ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P2.04
https://doi.org/10.26493/1855-3974.2968.d23
(Also available at http://amc-journal.eu)
Valuations and orderings
on the real Weyl algebra
Lara Vukšić *
Institute of Mathematics, Physics, and Mechanics, Ljubljana, Slovenia and
Faculty of Mathematics and Physics, University of Ljubljana, Slovenia
Received 19 September 2022, accepted 12 February 2023, published online 20 September 2023
Abstract
The first Weyl algebra A1(k) over a field k is the k-algebra with two generators x, y
subject to [y, x] = 1 and was first introduced during the development of quantum mechan-
ics. In this article, we classify all valuations on the real Weyl algebra A1(R) whose residue
field is R. We then use a noncommutative version of the Baer-Krull theorem to classify all
orderings on A1(R). As a byproduct of our studies, we settle two open problems in real
algebraic geometry. First, we show that not all orderings on A1(R) extend to an ordering
on a larger ring R[y; δ], where R is the ring of Puiseux series, introduced by Marshall and
Zhang in 2000, and characterize the orderings that do have such an extension. Second,
we show that for valuations on noncommutative division rings, Kaplansky’s theorem that
extensions by limits of pseudo-Cauchy sequences are immediate fails in general.
Keywords: Weyl algebra, noncommutative valuations, skew polynomial rings, orderings, extensions
of valuations, extensions of orderings.
Math. Subj. Class. (2020): 16W60, 06F25, 13J30, 14A22, 16S3
1 Introduction
Valuation theory was first developed for commutative fields in the context of number theory
and was first defined by József Kürschák [12] in 1913. For modern treatments, we refer to
the books of Engler and Prestel [5] or Kuhlmann [10]. Oscar Schilling wrote the first major
work on valuations on (noncommutative) division rings in 1945 [21].
A valuation on a division ring D is a map v : D → Γ ∪ {∞}, where Γ is an ordered
group written additively and ∞ ̸∈ Γ,∞ > γ for each γ ∈ Γ, with the following properties:
*I would like to thank my advisor Igor Klep for his guidance and many helpful comments and suggestions.
E-mail address: lara.vuksic@fmf.uni-lj.si (Lara Vukšić)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Ars Math. Contemp. 24 (2024) #P2.04
1. ∀x ∈ D : v(x) = ∞ ⇔ x = 0,
2. ∀x, y ∈ D : v(xy) = v(x) + v(y),
3. ∀x, y ∈ D : v(x+ y) ≥ min{v(x), v(y)}.
It follows that v is a homomorphism from D∗ to Γ. The set Ov := {x ∈ D | v(x) ≥ 0} is
called the valuation ring associated to v, and Mv := {x ∈ M | v(x) > 0} is its maximal
ideal. The division ring D := Ov/Mv is called the associated residue division ring. Since
v is a group homomorphism, the subgroup O∗v is normal in D∗. Several alternative ap-
proaches to noncommutative valuations, where v does not define a group homomorphism,
were introduced and studied recently by Nicolai Ivanovich Dubrovin in [7] and [4] (see
also [17] for a more thorough treatment), and by Jean-Pierre Tignol and Adrien Wadsworth
in [23].
Suppose F is a field. Then all valuations on the field or rational functions F (x) with
residue field F are well-known, namely, the p-adic valuations for irreducible polynomials
p(x) ∈ F [x], and the vdeg valuation, defined by
vdeg(
p
q
) := deg(q)− deg(p).
The description of all valuations on the field of rational functions in several variables with
residue field equal to the base field is much more involved. There are many descriptions
of constructions of such valuations in the literature. Among famous examples of such
descriptions are the one given by Saunders MacLane in [13] and the one given by Franz-
Viktor Kuhlmann in [11].
As valuations on Ore extensions uniquely extend to their quotient division ring, the
description of all valuations on Ore division rings is equivalent to the description of all
valuations on corresponding quotient division rings. The description of all valuations on
noncommutative Ore extensions R[x;σ, δ] where R is a domain, σ : R → R is a ring
homomorphism and δ : R → R a σ-derivation is even more complex than in the commu-
tative case. Additional difficulties arise from the fact that [f, g] = 0 does not hold for all
f, g ∈ R[x;σ, δ]. Granja, Martı́nez, and Rodrı́guez have shown in [6] that the set of all
real valuations extending to the skew polynomial ring has the structure of a parameterized
complete non-metric tree. Further recent progress on valuations on Ore extensions is given
by Onay in [18] and Rohwer in his PhD thesis [20].
1.1 Results
Our main goal is to classify all orderings and real valuations on the real Weyl algebra A1(R)
or, equivalently, its quotient division ring D1(R). The Weyl algebra is the noncommutative
algebra generated by two elements x, y satisfying [y, x] = 1. Hence its elements are all of
the form ∑
i,j
αi,jx
iyj , αi,j ∈ R.
Because of this, our approach to constructing valuations on A1(R) is inspired by classical
constructions of valuations on commutative rational functions in two unknowns mentioned
above. However, the relation [y, x] = 1 gives rise to additional constraints and far fewer
valuations than in the commutative case.
L. Vukšić: Valuations and orderings on the real Weyl algebra 3
As we will show, the valuations on A1(R) we are interested in all satisfy v[a, b] > v(ab)
for all nonzero a, b. We call such valuations strongly abelian. They have an abelian value
group and commutative residue field. In Section 2 we give some properties of strongly
abelian valuations. We show that if a valuation v on a division ring D satisfies D = Z(D)
and the value group is of rational rank one, then v is strongly abelian. Under additional
constraints on the residue field and the value group we extend this statement to valuations
of higher rational rank.
In Section 3 we give a characterization of all valuations v on the real Weyl algebra
A1(R) with residue field R in the spirit of MacLane [13]. The construction is inspired
by the outline given by Shtipelman in [22] for valuations on the complex Weyl algebra
A1(C). We also explicitly describe the associated value groups and show that they are all
isomorphic to subgroups of Q or Q× Z.
In their attempt to describe all orderings on A1(R) in [16], Murray Marshall and Yufei
Zhang introduced the Ore extensions R[y; δ] and R̃[y; δ], with
R := {
∑
k≥m
akx
− kn | ak ∈ R,m ∈ Z, n ∈ N},
R̃ := {
∑
q∈A
aqx
−q | A ⊂ Q is well-ordered}
and δ(p(x)) = p′(x). As is often done in real algebraic geometry, all orderings are de-
scribed by classifying all real valuations via the Baer-Krull theorem. Marshall and Zhang
described almost all valuations v on R[y; δ] with residue field R; in one case, they did not
prove that v is a valuation. In Section 4, we complete their characterization. Marshall
and Zhang also conjectured that all valuations on A1(R) with residue field R extend to a
valuation on R[y; δ] with the same residue field. We refute their conjecture in Section 4.
Further, we combine our classification of valuations on A1(R) with Marshall and Zhang’s
description of valuations on R[y; δ] to characterize the valuations on A1(R) with residue
field R that extend to a valuation R[y, δ] with the same residue field. All such extensions
are again strongly abelian.
In Section 5, we show that all valuations on R[y; δ] with residue field R uniquely extend
to a strongly abelian valuation on R̃[y; δ] with the same residue field. We also show that
the value group of such an extension is not of rational rank one.
As a byproduct of our investigations, we show that Kaplansky’s theorem that all ex-
tensions by limits of pseudo-Cauchy sequences are immediate (in particular, they do not
change the rational rank of the value group) fails for noncommutative division rings.
As Marshall and Zhang observe in [15], all strongly abelian valuations v on a division
ring D with a formally real residue field are compatible with an order on D. In Section 6,
we describe all v-compatible orders on A1(R) for every valuation v on A1(R) constructed
in Section 3 using a noncommutative version of the Baer-Krull theorem as given in [2]
(see also [1, 3, 24] and [9] for modern treatments and extensions). We also characterize
the v-compatible orders on A1(R) that extend to an order on R[y; δ] compatible with v’s
extension to R[y; δ].
2 Strongly abelian valuations
We present some properties of valuations on noncommutative division rings which we will
use later to describe order-compatible valuations on the real Weyl algebra A1(R) and some
of its ring extensions. First, we define a property of valuations on division rings.
4 Ars Math. Contemp. 24 (2024) #P2.04
Definition 2.1. Suppose v is a valuation on a division ring D. We say v is strongly abelian
if v[a, b] > v(ab) holds for all nonzero a, b ∈ D.
Any valuation on a field is strongly abelian. In this section, we describe a sufficient
condition for a valuation v to be strongly abelian. This property will be important for us
for two reasons. Firstly, it is obvious that if a valuation v on a division ring D is strongly
abelian, then the associated value group and residue division ring are commutative. Sec-
ondly, we are particularly interested in order-compatible valuations on A1(R); minimal
such have residue field R, as it was shown in [16]. It follows from Theorem 2.5 of [15]
that a strongly abelian valuation v on a division ring D with a formally real residue field is
compatible with an order on D by the noncommutative version of the Baer-Krull theorem
as given in [24].
Proposition 2.2. Let v be a valuation on a division ring D such that D = Z(D). Let
a, b ∈ D∗ be such that v(a) and v(b) are rationally dependent. Then v[a, b] > v(ab).
Proof. Since v(a) and v(b) are rationally dependent, v(ab) = v(ba) ≤ v[a, b]. Suppose
v[a, b] = v(ab). In particular, a and b do not commute. Let β := aba−1b−1 ∈ D. We have
β = aba−1b−1 = ([a, b] + ba)a−1b−1 = [a, b]a−1b−1 + 1 ̸= 1.
Let v(b) = − ℓkv(a) for ℓ, k ∈ Z, ℓ and k coprime. It follows that v[a, b] =
k−ℓ
k v(a).
Define γ := (ba)kaℓ−k ∈ D. Let β′, γ′ ∈ Z(D) be such that v(β′) = v(γ′) = 0, β′ = β
and γ′ = γ. Then on one hand,
v(a(ba)k − γ′ak−ℓ+1) = v(a((ba)kaℓ−k − γ′)ak−ℓ) > (k − ℓ+ 1)v(a),
and on the other,
v(a(ba)k − γ′ak−ℓ+1) = v((ab)ka− γ′ak−ℓ+1) = v(((ab)kaℓ−k − γ′)ak−ℓ+1)
= (k − ℓ+ 1)v(a)
since
(ab)kaℓ−k − γ′ = (aba−1b−1ba)kaℓ−k − γ = βk(ba)kaℓ−k − γ ̸= 0.
