that, with this valuation, ∆(tΓ) is a maximal field. Theorem 1.2. (Kaplansky, 1942) Let Kbe a maximal valued field with values group Γ and residue field ∆that satisfies the hypothesis A. Then Kis analytically isomorphic to the power series field ∆(tΓ). 1.1 Notation and definitions. Let K=k((X)) = k((X1,...,Xn)) be the quotient field of the formal power series ring R=k[[X]] = k[[X1,...,Xn]]. Remark 1.3.1) We shall write all series f∈Ras f=PA∈Zn 0fAXA, where, if A= (a1,...,an), then XAmeans Xa1 1···Xan n. We shall say E(f) = {A∈Zn 0|fA6= 0}. 2) In Zmwe’ll consider the lexicographic order, it’ll be denoted by ≤lex. This is a total order for the group structure. 3) Let 0 < m ≤nbe an integer and L={B1,...,Bn} ⊂ Zm 0\ {0} such that Lis a generator system of Zm. Each monomial XAof Rhas an element of Zm 0associated, that is called its L-degree, that is degreeL(XA) = n X i=1 aiBi, A = (a1,...,an). 2 Definition 1.4. Let 0 < m ≤nbe an integer and L={B1,...,Bn} ⊂ Zm 0\ {0} such that Lis a generator system of Zm. Let v:R→Zm∪ {∞} be the function such that v(0) = ∞and v(f) = min ≤lex degreeL(XA)|A∈ E(f), with f6= 0. The extension of vto K|k, which values group is Zm, is a rank mdiscrete valuation, called monomial valuation associated to L. Throughout this work, let vbe a rank mdiscrete valuation of K|kcentered in R. Let Rv,mvand Γ = Zmbe the ring, the maximal ideal and the values group of the valuation v, respectively. We’ll denote by ∆v=Rv/mvto the residue field of v. Since [1] we know that the dimension of v(the transcendence degree of k⊂∆v) is lesser or equal than n−m. We shall suppose that the dimension of vis the maximum, n−m. So in this work valuation means rank mdiscrete valuation of K|kcentered in Rand dimension n−m. Remark 1.5.Let b Kbe a maximal immediate extension of K. Since [4] we can suppose that b Kis the completion of Kwith respect of v. Let bvthe only extension of vto b K. We know that there exists an analytic isomorphism of b Kin ∆v(tΓ), so its restriction to the ring R gives an injective homomorphism ϕ:R=k[[X]] →∆v(tΓ) Xi7→ Pj≥1ai,jtαi,j with ai,j ∈∆vy{αi,j} ⊂ Γ well ordered. If we consider the extension of ϕto the quotient field Kand the valuation νtof ∆v(tΓ) previously defined, then v=νt◦ϕ. The purpose of this work is to construct ϕexplicitly, in order to obtain a parametric equation of vand, in consequence, a construction of the residue field of v, as an extension of the field k. Therefore we’ll prove that for all valuation vof K|k, there exists an immediate extension K⊂L=k((Y)) such that the valuation that extends vis monomial. In other words: Any valuation vcomes from a monomial valuation. This result generalizes the obtained in [2, 3] for rank one discrete valuations of k((X1, X2)). 1.2 Monoidal Transformation and immediate extension. Let vbe a valuation of K|k. Let us consider the next monoidal transformation in K: k((X)) →L=k((Y)) Xi7→ Yiif i6= 2 X27→ Y2Y1 with v(X2)>lex v(X1). Then we have the following theorem. 3 Theorem 1.6. With these conditions, the extension K⊂Lis immediate. Proof. Let us consider the rings R=k[[X]] and S=k[[Y]], and the diagram Rϕ// _  ∆v(tΓ) S ψ 77 o o o o o o o o o o o o o Where ψis the natural extension of ϕto L, i.e. ψ(Y2) = ϕ(X2)/ϕ(X1). If ψis injective, then v′=νt◦ψis a valuation of Lthat extends vand both has the same values group Γ and the same residue field ∆v. So we can suppose, by contradiction, that ψis not injective. Let pbe the implicit ideal ker(ψ) and L′the quotient field of the ring k[[Y]] p. Clearly the restriction of ψto L′is injective and