a Xi :ma h/0307238 1 [ma h.AC] 17 Jul 2003
Monomial disc e e alua ions in k[[X]]
F ancisco J. He e a Go an es Miguel ´
Angel Olalla Acos a
Jos´e Luis Vicen e C´o doba
July, 2003
Abs ac
[2, 3] p o e ha all he ank one disc e e alua ions o k((X1, X2)), cen e ed in he
ing k[[X1, X2]], come om he usual o de unc ion, i.e. he e exis s a ini e numbe o
ans o ma ions such ha we ob ain a new ield k((Y1, Y2)) whe e he li ing o is a
monomial alua ion gi en by (Y1) = (Y2) = 1.
In his wo k we gene alize his esul o he ank mdisc e e alua ion o K=k((X)),
cen e ed in R=k[[X]]. We p o e ha , i he dimension o is n−m, he maximum since
[1], hen he e exis s an inmedia e ex ension Lo Kwhe e he alua ion is monomial.
The e o e we compu e explici ly he esidue ield o he alua ion.
1 P elimina ies.
Rema k 1.1.In his ema k we emembe some de ini ions and esul s gi en by I. Kaplansky
in [4].
Le Kbe a alued ield, le be he alua ion.Le Lbe an ex ension o Kand ′a
alua ion ha ex ends . We say ha he ex ension K⊆Lis immedia e i he alues
g oup and he esidue ield o and ′a e he same. We say ha a alued ield Kis
maximal i i doesn’ admi p ope immedia e ex ensions.
A well o de ed se {ai} ⊂ K, wi hou las elemen in K, is called pseudo-con e gen i
(aj−ai)< (ak−aj)
o all i < j < k. Easily we ha e ha , i {ai}is pseudo-con e gen , hen (aj−ai) =
(ai+1 −ai),∀i < j. So we can use he abb e ia ion ωi o (aj−ai), wi h j > i. Le us
no e ha {ωi}is an inc easing se o elemen s o Γ.
An elemen a∈Kis called limi o he pseudo-con e gen se {ai}i (a−ai) = ωi o
all i.
[4] p o e he ollowing esul s:
•I K⊂Lis an immedia e ex ension, hen e e y elemen o L Kis a limi o any
pseudo-con e gen subse o Kwi hou limi in K.
•A alued ield Kis maximal i and only i he e exis s a limi o all i s pseudo-
con e gen subse s.
1
In [5], K ull shown he exis ence o , a leas , one maximal immedia e ex ension o all
alued ield. The e o e, i b
Kis he comple ion o Kby he alua ion and b is he unique
alua ion o b
K ha ex ends , hen K⊂b
Kis an immedia e ex ension and b
Kis a maximal
alued ield.
Kaplansky shown he unici y o he maximal immedia e ex ension o a alued ield K
ha sa is y a ce ain condi ion, ha he calls hypo hesis A. In he case o ze o cha ac e is ic
his hypo hesis A is emp y.
Finally, [4] gi es an use ul s uc u e heo em o all maximal alued ield. I ∆ is a ield
and Γ an o de ed abelian g oup, he se o all o mal se ies
Xai αicon ai∈∆, αi∈Γ y {αi}well o de ed,
is a ield wi h he usual sum and imes. We shall deno e his ield by ∆( Γ). In ∆( Γ) we
can conside he alua ion ν gi en by
ν 
X
i≥1
ai αi
=α1wi h a16= 0.
K ull shown in [5] ha , wi h his alua ion, ∆( Γ) is a maximal ield.
Theo em 1.2. (Kaplansky, 1942) Le Kbe a maximal alued ield wi h alues g oup Γ
and esidue ield ∆ ha sa is ies he hypo hesis A. Then Kis analy ically isomo phic o he
powe se ies ield ∆( Γ).
1.1 No a ion and de ini ions.
Le K=k((X)) = k((X1,...,Xn)) be he quo ien ield o he o mal powe se ies ing
R=k[[X]] = k[[X1,...,Xn]].
Rema k 1.3.1) We shall w i e all se ies ∈Ras =PA∈Zn
0 AXA, whe e, i A=
(a1,...,an), hen XAmeans Xa1
1···Xan
n. We shall say
E( ) = {A∈Zn
0| A6= 0}.
2) In Zmwe’ll conside he lexicog aphic o de , i ’ll be deno ed by ≤lex. This is a o al
o de o he g oup s uc u e.