In the last equation, we used that (aba−1b−1ba)kaℓ−k = (β′ba)kaℓ−k = β′k(ba)kaℓ−k =
βk(ba)kaℓ−k since D = Z(D) as presumed. This holds if βk ̸= 1. If βk = 1, we repeat
our calculations with roles of a and b interchanged. The new β will now be the inverse value
of the former and since gcd(ℓ, k) = 1, β−ℓ ̸= 1. In either case, we get a contradiction from
which we deduce v[a, b] > v(ab).
Remark 2.3. The condition D = Z(D) is fulfilled by every valuation on an algebra over
a field that is isomorphic to the residue field. In particular, this holds for minimal order-
compatible valuations on R-algebras.
Corollary 2.4. Let v be a valuation on a division ring D such that D = Z(D). If the value
group has rational rank one, then v is strongly abelian.
L. Vukšić: Valuations and orderings on the real Weyl algebra 5
Lemma 2.5. Let v be a valuation on a division ring D such that the value group is abelian
and D = Z(D). Then for all x, y ∈ D \ {0}:
(1) If v[x, y] > v(xy), then v[xm, y] > v(xmy) for all m ∈ Z.
(2) Suppose v[x, y] = v(xy). Then v[x−1, y] = v(x−1y) and for each m ∈ N,
v[xm, y] > v(xmy) holds if and only if α := y−1x−1yx satisfies 1 + α + · · · +
αm−1 = 0 in D.
Proof. To prove (1), first observe
[x−1, y] = x−1y − yx−1 = x−1(yx− xy)x−1,
so if v[x, y] > v(x, y), then
v[x−1, y] = v[x, y]− 2v(x) > v(xy)− 2v(x) = v(x−1y).
Suppose m ∈ N. Then
[xm, y] =
m∑
ℓ=1
xm−ℓ[x, y]xℓ−1
and since the value group is commutative, v(xm−ℓ[x, y]xℓ−1) = (m− 1)v(x) + v[x, y] >
v(xmy) for each 1 ≤ ℓ ≤ m. Item (1) is thus proved.
To prove (2), suppose v[x, y] = v(xy). Then v[x−1, y] = v(x−1y) is proved as for the
first case. Since the value group is abelian, v[xm, y] ≥ v(xmy) holds, so we can observe
y−1x−m[xm, y] = y−1x−m
m∑
ℓ=1
xm−ℓ[x, y]xℓ−1 =
m∑
ℓ=1
y−1x−ℓ[x, y]xℓ−1.
For each 1 ≤ ℓ ≤ m,
y−1x−ℓ[x, y]xℓ−1 = (αx−1)ℓ−1y−1x−1[x, y]xℓ−1
= αℓ−1x−ℓ+1y−1x−1[x, y]xℓ−1 = y−1x−1[x, y]αℓ−1.
We can change the order of α and x−1 by Proposition 2.2 since v(α) = 0. The last equation
follows from v(y−1x−1[x, y]) = 0 and Proposition 2.2. So now we have
y−1x−m[xm, y] = y−1x−1[x, y]
m−1∑
ℓ=0
αℓ,
which proves the equivalence in (2).
Proposition 2.6. Let v be a valuation on a division ring D such that the value group is
abelian and D = Z(D). Suppose the residue field is formally real and suppose v[x, y] =
v(xy) for some x, y ∈ D. Then v[xm, y] = v(xmy) for all odd m > 2. If v[xm, y] >
v(xmy) for some even m, then v[x2, y] > v(x2y) and there is no a ∈ D such that v(a2) =
v(x).
6 Ars Math. Contemp. 24 (2024) #P2.04
Proof. Suppose v[x, y] = v(xy). If m is odd,
∑m−1
ℓ=0 α
ℓ = 0 does not have a solution in
the residue field. By Lemma 2.5, it follows that v[xm, y] = v(xmy). Now consider the case
for even m. If v[xm, y] > v(xmy), then α = −1 by Lemma 2.5 and v[x2, y] > v(x2y).
Suppose a ∈ D satisfies v(a2) = v(x). We will first show that v[a2, y] = v(a2y). Assume
that v[a2, y] > v(a2y). Then on one hand,
a−2x = a−2xy−1y = y−1a−2xy,
since v(a−2x) = 0. On the other hand,
a−2x = y−1ya−2x = y−1a−2yx,
where the last equation follows from v[a2, y] > v(a2y), or, by Lemma 2.5 equivalently,
v[a−2, y] > v(a−2y). From x−1a−2(xy − yx) = 0 we conclude v[x, y] > v(xy), which
is a contradiction. So v[a2, y] = v(a2y) and v[a, y] = v(ay) follows from Lemma 2.5.
Now we show v[a4, y] > v(a4y). On one hand, we can write
a−4x2 = y−1a−4x2y = y−1a−4yx2
since v[x2, y] > v(x2y). On the other hand,
a−4x2 = y−1ya−4x2,
so we conclude v[a4, y] > v(a4y). But by Lemma 2.5, v[a4, x] > v(a4x) gives us
xax−1a−1 = −1. But then, again by Lemma 2.5, v[a2, x] > v(a2x). The proposition
is thus proved.
Proposition 2.7. Let v be a valuation on a division ring D such that such that D = Z(D),
the value group is abelian and 2-divisible and the residue field is formally real. Suppose
the value group of v is of rational rank 2 and suppose there are x, y ∈ D∗ such that v(x)
and v(y) are rationally independent with v[x, y] > v(xy). Then v is strongly abelian.
Proof. Suppose a, b ∈ D. Since the value group is abelian, v[a, b] ≥ v(ab). Suppose
v[a, b] = v(ab). Then v(ak1) = v(x−m1y−n1) and v(bk2) = v(x−m2y−n2) for some
ki,mi, ni ∈ Z, i = 1, 2. We conclude from Lemma 2.5 that v[ak1 , bk2 ] = v(ak1bk2).
This is immediate if k1 and k2 are both odd. If k1 or k2 is even, v[ak1 , bk2 ] = v(ak1bk2)
follows from the 2-divisibility of the value group and Proposition 2.6. Let c := xm1yn1
and d := xm2yn2 . Then on the one hand,
ak1cbk2d = cak1dbk2 = dcak1bk2
since v(ak1) and v(c) are rationally dependent, v(bk2) and v(d) are rationally dependent
and v(bk2d) = v(ak1c) = 0. On the other hand,
ak1cbk2d = bk2dak1c = dbk2cak1 = cdbk2ak1 = dcbk2ak1 .
Here, the last equality follows from v[x, y] > v(xy) and Lemma 2.5. Thus we have
v(dc(ak1bk2 − bk2ak1)) > 0, so we get v[ak1 , bk2 ] > v(ak1bk2) which contradicts our
assumption v[a, b] = v(a, b). We conclude v[a, b] > v(ab).
The proof of the following proposition is the same as the proof of Proposition 2.7.
L. Vukšić: Valuations and orderings on the real Weyl algebra 7
Proposition 2.8. Let v be a valuation on a division ring D such that D = Z(D), the
value group is abelian and the residue field is formally real. Suppose the value group
of v is of rational rank 2 and suppose there are x, y ∈ D∗ such that for every z ∈ D,
v(zk) = v(x−my−n) holds for some k,m, n ∈ Z where k is odd. Then v is strongly
abelian.
We will later use this result to show that all valuations v on A1(R) with residue field R
are strongly abelian. Propositions 2.7 and 2.8 can be easily generalized to higher rational
ranks of the value group. The proofs are analogous.
Corollary 2.9. Let v be a valuation on a division ring D such that D = Z(D). Suppose the
value group is abelian and 2-divisible of rational rank n and that there are x1, . . . , xn ∈ D
such that v(x1), . . . , v(xn) are rationally independent with v[xi, xj ] > v(xixj) for all i, j.
Then v is strongly abelian.
Corollary 2.10. Let v be a valuation on division ring D such that D = Z(D). Suppose
the value group is abelian and of rational rank n and that there are x1, . . . , xn ∈ D such
that for every z ∈ D, v(zk) = v(xm11 · · ·xmnn ) for some k,m1, . . . ,mn ∈ Z with k odd.
Then v is strongly abelian.
3 Valuations on A1(R)
We now describe the construction of all valuations on A1(R) with residue field R that was
sketched in [22] over the ground field of C. Since every f ∈ A1(R) can be written as∑
m,n≥0 αm,nx
myn, the construction will be similar to the construction of all valuations
on the field of rational functions R(x, y) with residue field R (examples of constructions of
such valuations can be found in [11] or [13]), but with some additional constraints arising
from the fact that the generators x, y ∈ A1(R) satisfy [y, x] = 1. We first note that
it follows from Theorem 5.3 of [16] that the value group of any valuation on A1(R) is
commutative. Also, since every valuation v on A1(R) can be uniquely extended to its
quotient division ring D1(R), our construction will take place in the quotient ring as we
will use inverses.
To construct a valuation v trivial on R with residue field R, we compare v(x) and v(y).
It is easy to show, as it was done in [16], that v(xy) = v(yx) < 0, so v(x) or v(y) will be
less than zero. Without loss of generality, we can set v(x) = −1 ∈ Q and compare it to
v(y). If v(y) ̸∈ Q, then we get
v(
∑
m,n≥0
αm,nx
myn) = min
m,n
{mv(x) + nv(y)}
for all elements of A1(R). Otherwise, v(y) = m1n1 ∈ Q. It follows that x
m1yn1 = β1 ∈ R,
so v(xm1yn1 − β1) > 0. Set
ω1 := x
m1yn1 − β1
and as before compare v(ω1) to v(x) in terms of rational dependence. If v(ω1) = m2n2 ∈ Q,
then xm2ωn21 = β2 for some β2 ∈ R. Hence
ω2 := x
m2ωn21 − β2
8 Ars Math. Contemp. 24 (2024) #P2.04
also has value greater than zero. We continue this procedure. If we additionally define
ω−1 = x and ω0 = y, we thus get a sequence
(ωi)i≥−1, ωi ∈ A1(R)
which ends with ωn for some n ∈ N if v(ωn) ̸∈ Q or is infinite otherwise.