its composition with νtdefine a valuation, let us put w1. So the extension K⊂L′is immediate. Let w2be the valuation p–´adic of L. This is a discrete rank one valuation. The composition of both valuations, let us put w, is a discrete rank m+ 1 valuation of L|kwhose residue field is ∆v. Then we have rank(w) + dim(w) = n+ 1 >dim k[[Y]]. Since Abhyankar’s theorem ([1], Theorem 1, p. 330), we know rank(w) + dim(w)≤dim k[[Y]], so there is a contradiction. Remark 1.7.Here we are used the existence of such injective homomorphism, proved by Kaplansky [4]. Later we’ll give an explicit construction of ψ. There isn’t circular reasoning. The following example shows that the condition of maximal dimension of vis necessary. Example 1.8. Let R=C[[X1, X2, X3]] and Kits quotient field. Let us consider the injective homomorphism ϕ:R→C(u)[[t]] X17→ t X27→ ut2 X37→ eut −1. The composition of ϕ(really its extension to the quotient fields) with the usual order function νtof C(u)((t)) is a rank one discrete valuation of K, named v. The dimension of vis 1, the transcendence degree of the extension C⊂C(u). 4 If we make the monoidal transformation R=C[[X1, X2, X3]] →S=C[[Y1, Y2, Y3]] Xi7→ Yiif i= 1,3 X27→ Y2Y1 Then we have an homomorfism ϕ:S→C(u)[[t]] X17→ t X27→ ut X37→ eut −1. that is not injective. 2 Constructing ∆v. 2.1 Basis of a subgroup of Zm. There are well known procedures to compute a basis of a subgroup Γ0⊂Zmknowing any set of generators. In this subsection we describe an algorithm that will be very useful in order to prepare the valuation. Let {A1,...,An}be the set of generators of Γ0⊂Zm, with Ai>lex 0for all i. We can suppose, without lost of generality, that Ai≤lex Aj∀i < j. Let A= (ai,j)∈ Mn×nbe the matrix whose rows are the elements Ai. We shall consider two transformations with the rows of the matrix A: (1) Fi,j(q): To change the row iby itself plus qtimes the row j, with q∈Z. (2) To interchange rows. Clearly the group generated by the rows of the matrix Ais equal to the group generated by the rows of any transformation of A. Remark 2.1.Algorithm. The the matrix Ahas the following echelon form 0 Let jbe the first column different of 0in A, let ibe the first such that ai,j 6= 0. Then we shall say that ai,j is a pivot. How Ai≤lex Al∀i < l, clearly ai,j ≤al,j ∀i < l. Let us put al,j =qlai,j +rl, by making the integer Euclidean division of al,j by ai,j. We’ll apply the following procedure with the first step, that is perfectly exportable to the other steps: 5 I) For each row l, with i < l ≤n, we make the following transformation: a) If rl6= 0 we do Fl,i(−ql). The new row l, that we denote by Alagain, is such that Al<lex Aiand 0 < al,j < ai,j. b) If rl= 0 there are two possible situations: 1) Al−qlAi>lex 0: In this case we make too the transformation Fl,i(−ql). The new row lis such that Al<lex Aiand al,j = 0. So, after a reordering Al raises a step. 2) Al−qlAi≤lex 0: Then we make the change Fl,i(1 −ql). The new row l, that we denote Alagain, is such that Al≤lex Ai and ml,j =mi,j. Let us remark that if ql= 1 then Al=Ai, because we have suppose at begin that Ai≤lex Al. II) After these transformations we reorder the rows. If every row of the first step of the matrix are equal, then we raise a step and we begin with the procedure.In other case, we apply again this algorithm. Clearly we are doing the Euclidean algorithm in order to compute the maximal common divisor of the elements {ai,j,...,an,j }. Let pjbe the maximal common divisor. By a finite number of transformations we obtain a transformed matrix, that we