3) Le 0 < m ≤nbe an in ege and
L={B1,...,Bn} ⊂ Zm
0 {0}
such ha Lis a gene a o sys em o Zm. Each monomial XAo Rhas an elemen o
Zm
0associa ed, ha is called i s L-deg ee, ha is
deg eeL(XA) =
n
X
i=1
aiBi, A = (a1,...,an).
2
De ini ion 1.4. Le 0 < m ≤nbe an in ege and
L={B1,...,Bn} ⊂ Zm
0 {0}
such ha Lis a gene a o sys em o Zm. Le :R→Zm∪ {∞} be he unc ion such ha
(0) = ∞and
( ) = min
≤lex deg eeL(XA)|A∈ E( ),
wi h 6= 0. The ex ension o o K|k, which alues g oup is Zm, is a ank mdisc e e
alua ion, called monomial alua ion associa ed o L.
Th oughou his wo k, le be a ank mdisc e e alua ion o K|kcen e ed in R. Le
R ,m and Γ = Zmbe he ing, he maximal ideal and he alues g oup o he alua ion ,
espec i ely. We’ll deno e by ∆ =R /m o he esidue ield o . Since [1] we know ha
he dimension o ( he anscendence deg ee o k⊂∆ ) is lesse o equal han n−m. We
shall suppose ha he dimension o is he maximum, n−m.
So in his wo k alua ion means ank mdisc e e alua ion o K|kcen e ed in Rand
dimension n−m.
Rema k 1.5.Le b
Kbe a maximal immedia e ex ension o K. Since [4] we can suppose ha
b
Kis he comple ion o Kwi h espec o . Le b he only ex ension o o b
K. We know
ha he e exis s an analy ic isomo phism o b
Kin ∆ ( Γ), so i s es ic ion o he ing R
gi es an injec i e homomo phism
ϕ:R=k[[X]] →∆ ( Γ)
Xi7→ Pj≥1ai,j αi,j
wi h ai,j ∈∆ y{αi,j} ⊂ Γ well o de ed.
I we conside he ex ension o ϕ o he quo ien ield Kand he alua ion ν o ∆ ( Γ)
p e iously de ined, hen =ν ◦ϕ.
The pu pose o his wo k is o cons uc ϕexplici ly, in o de o ob ain a pa ame ic
equa ion o and, in consequence, a cons uc ion o he esidue ield o , as an ex ension
o he ield k.
The e o e we’ll p o e ha o all alua ion o K|k, he e exis s an immedia e ex ension
K⊂L=k((Y)) such ha he alua ion ha ex ends is monomial.
In o he wo ds: Any alua ion comes om a monomial alua ion. This esul gene -
alizes he ob ained in [2, 3] o ank one disc e e alua ions o k((X1, X2)).
1.2 Monoidal T ans o ma ion and immedia e ex ension.
Le be a alua ion o K|k. Le us conside he nex monoidal ans o ma ion in K:
k((X)) →L=k((Y))
Xi7→ Yii i6= 2
X27→ Y2Y1
wi h (X2)>lex (X1). Then we ha e he ollowing heo em.
3
Theo em 1.6. Wi h hese condi ions, he ex ension K⊂Lis immedia e.
P oo . Le us conside he ings R=k[[X]] and S=k[[Y]], and he diag am
Rϕ//
_

∆ ( Γ)
S
ψ
77
o
o
o
o
o
o
o
o
o
o
o
o
o
Whe e ψis he na u al ex ension o ϕ o L, i.e. ψ(Y2) = ϕ(X2)/ϕ(X1).
I ψis injec i e, hen ′=ν ◦ψis a alua ion o L ha ex ends and bo h has he
same alues g oup Γ and he same esidue ield ∆ .
So we can suppose, by con adic ion, ha ψis no injec i e. Le pbe he implici ideal
ke (ψ) and L′ he quo ien ield o he ing
k[[Y]]
p.
Clea ly he es ic ion o ψ o L′is injec i e and i s composi ion wi h ν de ine a alua ion,
le us pu w1. So he ex ension K⊂L′is immedia e.
Le w2be he alua ion p–´adic o L. This is a disc e e ank one alua ion.
The composi ion o bo h alua ions, le us pu w, is a disc e e ank m+ 1 alua ion o
L|kwhose esidue ield is ∆ . Then we ha e
ank(w) + dim(w) = n+ 1 >dim k[[Y]].
Since Abhyanka ’s heo em ([1], Theo em 1, p. 330), we know
ank(w) + dim(w)≤dim k[[Y]],
so he e is a con adic ion.