By the end of this section, we will prove a necessary and sufficient condition for the
possibility to extend v from (ωi)i≥−1 to a valuation on A1(R) with residue field R. Every
such extension from (ωi)i≥−1 to A1(R) will be uniquely determined. We will also show
that every valuation on A1(R) with residue field R is strongly abelian.
3.1 Properties of the sequence (ωi)i≥−1 associated to a valuation on A1(R)
Thorough this subsection let v be a valuation on A1(R).
Lemma 3.1. Suppose (ωi)i≥−1 ⊆ A1(R) is a sequence as described above, with ω−1 = x,
ω0 = y, ωi = xmiωnii−1 − βi for all i ≥ 0. Then [ωi, x] equals
xmi
ni∑
ℓi=1
ωni−ℓii−1
(
xmi−1
ni−1∑
ℓi−1=1
ω
ni−1−ℓi−1
i−2(
· · ·
(
xm2
n2∑
ℓ2=1
ωn2−ℓ21 n1x
m1yn1−1ωℓ21
)
ωℓ32
)
· · ·
)
ωℓii−1
for each i ≥ 1.
Proof. We prove the lemma by induction on i. If i = 1,
[ω1, x] = [x
m1yn1 − β1, x] = xm1
n1∑
ℓ1=1
yn1−ℓ1 [y, x]yℓ1−1 = n1x
m1yn1−1.
Now suppose that the equality holds for [ωi, x]. Then we have
[ωi+1, x] = [x
mi+1ω
ni+1
i − βi+1, x] = x
mi+1 [ω
ni+1
i , x]
= xmi+1
ni+1∑
ℓi+1=1
ω
ni+1−ℓi+1
i [ωi, x]ω
ℓi+1−1
i
We can then proceed by the induction hypothesis.
Before proving the next lemma, we define an equivalence relation between nonzero
elements of A1(R) that have the same v-value, but their difference does not. For any
a, b ∈ A1(R) \ {0}, we write a ∼ b if v(a) = v(b) < v(a− b). This is also a congurence
relation, as ac ∼ bc and ca ∼ cb holds for all a, b, c ∈ A1(R) \ {0} with a ∼ b.
Lemma 3.2. Suppose v is a valuation on A1(R) with residue field R and suppose (ωi)i≥−1
is a sequence such that ω−1 = x, ω0 = y, ωi = xmiωnii−1 − βi, v(x) = −1 and v(ωi) =
mi+1
ni+1
for all i ≥ 0 up to either some n ≥ 0 in which case v(ωn) ̸∈ Q, or up to infinity.
Then v[ωj , ωi] > v(ωiωj) for all i, j ≤ k if and only if v(Πkℓ=−1ωℓ) < 0, where k ≤ n in
case v(ωn) ̸∈ Q for some n ≥ 0.
L. Vukšić: Valuations and orderings on the real Weyl algebra 9
If any and hence both sides of the equivalence hold, then
v[ωj , ωi] = −v(xyω1 · · ·ωi−1ωi+1 · · ·ωj−1)
for all i < j ≤ k.
Proof. Suppose v is a valuation on A1(R) and (ωi)i≥−1 is a sequence as described in the
lemma. So v(ωi) ∈ Q either for all i ≥ 0 or for all 0 ≤ i < n for some n ≥ 0 and
v(ωn) ̸∈ Q. It follows from Proposition 2.2 that v[ωi, ωj ] > v(ωiωj) for all i, j < n since
v(ωi) and v(ωj) are rationally dependent. We shall use this fact to evaluate v[ωi, ωj ] for all
i, j ≤ k ≤ n. It follows from Lemma 3.1 that [ωk, x] is a sum of products P , all equal to
yn1−1xm1ωn2−11 x
m2 · · ·ωnk−1k−1 xmk up to the order of factors. Since v[ωi, ωj ] > v(ωiωj)
for all i, j ≤ k − 1,
P ∼ yn1−1xm1ωn2−11 xm2 · · ·ω
nk−1
k−1 x
mk
holds for every product P of the sum. Since
v(yn1−1xm1ωn2−11 x
m2 · · ·ωnk−1k−1 x
mk) + v(yω1 · · ·ωk−1) =
k∑
i=1
v(xmiωnii−1) = 0,
we can conclude v[ωk, x] = v(yn1−1xm1ωn2−11 x
m2 · · ·ωnk−1k−1 xmk) = −v(yω1 · · ·ωk−1).
It follows that v[x, ωk] > v(xωk) if and only if v(xyω1 · · ·ωk) < 0.
We will now prove that v[ωi+k, ωi] = −v(xyω1 · · ·ωi−1ωi+1 · · ·ωi+k−1) by induction
on k, 1 ≤ k ≤ n − i. It will then follow that v[ωi, ωj ] > v(ωiωj) for all i, j ≤ n if and
only if v(Πnℓ=−1ωℓ) < 0. If k = 1, then
[ωi+1, ωi] = [x
mi+1ω
ni+1
i − βi+1, ωi] = [x
mi+1 , ωi]ω
ni+1
i
= (
mi+1∑
ℓ=1
xmi+1−ℓ[x, ωi]x
ℓ−1)ω
ni+1
i ,
and since [x, ωi] is a sum of products all equal to yn1−1xm1ωn2−11 x
m2 · · ·ωni−1i−1 xmi up to
the order of factors, we can, using v[ωi, ωj ] > v(ωiωj) for j < i, deduce that v[ωi+1, ωi] =
−v(xyω1 · · ·ωi−1) just like we did when evaluating v[ωi, x]. For k > 1, we have
[ωi+k, ωi] = [x
mi+kω
ni+k
i+k−1 − βi+k, ωi] = [x
mi+k , ωi]ω
ni+k
i+k−1 + x
mi+k [ω
ni+k
i+k−1, ωi]
= (
mi+k∑
ℓ=1
xmi+k−ℓ[x, ωi]x
ℓ−1)ω
ni+k
i+k−1+
xmi+k(
ni+k∑
ℓ=1
ω
ni+k−ℓ
i+k−1 [ωi+k−1, ωi]ω
ℓ
i+k−1)
∼ mi+kxmi+k−1ω
ni+k
i+k−1[x, ωi] + ni+kx
mi+kω
ni+k−1
i+k−1 [ωi+k−1, ωi]
and using both Lemma 3.1 and induction on k, we see that the first sum has v-value equal
to −v(xyω1 · · ·ωi−1) and the second has v-value equal to −v(xyω1 · · ·ωi+k−1). Since the
latter is smaller, it is equal to v[ωi+k, ωi]. This proves the lemma.
10 Ars Math. Contemp. 24 (2024) #P2.04
It follows that if v can be extended from (ωi)i≥−1 to a valuation on A1(R),∑k
i≥−1 v(ωi) must be strictly less than 0 for all k ≤ n in case v(ωn) ̸∈ Q for some n,
and for all k ≥ 0 if v(ωi) ∈ Q for all i ∈ N. We will now describe a necessary condition
for the residue field to be R and then proceed to show that if both conditions are fulfilled,
v can be extended from (ωi)i≥−1 to a valuation on A1(R) with residue field R.
To ensure that the residue field is R, it is obviously necessary that ωkii−1ω
kj
j−1 ∈ R holds
for all ki, kj ∈ Z with ki mini + kj
mj
nj
= 0. For given i, j ≥ 0, all solutions (ki, kj) ∈ Z2 to
the diophantine equation
kimjni + kjminj = 0 (3.1)
are integer multiples of the pair (Ki,j ,−Kj,i) with
Ki,j =
mjni
di,j
,
Kj,i =
minj
di,j
,
where di,j = gcd{mjni,minj}.
So for all ki, kj ∈ Z with ki mini + kj
mj
nj
= 0, we can write
ωkii−1ω
kj
j−1 = ω
nKi,j
i−1 ω
−nKj,i
j−1 = (ω
Ki,j
i−1 ω
−Kj,i
j−1 )
n
for some n ∈ Z, where we used Proposition 2.2 in the second equality. So for every
ki, kj ∈ Z satisfying 3.1, ωkii−1ω
kj
j−1 is uniquely determined by ω
Ki,j
i−1 ω
−Kj,i
j−1 . For each
i, j ≥ 0, we define αi,j = ω
Ki,j
i−1 ω
−Kj,i
j−1 . We immediately see that αj,i = α
−1
i,j and hence
αi,i = 1 for all i, j ≥ 0. As
α
di,j
i,j = ω
Ki,jdi,j
i−1 ω
−Kj,idi,j
j−1 = ω
mjni
i−1 ω
−minj
j−1 = β
mj
i β
−mi
j
for all i, j ≥ 0, αi,j is one of the possible di,j-th roots for β
mj
i β
−mi
j . If v is a valuation on
D1(R) with residue field R, αi,j must be real for all i, j ≥ 0. For every i, j ≥ 0 with even
di,j , this means that β
mj
i β
−mi
j > 0 must hold. In the next lemma, we present a necessary
condition on the sequence (βi)i≥1 so that αi,j ∈ R can be chosen for all i, j. We also prove
that if ni is odd, αi,j is uniquely determined for all j ≥ 0.
Lemma 3.3. Let v be a valuation on D1(R) as in Lemma 3.2. Then the following holds:
(1) If ni is odd, there is a unique possible choice for αi,j ∈ R for all j ≥ 0.
(2) Only if sgn(βi) is constant on the set of all i ≥ 0 for which ni is even can we choose
αi,j ∈ R for all i, j ≥ 0.
Proof. Suppose ni is odd. Then for any j ≥ 0, let d̃i,j be the highest odd number dividing
di,j . Since ni is odd, ℓ1 :=
d̃i,jmj
di,j
∈ Z. If nj is odd as well, ℓ2 := d̃i,jmidi,j ∈ Z holds too.
If nj is even, mj is odd, so di,j = ˜di,j as di,j divides mjni. In both cases, ℓ2 ∈ Z holds.
L. Vukšić: Valuations and orderings on the real Weyl algebra 11
Then for ℓ := ℓ1mi = ℓ2mj =
d̃i,jmimj
di,j
we can evaluate
α
d̃i,j
i,j = ω
Ki,j d̃i,j
i−1 ω
−Kj,id̃i,j
j−1 = x
ℓx−ℓωℓ1nii−1 ω
−ℓ2nj
j−1 = (x
miωnii−1)
ℓ1(xmjω
nj
j−1)
−ℓ2
= βℓ1i β
−ℓ2
j ,
and since d̃i,j is odd, αi,j ∈ R is uniquely determined. The first point of the lemma is thus
proven.