denote again by A, such that the pivot is equal to pjand the last one divides all al,j with l≥i. We have just arrived to situation I.b), so, by a finite number of transformations, necessarily we must obtain a new matrix Awhere all the rows down the pivot ai,j =pjare equals to Ai. Hence, by applying this algorithm for each step of the matrix, we obtain a matrix Bwith only con sdifferent rows Bi1,...,Biswhose pivots are pj1,...,pjs. Clearly Γ0is isomorphic to pj1Z× · · · × pjsZand {Bi1,...,Bis}is a basis of Γ0. The algorithm described in this section allows us to prove the following lemma, that we are going to use frequently for preparing our valuation conveniently. Lemma 2.2. With the usual conditions over K=k((X)) and v, rank mdiscrete valuation, if Γ0is the subgroup generated by the values of the elements Xi, the we can find, by a finite number of monoidal transformations and interchanges of variables, an immediate extension L=k((Y)) of Ksuch that each Yihas the value in a basis {B1,...,Bs}of Γ0. Proof. We have to apply the precedent algorithm to the matrix of the values v(Xi). Where Fl,i(−ql) means “to apply qlmonoidal transformations such that Xl7→ YlYi”, and reorder rows means “to reorder variables according to its values”. 2.2 Preparing v. As usually, we begin with a rank m≤ndiscrete valuation vof K|k, centered in the ring R=k[[X]]. Let Γ = Zm,Rvand mvbe the values group, the ring and the maximal ideal of the valuation. v, respectively. We shall denote, as usual, by ∆vto the residue field of the valuation. 6 We’ll suppose that the extension k⊂∆vis transcendent pure of degree dim v=n−m. First, we apply the lemma 2.2 to obtain an immediate extension L=k((Y)) of Ksuch that the set of values of the elements Yiis a basis {B1,...,Bs}of the subgroup Γ0generated by the values of the elements Xi. By convenience we reorder the elements Yiin such way that the first selements takes all the values of the basis (i.e. v(Yi) = Bifor i= 1,...,s). Remark 2.3.Let A∈Γ0be such that A=r1B1+...+rsBs, then we shall denote by RA= (r1,...,rs,0,...,0) to the n–uple such that v(YRA) = A. 2.3 The first transcendental residue. Our purpose is to give an explicit description of the injective homomorphism ψ:k[[Y]] → ∆v(tΓ), such that v=νt◦ψ. In order not to complicate the exposition of this construction, we are supposing that all the residues are in kor they are transcendental over the ground field. This condition seems too strong, but after the proof of theorem 2.8 we’ll explain why this situation is really close to the general one. 1) For the first elements we put ψ(Yi) = tBi i= 1,...,s. 2) Let us take Ys+1 the first element whose value is a linear combination of the elements Bi (in fact it must be equal to some Bi). Let us suppose that v(Ys+1) = Bs+1,1, then we have two possibilities: a) For all α∈k,v(Ys+1 +αYRBs+1,1) = Bs+1,1. This fact means that the residue of Ys+1/YRBs+1,1in ∆vis transcendental over k. Let us put us+1 =Ys+1 YRBs+1,1 +mv and ψ(Ys+1) = us+1tBs+1,1. b) There exists α∈ksuch that v(Ys+1 +αYRBs+1,1)>lex Bs+1,1. So the residue of Ys+1/YRBs+1,1in ∆vis in k. Let us put αs+1,1=αand v(Ys+1 +αs+1,1YRBs+1,1) = Bs+1,2> Bs+1,1. There are two possibilities again: i) The new value Bs+1,2/∈Γ0. In this case, we make the change Zs+1 =Ys+1 + YRBs+1,1,Zi=Yi∀i6=s+ 1 and go back to the beginning of the procedure by preparing the new valuation of k((Z)) with the lemma 2.2. ii) The value Bs+1,2∈Γ0. If there exists αs+1,2such that vYs+1 +αs+1,1YRBs+1,1+αs+1,2YRBs+1,2=Bs+1,3>lex Bs+1,2, then we ask again if Bs+1,3∈Γ0. In the affirmative case, if there exists αs+1,3∈k such that vYs+1 +αs+1,1YRBs+1,1+αs+1,2YRBs+1,2+αs+1,3YRBs+1,3=Bs+1,4>lex Bs+1,3, 7 we go back to the beginning of the procedure. We continue this procedure until we find a value Bs+1,l /∈Γ0. If we cannot, there does not exist αs+1,l ∈ksuch that v Ys+1 + l X k=1 αs+1,kYRBs+1,k !