Rema k 1.7.He e we a e used he exis ence o such injec i e homomo phism, p o ed by
Kaplansky [4]. La e we’ll gi e an explici cons uc ion o ψ. The e isn’ ci cula easoning.
The ollowing example shows ha he condi ion o maximal dimension o is necessa y.
Example 1.8. Le R=C[[X1, X2, X3]] and Ki s quo ien ield. Le us conside he injec i e
homomo phism
ϕ:R→C(u)[[ ]]
X17→
X27→ u 2
X37→ eu −1.
The composi ion o ϕ( eally i s ex ension o he quo ien ields) wi h he usual o de
unc ion ν o C(u)(( )) is a ank one disc e e alua ion o K, named . The dimension o
is 1, he anscendence deg ee o he ex ension C⊂C(u).
4
I we make he monoidal ans o ma ion
R=C[[X1, X2, X3]] →S=C[[Y1, Y2, Y3]]
Xi7→ Yii i= 1,3
X27→ Y2Y1
Then we ha e an homomo ism
ϕ:S→C(u)[[ ]]
X17→
X27→ u
X37→ eu −1.
ha is no injec i e.
2 Cons uc ing ∆ .
2.1 Basis o a subg oup o Zm.
The e a e well known p ocedu es o compu e a basis o a subg oup Γ0⊂Zmknowing any
se o gene a o s. In his subsec ion we desc ibe an algo i hm ha will be e y use ul in
o de o p epa e he alua ion.
Le {A1,...,An}be he se o gene a o s o Γ0⊂Zm, wi h Ai>lex 0 o all i. We can
suppose, wi hou los o gene ali y, ha Ai≤lex Aj∀i < j. Le A= (ai,j)∈ Mn×nbe he
ma ix whose ows a e he elemen s Ai.
We shall conside wo ans o ma ions wi h he ows o he ma ix A:
(1) Fi,j(q): To change he ow iby i sel plus q imes he ow j, wi h q∈Z.
(2) To in e change ows.
Clea ly he g oup gene a ed by he ows o he ma ix Ais equal o he g oup gene a ed
by he ows o any ans o ma ion o A.
Rema k 2.1.Algo i hm. The he ma ix Ahas he ollowing echelon o m
0
Le jbe he i s column di e en o 0in A, le ibe he i s such ha ai,j 6= 0. Then
we shall say ha ai,j is a pi o .
How Ai≤lex Al∀i < l, clea ly ai,j ≤al,j ∀i < l. Le us pu al,j =qlai,j + l, by making
he in ege Euclidean di ision o al,j by ai,j.
We’ll apply he ollowing p ocedu e wi h he i s s ep, ha is pe ec ly expo able o
he o he s eps:
5

I) Fo each ow l, wi h i < l ≤n, we make he ollowing ans o ma ion:
a) I l6= 0 we do Fl,i(−ql). The new ow l, ha we deno e by Alagain, is such
ha Al<lex Aiand 0 < al,j < ai,j.
b) I l= 0 he e a e wo possible si ua ions:
1) Al−qlAi>lex 0: In his case we make oo he ans o ma ion Fl,i(−ql). The
new ow lis such ha Al<lex Aiand al,j = 0. So, a e a eo de ing Al
aises a s ep.
2) Al−qlAi≤lex 0: Then we make he change
Fl,i(1 −ql). The new ow l, ha we deno e Alagain, is such ha Al≤lex Ai
and ml,j =mi,j. Le us ema k ha i ql= 1 hen Al=Ai, because we ha e
suppose a begin ha Ai≤lex Al.
II) A e hese ans o ma ions we eo de he ows. I e e y ow o he i s s ep o he
ma ix a e equal, hen we aise a s ep and we begin wi h he p ocedu e.In o he case,
we apply again his algo i hm.
Clea ly we a e doing he Euclidean algo i hm in o de o compu e he maximal common
di iso o he elemen s {ai,j,...,an,j }. Le pjbe he maximal common di iso . By a ini e
numbe o ans o ma ions we ob ain a ans o med ma ix, ha we deno e again by A,
such ha he pi o is equal o pjand he las one di ides all al,j wi h l≥i.
We ha e jus a i ed o si ua ion I.b), so, by a ini e numbe o ans o ma ions, nec-
essa ily we mus ob ain a new ma ix Awhe e all he ows down he pi o ai,j =pja e
equals o Ai.
Hence, by applying his algo i hm o each s ep o he ma ix, we ob ain a ma ix Bwi h
only con sdi e en ows Bi1,...,Biswhose pi o s a e pj1,...,pjs. Clea ly Γ0is isomo phic
o pj1Z× · · · × pjsZand {Bi1,...,Bis}is a basis o Γ0.