To prove the second point of the lemma, suppose i, j ≥ 0 are such that ni and nj are
both even. As a consequence, both mi and mj are odd while di,j is even. So, provided
αi,j ∈ R, we compute
1 = sgn(α
di,j
i,j ) = sgn(β
mj
i β
−mi
j ) = sgn(βiβj),
which proves the second part of the lemma.
For even di,j we have seemingly two choices for αi,j ∈ R – a positive and a nega-
tive one. We will show that in most cases, we cannot choose sgn(αi,j) for all i, j ≥ 0
independently of each other.
Before that, we observe that for any i, j ≥ 0, at most one of Ki,j and Kj,i is even. In
fact, if at most one of ni and nj is odd, Ki,j is odd if and only if ni is divisible by the
greatest power of two that divides nj . For each i ≥ 1, let 2hi be the biggest power of two
that divides ni. Define also m0 = 1 and n0 = −1.
Proposition 3.4. Let v be a valuation on D1(R) associated to a sequence (ωi)i≥−1 with
v(ω−1) = v(x) = −1, v(ωi−1) = mini with gcd(mi, ni) = 1 and x
miωnii−1 = βi ∈ R
for each i ≥ 1. Suppose sgn(βi) is constant on the set of all i ≥ 0 for which ni is
even. Suppose αi,j ∈ R is determined for all i, j ≥ 0. Then Πri=0ω
ki
i−1 ∈ R is uniquely
determined for each set of integers k0, k1, . . . , kr ∈ Z with
∑r
i=0 ki
mi
ni
= 0 if and only if
for each a, b, c ≥ 0, αa,bαa,cαb,c > 0 whenever ha = hb ≤ hc holds.
Proof. To prove the necessity of the condition, suppose a, b, c ≥ 0 are such that ha =
hb ≤ hc holds. Suppose Ka,b and Kb,c are both odd. Choose ka, kb, kc ∈ Z \ {0} such
that kamana + kb
mb
nb
+ kc
mc
nc
= 0 and that ka and kb are odd while kc is even. Then on one
hand,
sgn(ωkaa−1ω
kb
b−1ω
kc
c−1) = sgn(ω
ka
a−1ω
kb
b−1ω
kc
c−1
Ka,b
)
= sgn(ωkaa−1ω
kb
b−1ω
kc
c−1
Ka,b
ω
−kaKb,a
b−1 ω
kaKb,a
b−1 )
= sgn(αkaa,bω
ℓb
b−1ω
ℓc
c−1)
with
ℓb = kbKa,b + kaKb,a,
ℓc = kcKa,b.
As v(ωℓbb−1ω
ℓc
c−1) = 0, (ℓb, ℓc) = ℓ(Kb,c,−Kc,b) for some ℓ ∈ Z with −ℓKc,b =
ℓc = kcKa,b. So we can conclude
sgn(ωkaa−1ω
kb
b−1ω
kc
c−1) = sgn(α
ka
a,bα
ℓ
b,c).
12 Ars Math. Contemp. 24 (2024) #P2.04
On the other hand, we see by analogous computations that
sgn(ωkaa−1ω
kb
b−1ω
kc
c−1) = sgn(ω
ka
a−1ω
kb
b−1ω
kc
c−1
Ka,c
)
= sgn(ωkaa−1ω
kb
b−1ω
kc
c−1
Ka,c
ω
−kaKc,a
c−1 ω
kaKc,a
c−1 )
= sgn(αkaa,cω
ℓ′b
b−1ω
ℓ′c
c−1) = sgn(α
ka
a,cα
ℓ′
b,c)
for some ℓ′b, ℓ
′
c, ℓ
′ ∈ Z with
ℓ′b = kbKa,c = ℓ
′Kb,c,
ℓ′c = kcKa,c + kaKc,a.
We have chosen ka, kb odd and kc even. In this case, the greatest power of two that divides
kc is 2hc−ha+1. On the other hand, the greatest power of two that divides Kc,a and Kc,b is
2hc−ha . We can thus conclude from ℓKc,b = −kcKa,b and ℓ′Kb,c = kbKa,c that ℓ is even
while ℓ′ is odd since Ka,b,Ka,c and Kb,c are all odd. So we see that
sgn(ωkaa−1ω
kb
b−1ω
kc
c−1) = sgn(αa,b) = sgn(αa,cαb,c),
which proves the necessity of the condition.
Now suppose sgn(αa,bαa,cαb,c) = 1 for all a, b, c ≥ 0 with ha = hb ≤ hc. Let
K := (k0, . . . , kr) ∈ Zr+1 be such that
∑r
i=0 ki
mi
ni
= 0. Let suppK := {i | ki ̸= 0}
and n := | suppK|. We prove that Πri=0ω
ki
i−1 ∈ R is uniquely determined by induction on
n ≥ 2. We first suppose 0 ̸∈ suppK. We will deal with the case 0 ∈ suppK at the end of
our proof.
If n = 2, then Πri=0ω
ki
i−1 = ω
ki
i−1ω
kj
j−1 for some i, j > 0, and its value in the residue
field is a power of αi,j .
Now suppose n > 2. Take two distinct a, b ∈ suppK. As at least one of Ka,b and
Kb,a is odd, so suppose Ka,b is odd. Then
Πri=1ω
ki
i−1
Ka,b
= Πri=1ω
ki
i−1
Ka,b
ω
−kaKb,a
b−1 ω
kaKb,a
b−1 = Π
r
i=1ω
ℓi,1
i−1α
ka
a,b
with ℓa,1 = 0, ℓb,1 = kaKa,b + kbKb,a and ℓi,1 = kiKa,b for i ̸= a, b. Since |{i | ℓi,1 ̸=
0}| is strictly smaller than n, Πri=0ω
ℓi,1
i−1 ∈ R is uniquely determined by the induction
hypothesis. So we have determined Πri=0ω
ki
i−1
Ka,b
∈ R. As Ka,b is odd, Πri=0ω
ki
i−1 ∈ R is
determined as well.
We now need to show that in this way, Πri=1ω
ki
i−1 is uniquely determined, that is, if we
choose another a′, b′ ∈ suppK instead of a, b, we get the same value for Πri=1ω
ki
i−1 ∈ R.
We will show this by choosing c ∈ suppK \ {a, b} and proving that the evaluated value of
Πri=1ω
ki
i−1 is the same whether we factor a power of αa,b as above, or αa,c or αb,c instead.
By transitivity of the equality relation, this will imply that the obtained value of Πri=1ω
ki
i−1
is independent of the choice of a, b ∈ suppK. Suppose without loss of generality that
ha ≤ hb ≤ hc and that Ka,b,Ka,c and Kb,c are odd. Above, we have evaluated
Πri=1ω
ki
i−1
Ka,b
= Πri=1ω
ℓi,1
i−1α
ka
a,b
L. Vukšić: Valuations and orderings on the real Weyl algebra 13
with ℓa,1 = 0, ℓb,1 = kaKa,b + kbKb,a and ℓi,1 = kiKa,b for i ̸= a, b. We proceed by
evaluating, in the same way as before,
Πri=1ω
ℓi,1
i−1
Kb,c
= Πri=1ω
pi,1
i−1α
ℓb,1
b,c
with pa,1 = pb,1 = 0, pc,1 = ℓc,1Kb,c + ℓb,1Kc,b and pi,1 = ℓi,1Kb,c for i ̸= a, b, c. So
Πri=1ω
ki
i−1
Ka,bKb,c
= Πri=1ω
pi,1
i−1α
kaKb,c
a,b α
ℓb,1
b,c (3.2)
with ℓi,1 and pi,1 for all 0 ≤ i ≤ r as above. In particular, we see that for i ̸= a, b, c,
pi,1 = ℓi,1Kb,c = kiKa,bKb,c. Similarly, we can compute
Πri=1ω
ki
i−1
Ka,cKb,c
= (Πri=1ω
ℓi,2
i−1α
ka
a,c)
Kb,c = Πri=1ω
pi,2
i−1α
kaKb,c
a,c α
ℓb,2
b,c (3.3)
with
1. ℓa,2 = 0, ℓc,2 = kaKa,c + kcKc,a ℓi,1 = kiKa,c for i ̸= a, c, and
2. pa,2 = pb,2 = 0, pc,2 = ℓb,2Kc,b + ℓc,2Kb,c, pi,2 = kiKa,cKb,c for i ̸= a, b, c.
Let N := Ka,bKa,cKb,c. On one hand, we see from 3.2 that
Πri=1ω
ki
i−1
N
= Πri=1ω
pi,1
i−1
Ka,c
α
kaKb,cKa,c
a,b α
ℓb,1Ka,c
b,c , (3.4)
and on the other hand, we see from 3.3 that
Πri=1ω
ki
i−1
N
= Πri=1ω
pi,2
i−1
Ka,b
α
kaKb,cKa,b
a,c α
ℓb,2Ka,b
b,c . (3.5)
We need to show that in both equations, we get the same value. We first see that for all
i ̸= c, pi,1Ka,c = pi,2Ka,b. So, given that
r∑
i=1
pi,1
mi
ni
=
r∑
i=1
pi,2
mi
ni
= 0,
we can see pc,1Ka,c = pc,2Ka,b holds as well, and thus we conclude
Πri=1ω
pi,1
i−1
Ka,c
= Πri=1ω
pi,2
i−1
Ka,b
.
It then follows that
α
kaKa,cKb,c
a,b α
ℓb,1Ka,c
b,c = α
kaKb,cKa,b
a,c α
ℓb,2Ka,b
b,c ,
since both sides of the equation are equal to ωNkaa−1 ω
Nkb
b−1 ω
Nkc−pc,1Ka,c
c−1 and the signs of
αa,b, αa,c and αb,c were chosen so that the signs of both sides of the equality match. We
conclude that the value of Πri=1ω
ki
i−1
N
is the same in both 3.4 and 3.5. As N is odd (since
Ka,b, Ka,c and Kb,c are all odd), we conclude that Πri=1ω
ki
i−1 is the same whether we factor
a power of αa,b or αa,c. If we factored a power of αb,c, we would, as similar computations
as above would show, get the same value for Πri=1ω
ki
i−1.