=Bs+1,l+1 > Bs+1,l. In the first case we make the change Zs+1 =Ys+1 + l−1 X k=1 αs+1,kYRBs+1,k , Zi=Xi∀i6=s+ 1 and move to the lemma 2.2. In the second case we have that the residue us+1 =Ys+1 +Pl−1 k=1 αs+1,kYRBs+1,k YRBs+1,l +mv is transcendental over k. In this case we put ψ(Ys+1) = l−1 X k=1 αs+1,ktBs+1,k +us+1tBs+1,l . We have to prove that this procedure ends up finding a new value that it is not in Γ or a transcendental residue. Lemma 2.4. The situation described in 2.b.i) only occurs a finite number of times. Proof. There are two possible situations to find an element of value B /∈Γ0: 1) The value Bis not a rational linear combination of the elements of Γ0. In this case, the rank of the new subgroup of Zmincreases by 1. Trivially this fact only occurs a finite number of times. 2) The new value Bis a non integer rational linear combination of the elements of the basis of Γ0. Let {p1,...,ps}be the pivots that appears in the construction of the basis of Γ0, let {q1,...,qs}be the ones of the new subgroup, Γ1. As Γ0⊂Γ1, then qi≤pi∀iand, at least, one inequality is strict. As the pivotes are greater or equal than 1, this only occurs a finite number of times. Remark 2.5.Let us suppose that we have a pseudo convergent set {fj}of elements of a valued field K, with value group Zmwith lexicographic order. The set of values {ωj}of the elements fjis a strictly increasing sequence of elements of the group Zm. Let us suppose that the set {ωj}is bounded. Then, from an index jsufficiently big, we have ωj= (a1,...,al, al+1,j,...,am,j), in such way that the first lcoordinates of the values ωjare stabilized. Let us suppose that lis the greatest integer between 1 and msuch that this fact occurs. Let fbe a limit of {fj}and ω=v(f) its value. Then, if ω= (b1,...,bn), as ω >lex ωj for all j, then there exists an l0≤lsuch that bi=aifor all i= 1,...,l0−1 and bl0> al0. 8 Lemma 2.6. The procedure described in the situation 2.b.ii) finds a value that is not in Γ0 or a transcendental residue over k. Proof. If the procedure is not finite, then we have a set of elements {fj}j≥1such that fj=Ys+1 + j−1 X l=1 αs+1,lYRBs+1,l . So v(fj+1) = Bs+1,j+1 >lex v(fj) = Bs+1,j for all j. Therefore, if i < j < k, we have v(fj−fi) = Bs+1,i <lex v(fk−fj) = Bs+1,j, so {fj}is a pseudo-convergent set. Let us suppose that the set of values {Bs+1,j}is bounded. Like in the remark 2.5, let us suppose that, since an index jsufficiently great, the first lcoordinates of the values are stabilized, where lis the greatest integer between 1 and msuch that this fact occurs. Let us put v(fj) = (a1,...,al, al+1,j,...,am,j). Again by remark 2.5, we know that any limit of {fj}is such that its value is something like (a1,...,ak, bk+1,...,bm), with k < l and bk+1 > ak+1. Let us take the series Ys+1 + ∞ X j=1 αs+1,jYRBs+1,j . As limit, by convenience we put g1= ∞ X j=1 αs+1,jYRBs+1,j and v(Ys+1 +g1) = B1 s+1,1. If B1 s+1,1/∈Γ0, the we have finished. In other case, if there exists α1 s+1,1such that vYs+1 +g1+α1 s+1,1YRB1 s+1,1=B1 s+1,2>lex B1 s+1,1, then we continue with our procedure. If it is infinite again and the values contained are bounded, then we shall have a pseudo-convergent set with limit Ys+1 +g1+ ∞ X j=1 α1 s+1,jYRB1 s+1,j . 9