The algo i hm desc ibed in his sec ion allows us o p o e he ollowing lemma, ha we
a e going o use equen ly o p epa ing ou alua ion con enien ly.
Lemma 2.2. Wi h he usual condi ions o e K=k((X)) and , ank mdisc e e alua ion,
i Γ0is he subg oup gene a ed by he alues o he elemen s Xi, he we can ind, by a ini e
numbe o monoidal ans o ma ions and in e changes o a iables, an immedia e ex ension
L=k((Y)) o Ksuch ha each Yihas he alue in a basis {B1,...,Bs}o Γ0.
P oo . We ha e o apply he p eceden algo i hm o he ma ix o he alues (Xi). Whe e
Fl,i(−ql) means “ o apply qlmonoidal ans o ma ions such ha Xl7→ YlYi”, and eo de
ows means “ o eo de a iables acco ding o i s alues”.
2.2 P epa ing .
As usually, we begin wi h a ank m≤ndisc e e alua ion o K|k, cen e ed in he ing
R=k[[X]]. Le Γ = Zm,R and m be he alues g oup, he ing and he maximal ideal o
he alua ion. , espec i ely. We shall deno e, as usual, by ∆ o he esidue ield o he
alua ion.
6
We’ll suppose ha he ex ension k⊂∆ is anscenden pu e o deg ee dim =n−m.
Fi s , we apply he lemma 2.2 o ob ain an immedia e ex ension L=k((Y)) o Ksuch
ha he se o alues o he elemen s Yiis a basis {B1,...,Bs}o he subg oup Γ0gene a ed
by he alues o he elemen s Xi.
By con enience we eo de he elemen s Yiin such way ha he i s selemen s akes
all he alues o he basis (i.e. (Yi) = Bi o i= 1,...,s).
Rema k 2.3.Le A∈Γ0be such ha A= 1B1+...+ sBs, hen we shall deno e by
RA= ( 1,..., s,0,...,0) o he n–uple such ha (YRA) = A.
2.3 The i s anscenden al esidue.
Ou pu pose is o gi e an explici desc ip ion o he injec i e homomo phism ψ:k[[Y]] →
∆ ( Γ), such ha =ν ◦ψ. In o de no o complica e he exposi ion o his cons uc ion,
we a e supposing ha all he esidues a e in ko hey a e anscenden al o e he g ound
ield. This condi ion seems oo s ong, bu a e he p oo o heo em 2.8 we’ll explain why
his si ua ion is eally close o he gene al one.
1) Fo he i s elemen s we pu
ψ(Yi) = Bi i= 1,...,s.
2) Le us ake Ys+1 he i s elemen whose alue is a linea combina ion o he elemen s Bi
(in ac i mus be equal o some Bi). Le us suppose ha (Ys+1) = Bs+1,1, hen we ha e
wo possibili ies:
a) Fo all α∈k, (Ys+1 +αYRBs+1,1) = Bs+1,1. This ac means ha he esidue o
Ys+1/YRBs+1,1in ∆ is anscenden al o e k. Le us pu
us+1 =Ys+1
YRBs+1,1
+m
and ψ(Ys+1) = us+1 Bs+1,1.
b) The e exis s α∈ksuch ha (Ys+1 +αYRBs+1,1)>lex Bs+1,1. So he esidue o
Ys+1/YRBs+1,1in ∆ is in k. Le us pu αs+1,1=αand (Ys+1 +αs+1,1YRBs+1,1) =
Bs+1,2> Bs+1,1. The e a e wo possibili ies again:
i) The new alue Bs+1,2/∈Γ0. In his case, we make he change Zs+1 =Ys+1 +
YRBs+1,1,Zi=Yi∀i6=s+ 1 and go back o he beginning o he p ocedu e by
p epa ing he new alua ion o k((Z)) wi h he lemma 2.2.
ii) The alue Bs+1,2∈Γ0. I he e exis s αs+1,2such ha
Ys+1 +αs+1,1YRBs+1,1+αs+1,2YRBs+1,2=Bs+1,3>lex Bs+1,2,
hen we ask again i Bs+1,3∈Γ0. In he a i ma i e case, i he e exis s αs+1,3∈k
such ha
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 o he beginning o he p ocedu e.
We con inue his p ocedu e un il we ind a alue Bs+1,l /∈Γ0. I we canno , he e
does no exis αs+1,l ∈ksuch ha
Ys+1 +
l
X
k=1
αs+1,kYRBs+1,k !=Bs+1,l+1 > Bs+1,l.