14 Ars Math. Contemp. 24 (2024) #P2.04
We have now shown that if the condition of the proposition is fulfilled, Πri=1ω
ki
i−1 is
uniquely determined for all k1, . . . , kr ∈ Z with
∑r
i=1 ki
mi
ni
= 0.
Now we consider the case 0 ∈ suppK. Let K = (k0, . . . , kr) ∈ Zr+1 be such that
k0 ̸= 0 and
∑r
i=0 ki
mi
ni
= 0. Let N := gcd{ni | i ∈ suppK}. If N is odd, i.e., if ni is
odd for every i ∈ suppK, then Πri=0ω
ki
i−1 is uniquely determined. This is because
Πri=0ω
ki
i−1
N
= (Πri=0x
ki
mi
ni Πri=0ω
ki
i−1)
N = Πi=1(xmiωnii−1)
kici = Πri=1β
kici
i
where ci := Nni for each i ≤ i ≤ r. We thus conclude Π
r
i=0ω
ki
i−1 ∈ R is the uniquely
determined N -the real root of Πri=1β
kici
i . Now suppose nj is even for some j ∈ suppK.
Then mj must be odd since gcd(mj , nj) = 1. Let k′j := kj − k0nj and k′i := ki for all
i ∈ suppK \ {0, j}. Then
∑r
i=1 k
′
i
mi
ni
= 0 and
Πri=0ω
ki
i−1
mj
= (xmjωnj )k0(Πri=1ω
k′i
i−1)
mj = βk0j (Π
r
i=1ω
k′i
i−1)
mj .
We evaluate (Πri=1ω
k′i
i−1)
mj as above k0 = 0 and conclude that Πri=0ω
ki
i−1 ∈ R is the
unique mj-th real root of βk0j (Π
r
i=1ω
k′i
i−1)
mj .
This concludes the proof of our proposition.
In Lemma 3.6, we suppose that v is a valuation on D1(R) extended from (ωi)i≥−1 to
D1(R) and compute the value of certain elements of D1(R) in this case.
Lemma 3.5. Let D be a division ring endowed with a valuation v with an abelian value
group and a commutative residue field with characteristic zero. Let a, b ∈ D be such that
a ∼ b, v(a) = v(b) = 0 and v(ab) < v[a, b]. Then v(an − bn) = v(a − b) for all
n ∈ Z \ {0}. If there exist c, d ∈ D such that cn = a, dn = b, c = d for some n ∈ N, then
v(cm − dm) = v(a− b) for all m ∈ Z.
Proof. For n ∈ N, write an − bn = (a− b)
∑n−1
i=0 a
n−1−ibi + terms with higher v-value.
Since an−1−ibi is the same for all 0 ≤ i ≤ n − 1, the v-value of the sum is equal to
zero, proving the statement for positive integers n. For negative n ∈ Z, the statement
follows from an − bn = −an(a−n − b−n)bn. The last statement of the lemma follows
from a− b ∼ (c− d)
∑n−1
i=0 c
n−1−ibi.
Lemma 3.6. Suppose v is a valuation on A1(R) and suppose i1, i2, . . . , ir ∈ N and
k0 ∈ Z, ki1 , ki2 , . . . , kir ∈ Z \ {0} are such that v(xk0ω
ki1
i1−1 · · ·ω
kir
ir−1) = 0. If
min1≤j≤r{v(ωij )} is achieved at exactly one j, then
v(xk0ω
ki1
i1−1 · · ·ω
kir
ir−1 − xk0ω
ki1
i1−1 · · ·ω
kir
ir−1) = min{v(ωij ) | 1 ≤ j ≤ r}.
Proof. Let n be the least common multiple of ni1 , . . . , nir and cij =
n
nij
for each ij . Since
(xk0ω
ki1
i1−1 · · ·ω
kir
ir−1)
n = Πrj=1x
nkij
mij
nij (ω
ki1
i1−1 · · ·ω
kir
ir−1)
n
= Πrj=1(x
mijω
nij
ij−1)
kij cij = Πrj=1β
kij cij
ij
,
L. Vukšić: Valuations and orderings on the real Weyl algebra 15
by Proposition 2.2, we can compute
(xk0ω
ki1
i1−1 · · ·ω
kir
ir−1)
n −Πrj=1β
kij cij
ij
∼
r∑
j=1
(xmi1ω
ni1
i1−1)
ki1ci1 · · · (xmij−1ω
nij−1
ij−1−1)
kij−1cij−1 ·
((xmijω
nij
ij−1)
kij cij − β
kij cij
ij
)β
kij cij
ij+1
· · ·β
kij cij
ir
.
For each j such that kij > 0,
(xmijω
nij
ij−1)
kij cij − β
kij cij
ij
∼ ωij
kij cij−1∑
i=0
(xmijω
nij
ij−1)
kij cij−iβiij ,
which gives us v((xmijω
nij
ij−1)
kij cij − β
kij cij
ij
) = v(ωij ). In case kij < 0, we see that
(xmijω
nij
ij−1)
kij cij − β
kij cij
ij
=
− (xmijω
nij
ij−1)
kij cij β
kij cij
ij
((xmijω
nij
ij−1)
−kij cij − β
−kij cij
ij
),
which again implies v((xmijω
nij
ij−1)
kij cij − β
kij cij
ij
) = v(ωij ). We conclude, using
Lemma 3.5,
v(xk0ω
ki1
i1−1 · · ·ω
kir
ir−1 − xk0ω
ki1
i1−1 · · ·ω
kir
ir−1)
= v((xk0ω
ki1
i1−1 · · ·ω
kir
ir−1)
n − (xk0ωki1i1−1 · · ·ω
kir
ir−1)
n)
= min{v(ωij ) | 1 ≤ j ≤ r}.
With the help of Lemma 3.6, we will evaluate v(xk0ωki1i1−1 · · ·ω
kir
ir−1 −
xk0ω
ki1
i1−1 · · ·ω
kir
ir−1) when v(x
k0ω
ki1
i1−1 · · ·ω
kir
ir−1) = 0 in general. As in Lemma 3.6, we
assume k0 ∈ Z, kij ∈ Z \ {0} for all 1 ≤ j ≤ r. This will be helpful when we will
later construct a valuation v associated to a sequence (ωi)i≥−1. Let us assume for now that
i1 < i2 < · · · < ir; at the end of the calculation we will see that the order of ij does not
affect the v-value.
To start, we introduce some abbreviations to make the written equations easier to read.
Let n and cij for all j be as in the proof of Lemma 3.6,
A0 := x
k0ω
ki1
i1−1 · · ·ω
kir
ir−1
B0 := A0
A
(n)
0 := x
nk0ω
nki1
i1−1 · · ·ω
nkir
ir−1
B
(n)
0 := A
(n)
0 = Π
r
j=1β
kij cij
ij
.
Since B0 is in R, we can write
A0 −B0 = (An0 −Bn0 )(
n−1∑
i=0
An−i−10 B
i
0)
−1 ∼ (A(n)0 −B
(n)
0 )(
n−1∑
i=0
An−i−10 B
i
0)
−1. (3.6)
16 Ars Math. Contemp. 24 (2024) #P2.04
Since v(
∑n−1
i=0 A
n−i−1
0 B
i
0) = v(A
n−i−1
0 B
i
0) = 0 for all i, which holds due to A
n−i−1
0 B
i
0
= Bn−10 for all i, v(A0 −B0) = v(A
(n)
0 −B
(n)
0 ). To evaluate the right-hand side of (3.6),
we first proceed as we have done in the proof of Lemma 3.6, so
A
(n)
0 −B
(n)
0 ∼
r∑
j=1
Πj−1ℓ=1(x
miℓω
niℓ
iℓ−1)
kiℓciℓ · ((xmijω
nij
ij−1)
kij cij − β
kij cij
ij
)·
Πrℓ=j+1β
kij cij
iℓ
.
If kij > 0, we proceed by
(xmijω
nij
ij−1)
kij cij − β
kij cij
ij
∼ ωij
kij cij−1∑
p=0
(xmijω
nij
ij−1)
kij cij−p−1βpij .
For ij > 0, kij < 0, we can on the other hand write
(xmijω
nij
ij−1)
kij cij − β
kij cij
ij
∼ −ωij (x
mijω
nij
ij−1)
kij cij β
kij cij
ij
−kij cij−1∑
p=0
(xmijω
nij
ij−1)
−kij cij−p−1βpij .
We now define, if kj > 0,
Cj :=
kij cij−1∑
p=0
Πj−1ℓ=1(x
miℓω
niℓ
iℓ−1)
kiℓciℓ · (xmijω
nij
ij−1)
kij cij−p−1βpij ·Π
r
ℓ=j+1β
kiℓciℓ
iℓ
,
and, if kj < 0,
Cj := −(xmijω
nij
ij−1)
kij cij β
kij cij
ij
−kij cij−1∑
p=0
Πj−1ℓ=1(x
miℓω
niℓ
iℓ−1)
kiℓciℓ ·
(xmijω
nij
ij−1)
−kij cij−p−1βpij ·Π
r
ℓ=j+1β
kiℓciℓ
iℓ
for each 1 ≤ j ≤ r. Further, we define
A1,j = ωijCj
A
(n)
0 −B
(n)
0 ∼
r∑
j=1
ωijCj =
r∑
j=1
A1,j
for each 1 ≤ j ≤ r. Thus we can conclude that v(A1,j) = v(ωij ), since the image of
Cj in the residue field is equal kijcij · Πrℓ=1β
kiℓciℓ
iℓ
̸= 0 if kij > 0 and −β
2kij cij
ij
|kijcij | ·
Πrℓ=1β
kiℓciℓ
iℓ
̸= 0 if kij < 0 , making v(Cj) = 0. We can now write
A
(n)
0 −B
(n)
0 =
r∑
j=1
A1,j +A =
∑
v(A1,j) is minimal
A1,j +
∑
v(A1,j) is not minimal
A1,j .
L. Vukšić: Valuations and orderings on the real Weyl algebra 17
Here we note that the second of both finite sums on the right-hand side of this equation
includes A, which denotes the sum of all terms obtained by changing the order of factors
of the form ωiℓ (which was not explicitly written above). The fact that the v-value of these
terms is higher than the v-value of the terms of the first sum (the ones with minimal v-value)
follows from Lemma 3.2.
If min1≤j≤r{v(ωij )} is achieved at more than one j, we take the sum of all ωijCj
that have the minimal v-value, i.e.,
∑
v(A1,j) is minimal A1,j , then factor ωi1 , so the sum now
looks like∑
v(A1,j) is minimal
A1,j = ωi1
∑
v(A1,j) is minimal
ω−1i1 A1,j = ωi1
∑
v(A1,j) is minimal
ω−1i1 ωijCj
and, since v(ω−1i1 ωijCj) = 0, for each j, we can evaluate the sum of their images in the
residue field. If this sum is not equal to zero, then v(A0 − B0) = min1≤j≤r{v(A1,j)}.
Otherwise write
ωi1
∑
j
ω−1i1 ωijCj =
∑
j
ωi1(ω
−1
i1
ωijCj − ω−1i1 ωijCj).
For every j, we write ω−1i1 ωijCj − ω
−1
i1
ωijCj as an R-linear sum of terms of the form
Πℓω
nℓ
iℓ−1 (in the same way we did with A0 −B0). We sum all of the newly obtained terms,
as well as the terms in
∑
v(A1,j) is not minimal A1,j , and relabel them as A2,j where j goes
from 1 to the number of all terms.
As A0 −B0 can be written in the form
A0 −B0 = (
∑
v(A2,j) is minimal
A2,j +
∑
v(A2,j) is not minimal
A2,j)D,
where we use D as the label of the product of all terms of the form (
∑
i A
k−iBi)−1 where
A = xk
′
0Πrj=1ω
k′ij
ij−1 for some i1, . . . , ir ∈ N, k
′
0, k
′
i1
, . . . , k′ir ∈ Z and B = A, that we
factor out when we evaluate ω−1i1 ωijCj −ω
−1
i1
ωijCj for each j. All terms (
∑
i A
k−iBi)−1
have v-value equal to zero and their image in the residue field, which is of the form
(mBm−1)−1 ∈ R for some m ∈ N, is easy to determine.
We repeat the described procedure, writing A0 − B0 = (
∑
j Ak,j)D for increasing k.
We stop when for some k,
∑
j,v(Ak,j) is minimal Ak,j is either composed of one single term
or, after factoring out one of the terms, the image of the sum in the residue field is not zero.
In this case, we conclude that v(A0 − B0) is equal to v(Ak,j) for any term of the sum∑
j,v(Ak,j) is minimal Ak,j .
We must show that the process ends at some point even if the number of terms whose v-
value we evaluate at each step is growing. We see that whenever we write xk0Πrℓ=1ω
kℓ
iℓ−1−
xk0Πrℓ=1ω
kℓ
iℓ−1 as a sum of terms with strictly positive v-value, the value of each of these
terms is v(ωiℓ) for some ℓ = 1, . . . , r. It follows that v(A0 − A0) is a sum of v(ωℓ) for
some ℓ ≥ 1.
If v(ωN ) is irrational for some N ≥ 0, the process either stops beforehand or, after
k ≥ N − ir steps we get a unique term Ak,j that has v-value equal to v(ωirωir+1 · · ·ωN ).
This is the term we get when we take the last term of A(n)0 − B
(n)
0 , written as a sum of
terms A1,j with higher v-value and in each of the following steps whenever the v-value of
18 Ars Math. Contemp. 24 (2024) #P2.04
this term is minimal, take the last term when Ak,j is written as a sum of terms with higher
v-value.
If on the other hand, v(ωk) ∈ Q for an infinite sequence (ωk)k≥−1, then
limk→∞ v(ωk) = 0 since by Lemma 3.2,
∑k
i≥−1 v(ωk) < 0 for all k ≥ −1 and v(ωi) > 0
for i ≥ 1. Then for some N ≥ 1, v(ωN ) < v(ωi) for all 1 ≤ i < N . The evaluation of
v(A0 − B0) again either stops beforehand or we get a unique term Ak,j that has v-value
equal to v(ωirωir+1 · · ·ωN ). As in the first case, this term is the one we get when we take
the last term of A(n)0 − B
(n)
0 , written as a sum of terms A1,j with higher v-value and in
each of the following steps whenever the v-value of this term is minimal, take the last term
when Ak,j is written as a sum of terms with higher v-value.
In both cases, the value of this term, v(ωirωir+1 · · ·ωN ) is strictly smaller than the
value of all additional terms we get when we change the order of the factors in a product.
It follows that given the v-values of (ωi)i≥−1, v(A0 − B0) is the same as it would be if
all elements of the sequence (ωi)i≥−1 commuted. This follows from the fact that when we
change the order of factors ωi and ωj in some Ak,j1 , the term we obtain has v-value greater
by v[ωi, ωj ] − v(ωiωj) = −v(xyω1 · · ·ωj) for −1 ≤ i < j ≤ N while v(Ak+1,j1) −
v(Ak,j2) is for any j1, j2 equal to v(ωiℓ+1) for some iℓ such that Ak,j2 contains a power
of ωiℓ . Since, as we have shown in the proof of Lemma 3.2,
∑N
i=−1 v(ωi) < 0, it follows
that the terms we obtain by changing the order of the factors have v-value greater than
v(A0 − B0). And since v(xk0ω
ki1
i1−1 · · ·ω
kir
ir−1 − xk0ω
ki1
i1−1 · · ·ω
kir
ir−1) is higher than the
v-value of any terms we get when we change the order of factors, the order of ωi1 , . . . , ωir
does not matter.
3.2 Extending v from the sequence (ωi)i≥−1 to A1(R)
We can now prove that every v associated to either a finite or an infinite sequence (ωi)i≥−1
can be extended to a valuation on D1(R).
Lemma 3.7. For every r ≥ 0, there exists a finite number M of elements of the form
ai := ω
ki,−1
−1 ω
ki,0
0 ω
ki,1
1 · · ·ω
ki,r−1
r−1 for some ki,1, . . . , ki,r−1 ∈ Z such that v(ai) = 0 for
all 1 ≤ i ≤ M and every ωℓ−1−1 · · ·ω
ℓr−1
r−1 ∈ D1(R) with v-value zero is ∼-equivalent to a
product of positive integer powers of a1, . . . , aM .
Proof. Since v(ω−1) = −1 and v(ωj) = mj+1nj+1 ∈ Q for j ≥ 0, the problem translates to
finding general classes of solutions to the diophantine equation x0a0 + · · ·xkak = 0 with
a0 = −Πkj=1ni and ai = mini Π
k
j=1ni for all 0 ≤ i ≤ k.
Theorem 3.8. Let v and (ωi)i≥−1 be as described in the beginning of the section, i.e.,
ω−1 = x, ω0 = y, v(ω−1) = −1, v(ωi) = mi+1ni+1 ∈ Q, x
mi+1ω
ni+1
i = βi+1 ∈ R,
ωi+1 = x
mi+1ω
ni+1
i − βi+1 for i, 0 ≤ i ≤ N − 1 and v(ωN ) ̸∈ Q for some N ≥ 0 or
v(ωi) ∈ Q for infinitely many i, that
∑k
i=−1 v(ωi) < 0 for all k ≥ −1. Suposse that sgnβi
is constant on the set of all i for which ni is even. Then v can be extended to a valuation
on D1(R) with residue field R. The valuation is unique for every choice of {αi,j}i,j≥0
where αi,j = ω
Ki,j
i−1 ω
Kj,i
j−1 ,Ki,j =
mjni
di,j
,Kj,i = −minjdi,j with di,j = gcd{mjni,minj}.
The associated value group is group-isomorphic to a subgroup of Q × Z generated by
{v(ωi)}i≥−1.
L. Vukšić: Valuations and orderings on the real Weyl algebra 19
Proof. The following construction of the valuation v associated to the sequence (ωi)i≥−1
was first sketched in [22]. Here we present it in full detail.
Before we begin with the construction of the v-value for an arbitrary element of D1(R),
we define it for some specific elements of D1(R).
1. Since we have defined v(ωi) for all −1 ≤ i ≤ N , v(ΠNi=−1ω
ki
i ) =
∑N
i=−1 kiv(ωi)
must hold for all k−1, . . . , kN ∈ Z.
2. Since we supposed
∑k
i=−1 v(ωi) < 0 for all k ≥ −1, it follows from Lemma 3.2
that
v[ωi, ωj ] = −v(ω−1ω0 · · ·ωi−1ωi+1 · · ·ωj−1),
which must be strictly greater than v(ωiωj) for all i < j.
3. We also see that if v(ωk0−1 · · ·ω
kr
r−1) = 0 for some k0, . . . , kr ∈ Z, then ω
k0
−1 · · ·ω
kr
r−1
is uniquely determined by {αi,j}i,j≥0 as in shown in Proposition 3.4. In this case,
v(ωk0−1 · · ·ω
kr
r−1−ω
k0
−1 · · ·ω
kr
r−1) must be equal to the value determined in Lemma 3.6
and the discussion following it.
In all three cases, the chosen values were the only possible extensions of v from (ωi)i≥−1
if we want v to be a valuation.
To determine v(F ) for any F ∈ A1(R), we first note that F can be written as a finite
sum
F =
∑
ℓ
αℓx
iℓyjℓ , αℓ ∈ R.
Let F1 be the sum of all terms αℓxiℓyjℓ such that iℓv(x) + jℓv(y) is equal to u :=
minℓ{iℓv(x) + jℓv(y)}.
If F1 consists of only one such term, then we define v(F ) = u; this is obviously always
the case whenever v(y) ̸∈ Q. Otherwise, we factor out xi1yj1 with the smallest power of x
and get
F1 ∼ xi1yj1
∑
ℓ, iℓv(x)+jℓv(y)=u
αℓx
iℓ−i1yjℓ−j1 .
Since
(iℓ − i1)v(x) + (jℓ − j1)v(y) = 0,
for each ℓ in the sum, iℓ−i1jℓ−j1 = Kℓ
m1
n1
for some Kℓ ∈ Z. We can write
F1 ∼ xi1yj1f(xm1yn1)
where f(t) is a polynomial in R[t]. Since we know v(ω1) > 0 and v(α) = 0 for α ∈ R∗, v
is uniquely determined on R[ω1]. From this, it follows that v(f(xm1yn1)) = 0 if and only
if f(β1) ̸= 0 since xm1yn1 = ω1 + β1.