In he i s case we make he change
Zs+1 =Ys+1 +
l−1
X
k=1
αs+1,kYRBs+1,k , Zi=Xi∀i6=s+ 1
and mo e o he lemma 2.2. In he second case we ha e ha he esidue
us+1 =Ys+1 +Pl−1
k=1 αs+1,kYRBs+1,k
YRBs+1,l
+m
is anscenden al o e k. In his case we pu
ψ(Ys+1) =
l−1
X
k=1
αs+1,k Bs+1,k +us+1 Bs+1,l .
We ha e o p o e ha his p ocedu e ends up inding a new alue ha i is no in Γ o
a anscenden al esidue.
Lemma 2.4. The si ua ion desc ibed in 2.b.i) only occu s a ini e numbe o imes.
P oo . The e a e wo possible si ua ions o ind an elemen o alue B /∈Γ0:
1) The alue Bis no a a ional linea combina ion o he elemen s o Γ0. In his case,
he ank o he new subg oup o Zminc eases by 1. T i ially his ac only occu s a ini e
numbe o imes.
2) The new alue Bis a non in ege a ional linea combina ion o he elemen s o he
basis o Γ0. Le {p1,...,ps}be he pi o s ha appea s in he cons uc ion o he basis o
Γ0, le {q1,...,qs}be he ones o he new subg oup, Γ1. As Γ0⊂Γ1, hen qi≤pi∀iand,
a leas , one inequali y is s ic . As he pi o es a e g ea e o equal han 1, his only occu s
a ini e numbe o imes.
Rema k 2.5.Le us suppose ha we ha e a pseudo con e gen se { j}o elemen s o a
alued ield K, wi h alue g oup Zmwi h lexicog aphic o de . The se o alues {ωj}o he
elemen s jis a s ic ly inc easing sequence o elemen s o he g oup Zm. Le us suppose
ha he se {ωj}is bounded. Then, om an index jsu icien ly big, we ha e
ωj= (a1,...,al, al+1,j,...,am,j),
in such way ha he i s lcoo dina es o he alues ωja e s abilized. Le us suppose ha
lis he g ea es in ege be ween 1 and msuch ha his ac occu s.
Le be a limi o { j}and ω= ( ) i s alue. Then, i ω= (b1,...,bn), as ω >lex ωj
o all j, hen he e exis s an l0≤lsuch ha bi=ai o all i= 1,...,l0−1 and bl0> al0.
8
Lemma 2.6. The p ocedu e desc ibed in he si ua ion 2.b.ii) inds a alue ha is no in Γ0
o a anscenden al esidue o e k.
P oo . I he p ocedu e is no ini e, hen we ha e a se o elemen s { j}j≥1such ha
j=Ys+1 +
j−1
X
l=1
αs+1,lYRBs+1,l .
So ( j+1) = Bs+1,j+1 >lex ( j) = Bs+1,j o all j.
The e o e, i i < j < k, we ha e
( j− i) = Bs+1,i <lex ( k− j) = Bs+1,j,
so { j}is a pseudo-con e gen se .
Le us suppose ha he se o alues {Bs+1,j}is bounded. Like in he ema k 2.5, le
us suppose ha , since an index jsu icien ly g ea , he i s lcoo dina es o he alues a e
s abilized, whe e lis he g ea es in ege be ween 1 and msuch ha his ac occu s. Le
us pu
( j) = (a1,...,al, al+1,j,...,am,j).
Again by ema k 2.5, we know ha any limi o { j}is such ha i s alue is some hing
like
(a1,...,ak, bk+1,...,bm),
wi h k < l and bk+1 > ak+1. Le us ake he se ies
Ys+1 +
∞
X
j=1
αs+1,jYRBs+1,j .
As limi , by con enience we pu
g1=
∞
X
j=1
αs+1,jYRBs+1,j
and (Ys+1 +g1) = B1
s+1,1.
I B1
s+1,1/∈Γ0, he we ha e inished. In o he case, i he e exis s α1
s+1,1such ha
Ys+1 +g1+α1
s+1,1YRB1
s+1,1=B1
s+1,2>lex B1
s+1,1,
hen we con inue wi h ou p ocedu e. I i is in ini e again and he alues con ained a e
bounded, hen we shall ha e a pseudo-con e gen se wi h limi
Ys+1 +g1+
∞
X
j=1
α1
s+1,jYRB1
s+1,j .
9