In this case, v(F1) = u and since all terms in F − F1 have v-value strictly greater than
u, v(F ) = u must hold. Since v(u) is a sum of integer powers of v(x) and v(y), v(F ) is in
20 Ars Math. Contemp. 24 (2024) #P2.04
the abelian group, generated by {v(ωi)}i≥−1. If u = 0, F = βk1f(β1) ∈ R. If on the other
hand f(β1) = 0, write f(t) = g1(t)(t− β1)k1 with g1(β1) ̸= 0 and we have
F1 ∼ xi1yj1g1(xm1yn1)ωk11 .
We set v(F1) = u + k1v(ω1) and add all terms we get from exchanging the order of
factors, whose v-value can be lower than the newly set v(F1), although still strictly higher
than u due to v[x, y] > v(xy), to F − F1. It is immediate that v(F1) is in the subgroup
of Γ, generated by v(x), v(y) and v(ω1) and that if v(F1) = 0, F1 ∈ R. It is important
to note that in both cases, we consider F1 as a single term. It follows that during our
transformation, the number of terms (if we ignore the ones we got when we changed the
order of factors in a product) is strictly smaller than before (unless, of course, F1 was just
a single term in the beginning and we get v(F ) = v(F1)).
We now consider the values of the terms in F − F1. If all of them have v-value strictly
greater than that of F1, we conclude v(F ) = v(F1). Otherwise, we take all terms of
F − F1 =
∑
ℓ′, iℓ′v(x)+jℓ′v(y)>u
αℓ′x
iℓ′ yjℓ′ , αℓ′ ∈ R
for which iℓ′v(x) + jℓ′v(y) = u′ := minℓ′{iℓ′v(x) + jℓ′v(y)} and then as before define
F2 = x
i2yj2
∑
ℓ′, iℓ′v(x)+jℓ′v(y)=u
′
αℓ′x
iℓ′−i2yjℓ′−j2 .
As above, we write F2 ∼ xi2yj2g2(xm1yn1)(x − β1)k2 , k2 ≥ 0, g2(β1) ̸= 0 and add all
the terms we get when we change the order of factors to F − F1 − F2. Their v-value is
strictly greater than u′ due to v[x, y] > v(xy).
We continue this process, defining F1, F2, . . . , Fk until all terms in F − F1 − · · · − Fk
have v-value strictly greater than min{F1, . . . , Fk}. Note that it is possible that Fk consists
of only one term from F − F1 − · · · − Fk−1.
Afterwards, we sum together all those Fi for 1 ≤ i ≤ k for which v(Fi) = u1 :=
min{v(F1), . . . , v(Fk)}. If the minimum is achieved at exactly one such Fi, we set v(F ) =
v(Fi). This is always the case whenever v(ω1) ̸∈ Q. Otherwise we can relabel the terms so
the minimum is achieved at F1, . . . , Fr for some r ≤ k. As we have shown, each Fi can be
written as Fi = xiiyjigi(xm1yn1)ωki1 . We sum the terms together, factor out x
i1yj1ωk11 ,
the term that has, written as a polynomial in x and y, the lowest power of x, and label the
new sum F1,1.
To evaluate v(F1,1), we follow a procedure similar to the one evaluating v(F1). After
factoring xi1yj1ωk11 , we are left with
F1,1 ∼ xi1yj1ωk11 (g1(xm1yn1) + g2(xm1yn1)xi2−i1yj2−j1ω
k2−k1
1 + · · ·
+ gr(x
m1yn1)xir−i1yjr−j1ωkr−k11 )
∼ xi1yj1ωk11
R∑
J=1
αJx
iJ yjJωkJ1 , with αJ ∈ R, iJ , jJ , kj ∈ Z, R ≥ 1.
Each term in the sum has v-value zero. Let a1, a2, · · · , aℓ be the terms such that each
product of the form xiyjωk1 , i, j, k ∈ Z that fulfills the condition v(xiyjωk1 ) = 0 is a
L. Vukšić: Valuations and orderings on the real Weyl algebra 21
product of positive integer powers of some of ai up to the order of factors x, y and ω1. The
existence of a1, . . . , aℓ is assured by Lemma 3.7. We can then write
F1,1 ∼ xi1yj1ωk11
R∑
J=1
γJa
m1,J
1 a
m2,J
2 · · · a
mℓ,J
ℓ
∼ xi1yj1ωk11 g(a1, · · · , aℓ), g ∈ R[t1, · · · , tℓ] (3.7)
with γJ ∈ R, mI,J ∈ Z for all 1 ≤ I ≤ ℓ and 1 ≤ J ≤ R. As before, we add all terms
we get when we change the order of multiplication of x, y or ω1 in a product to F − F1,1
since the value of its terms is strictly greater than u1. Since, as we have determined in the
beginning, each term in the sum (3.7) has v-value equal to zero and we know what ai ∈ R
is for each 1 ≤ i ≤ ℓ, v(g(a1, . . . , aℓ)) will have to be greater than or equal to zero, we can
define g(a1, . . . , aℓ) = g(a1, . . . , aℓ).
If g(a1, . . . , aℓ) ̸= 0, then we set v(g(a1, . . . , aℓ)) = 0 and v(F ) = v(F1,1) =
i1v(x) + j1v(y) + k1v(ω1). Otherwise write
g(t1, . . . , tℓ) = h(t1 − a1, . . . , tℓ − aℓ) =
L∑
i=1
Πℓj=1(tj − aj)mi,jhi(t1, . . . , tℓ)
with h, h1, . . . , hL ∈ R[t1, . . . , tn] and hi(a1, . . . , aℓ) ̸= 0 for all i. We factor out the
Πj(tj − aj)mi,j for those i for which
∑
j v(aj − aj)mi,j is minimal. Then
g(t1, . . . , tℓ) = Πj(tj − aj)mi,j g̃(t1, . . . , tk), with g̃ ∈ R[t1, . . . , tℓ].
If g̃(a1, . . . , ak) ̸= 0, we set
v(F1,1) = v(x
i1yj1) +
∑
j
mi,jv(aj − aj).
If on the other hand, g̃(a1, . . . , ak) = 0, we do the same thing as we did with g. The
process cannot go on indefinitely since g is a polynomial and hence of finite degree. All
terms we get when we exchange the order of x, y and ω1 are added to F − F1,1. Their
v-value must be strictly greater than u1. It follows from the construction that v(F1,1) must
be in the group generated by {v(ωi)}i≥−1 since this holds for v(ai − ai) for all i and that
if v(F1,1) = 0, F1,1 ∈ R.
Since v(ai − ai) is, as we have shown in Lemma 3.6 and the discussion following it, a
sum of v(ωj) and thus v(Πjω−1j (ai − ai)) = 0, we can write F1,1 as one term of the form
Πni=−1ω
ki
i g(a1, a2, . . . , aℓ) with n ∈ N and g ∈ R[t1, . . . , tℓ] and g(a1, a2, . . . , aℓ) ̸= 0.
After v(F1,1) is set, we compare it to both v(Fi) for all Fi that are not part of F1,1
and the terms of F − F1 − · · · − Fk − F1,1. If all of these terms have v-value strictly
greater than v(F1,1), then we can set v(F ) = v(F1,1). Otherwise, we collect all terms with
minimal v-value in a sum which we label F1,2. We determine v(F1,2) in the same way we
determined F1,1 and then sum all of the remaining terms that have v-value less or equal to
min{v(F1,1), v(F1,2)} to a sum labeled F1,3.
We repeat the process until for some k, min{v(F1,1), · · · , v(F1,k)} is strictly smaller
than the v-value of any of the remaining terms.
If min{v(F1,1), · · · , v(F1,k)} is achieved at exactly one i, we set v(F ) = v(F1,i).
Otherwise we sum all the terms with the minimal v-value and label the sum F2,1. We
22 Ars Math. Contemp. 24 (2024) #P2.04
evaluate v(F2,1) in the same way we evaluated v(F1,1). We repeat the process, defining
Fi,j and determining its v-value in the same way as above. We point out that after v(Fi,j)
is defined, we regard Fi,j as one single term in future evaluations.
Now we must show that at one point, the process ends, i.e., that for some i, j, v(Fi,j) is
strictly smaller than the v-value of all other terms. This holds because each time we define
Fi,j for some i, j, we sum a number of different terms into one single term and because
whenever we change the order of factors in a term, the degree of x and y in the difference
is strictly smaller. This means that we eventually run out of terms. We have thus defined v
for an arbitrary polynomial F ∈ A1(R). What we essentially did was that we wrote
F = F̃ + F̃1
where F̃ is written as a single term, v(F̃ ) is computed as if x and y commuted and the
v-value of each term of F̃1 is strictly greater than v(F̃ ). For another G ∈ A1(R), we can
write
FG = F̃G+ ˜(FG)1
and since we evaluate v(F̃ ) and v(G̃) as if x and y commuted, v(F̃G) = v(F̃ ) + v(G̃).
We use the same reasoning to show v(F +G) ≥ min{v(F ), v(G)}.
It follows from the construction that for each i, j, v(Fi,j) is a linear combination of
{v(ωi)}i≥−1 and that in case v(Fi,j) = 0, Fi,j ∈ R.
Theorem 3.9. Let v be a valuation on A1(R) trivial on R with residue field R. Then v is
strongly abelian.
Proof. If v’s value group is Q, then the theorem follows from Corollary 2.4. Otherwise,
v(ωN ) ̸∈ Q for some N by our construction. But as we have shown in Lemma 3.2,
v[ωN , x] = −v(yω1 · · ·ωN−1). If v(ωNx) = v[ωN , x], it follows that v(ωN ) ∈ Q, a
contradiction. Since the value group is generated by {v(ωi)}i≥−1, it follows from Propo-
sition 2.7 that v is strongly abelian.
4 Valuations on R[y; δ]
In this section, we explain a construction of valuations on the ring R[y; δ] with
R := {
∑
k≥m
akx
− kn | ak ∈ R,m ∈ Z, n ∈ N}
and δ(p(x)) = p′(x). This construction, which was first introduced in [16], will, as we
will see in this section, give us all valuations on R[y; δ] with residue field R. Then, we
will prove exactly which valuations on A1(R) with residue field R extend to a valuation on
R[y; δ] with the same residue field, answering the question posed by Marshall and Zhang
in [16]. We will see the extensions of valuations on R[y; δ] are strongly abelian.
Every valuation on R[y; δ] can be uniquely extended to its quotient ring, which we label
as D, because R[y; δ] is an Ore domain. Since [y, x] = 1 as before, v(xy) < 0 must hold.
We set v(x) = −1, z0 := y and consider v(y). If v(y) ̸∈ Q, then
v(
n∑
i=0
pi(x)y
i) = min
0≤i≤n
{v(pi(x)) + iv(y)}
L. Vukšić: Valuations and orderings on the real Weyl algebra 23
for any
∑n
i=0 pi(x)y
i ∈ R[y; δ]. Otherwise v(y) = r1 ∈ Q and hence v(y − γ1x−r1) >
v(y) for some γ1 ∈ R. If v(z1) = r2 ∈ Q for z1 := y − γ1x−r1 , we proceed to find
z2 = z1 − γ2x−r2 such that v(z2) is greater than r2. We repeat this process to construct a
sequence (zi)i≥0.
If v(zk) ̸∈ Q for some k ∈ N, then we can write every f ∈ R[y; δ] as
∑n
i=0 pi(x)z
i
k
and deduce
v(f) = min
0≤i≤n
{v(pi(x)) + iv(zk)}.
The value group is then group-isomorphic to Q×Z. Since v[x, zk] = v[x, y] = 0 > v(xzk),
v is strongly abelian by Proposition 2.7. Otherwise, the sequence (zi)i≥0, v(zi) = ri+1 ∈
Q is infinite. We take note of the fact that v(zi+1) > v(zi) and since [zi, x] = [y, x] = 1
for all i, v(zi) < 1 for all i. We define r := limi→∞ ri ≤ 1.
4.1 Case r < 1
If r < 1, it has been shown in [16] that v can be extended to a valuation on R[y; δ] with
residue field R. We first extend v from R to
R̃ = {
∑
q∈A
aqx
−q | aq ∈ R, A ⊂ Q is well-ordered}
in a natural way, i.e., by defining
v(
∑
q∈A
aqx
−q) = minA
for each
∑
q∈A aqx
−q ∈ R̃[y; δ]. Then for every f(t) ∈ R[t], define v(f(y)) = v(f(z))
with z :=
∑∞
i≥1 aix
−ri and f(y) =
∑n
i=0 piy
i for f(t) =
∑n
i=0 pit
i. This gives rise to a
valuation on R[y; δ].
However if r = 1, we cannot define a valuation in this way. Let k ∈ N be such that
2rk > 1 + r1, which exists since r = 1, and ak = y − zk =
∑k
i=1 γix
−ri . Let
f(t) = (t− ak)(t− ak) = t2 − 2tak + a2k ∈ R[t].
On one hand,
v(f(y)) = v(f(z)) = 2v(z − ak) = 2rk+1.
On the other,
2rk+1 = v((y−ak)(y−ak)) = v(y2−2yak+a2k+[y, ak]) = min{2rk+1, 1+r1} = 1+r1,
contradicting the assumption that v is a valuation, as shown in [16].
Of course, even in case r < 1, there may also exist a k such that 2rk > 1 + r1. But the
important difference between the two cases is that if r < 1, there is always an ℓ ∈ N such
that 1 + rℓ > 2rk for all k ∈ N, which does not hold in case r = 1. Then, since
f : R[y; δ] → R[y; δ], f(y) = zℓ, f(a) = a for a ∈ R
is a real algebra automorphism of R[y; δ], we can translate the sequence by replacing y
with zℓ.
We see that since the associated value group is Q, v is a strongly abelian valuation by
Corollary 2.4.
24 Ars Math. Contemp. 24 (2024) #P2.04
4.2 Case r = 1
The question whether in case r = 1, v can be extended from a sequence (zi)i≥0 to a
valuation on R[y; δ] was left open in [16]. In this subsection, we show that it can be done
using model theory (for reference, see for example [19]). We also show that the valuation
we get in this way is uniquely determined.
Suppose we have infinite sequences (zi)i≥0 ⊆ R[y; δ], (ri)i≥1 ⊆ Q and (γi)i≥1 ⊆ R
and v : (zi)i≥0 → Q with z0 = y, zi+1 = zi − γi+1x−ri+1 and v(zi) = ri+1 ∈ Q with
(ri)i≥1 a strictly increasing sequence with r = limi→∞ ri = 1. Then for each n ≥ 0, there
is a valuation vn on R[y; δ] such that vn(zi) = ri+1 for all 0 ≤ i ≤ n− 1 and vn(zn) ̸∈ Q.
We now present the first-order theory that the valuation associated to the infinite se-
quence we wish to prove exists is a model of. The theory will be a union of the theory of
D, the quotient division ring of R[y; δ] and the theory of valuations. We will see that each
finite subset of this theory has a model. By compactness, so does the whole theory.
The language of our theory will be
F ∪ {+,−, ·,−1 , O,<} ∪ {czi | i ≥ −1}
where F is the set of all constants ca for each a ∈ D, +, · and < are binary function
symbols, − and −1 are unary function symbols, O is an unary relation symbol and czi is a
constant for all i ≥ 0. Let A be the theory of the quotient division ring of the ring R[y; δ].
By B we will denote the set of axioms for valuation rings O on division rings:
1. O(0) ∧O(1)
2. ∀a : O(a) ∨O(a−1)
3. ∀a, b : O(a) ∧O(b) → O(a+ b) ∧O(ab) ∧O(ba)
We add all sentences C that will give proper meaning to the constants czi for all i ≥ −1:
4. cz0 = cy
5. czi+1 = czi − cγi+1x−ri+1
6. O(cxri+1zi) ∧O(c(xri+1zi)−1)
Our theory is then the union of all the above axioms from A to C. Since all finite
subsets of the theory have a model, namely, the valuation vn described in the beginning of
this subsection, so does, by compactness, the whole theory. Since the theory contains F ,
the set of all constants ca for each a ∈ D, the models are valued division rings which all
contain D. We pick a model of the theory, a pair (D1, v), where D1 is a division ring with
valuation v.
We now show that the v-value is uniquely determined for every f ∈ R[y; δ]. It will
then follow that v is uniquely determined on the whole quotient ring D. Every f ∈ R[y; δ]
can be written as
f =
n∑
i=0
p
(0)
i (x)y
i, with p(0)i (x) ∈ R for each 0 ≤ i ≤ n.
L. Vukšić: Valuations and orderings on the real Weyl algebra 25
For the time being, we ignore the terms we get when we change the order of multipli-
cation. At the end of this subsection, we will see that they do not influence v(f). For each
k ≥ 1, we define
ak := y − zk =
k∑
i=1
γix
−ri
and write
f =
n∑
i=0
p
(0)
i (x)y
i =
n∑
i=0
p
(0)
i (x)(ak + zk)
i
∼
n∑
i=0
p
(0)
i (x)
i∑
j=0
(
i
j
)
ai−jk z
j
k =
n∑
j=0
p
(k)
j (x)z
j
k
with
p
(k)
j (x) :=
n∑
i=j
(
i
j
)
p
(0)
i (x)a
i−j
k
for each 0 ≤ j ≤ n and k ≥ 1. For each 0 ≤ j ≤ n, gj(t) :=
∑n
i=j
(
i
j
)
p
(0)
i (x)t
i is a
polynomial in R[t]. The quotient field of
C := {
∑
k≥m
akx
− kn | ak ∈ C,m ∈ Z, n ∈ N}
is the algebraic closure of the quotient field of R, as shown in for example [14]. Since∑∞
i=1 γix
−ri is not in the quotient field of C, it is not a root of gj(t) for any 0 ≤ j ≤ n.
We conclude that for some K ∈ N, v(p(k)j (x)) = v(p
(K)
j (x)) for all k ≥ K and all
0 ≤ j ≤ n. We can then write
f =
n∑
i=0
p
(0)
i (x)y
i =
n∑
i=0
p
(K)
i (x)z
i
K =
n∑
i=0
p
(k)
i (x)z
i
k
with v(p(k)i (x)) = v(p
(K)
i (x)) for all k ≥ K.
We now show that from some K ′ ≥ K, v(p(k)0 ) < v(p
(k)
i z
i
k) for all 1 ≤ i ≤ n and all
k > K ′. For all k > K,
p
(k+1)
0 (x) =
n∑
i=0
p
(k)
i (x)(ak+1 − ak)
i =
n∑
i=0
p
(k)
i (x)(γk+1x
−rk+1)i
with v(p(k)i (x)) = v(p
(K)
i (x)). Since (ri)i≥1 is an increasing sequence with limi→∞ ri =
1, there exists K ′ ≥ K such that
min
i=0,...,n
{v(p(K
′)
i (x)) + irK′+1} = min
i=0,...,n
{v(p(K)i (x)) + irK′+1}
is achieved at exactly one 0 ≤ i ≤ n. We conclude v(p(K
′+1)
0 ) = v(p
(K)
0 ). Then for each
1 ≤ i ≤ n,
v(p
(k)
i (x)z
i
k) = v(p
(k)
i (x))+irk+1 = v(p
(K)
i (x))+irk+1 > v(p
(K)
i (x))+irk ≥ v(p
(K)
0 ).
26 Ars Math. Contemp. 24 (2024) #P2.04
To show that v(f) is equal to v(p(K)0 (x)) ∈ Q, we must show that the v-value of all
the terms we get when we change the order of multiplication must be strictly greater than
v(p
(K)
0 (x)). For all k ≥ 1, we write
p
(k)
0 (x) =
n∑
i=0
p
(0)
i (x)a
i
k =
n∑
i=0
p
(0)
i (x)(
k∑
j=1
γjx
−rj )i =
n∑
i=0
ki∑
j=1
p
(0)
i (x)αix
−qj ,
with qj ∈ Q for all 1 ≤ j ≤ ki and 1 ≤ i ≤ n. Since for all 0 ≤ i ≤ n, p(0)i (x) ∈ R and
(ri)i≥1 is an increasing sequence with limi→∞ ri = 1, there exists some k ≥ 1 such that
the term of the sum with v-value
min
i=1,...,n
{v(p(0)i (x)) + (i− 1)r1 + rk}
is the only term in the sum with its v-value. We conclude
v(p
(K)
0 (x)) = v(p
(k)
0 (x)) ≤ v(p
(0)
i (x)) + (i− 1)r1 + rk (4.1)
for all 1 ≤ i ≤ n. On the other hand, we can write
f =
n∑
i=0
p
(0)
i (x)y
i =
n∑
i=0
p
(0)
i (x)(ak + zk)
i.
For all 0 ≤ i ≤ n, all terms of (ak + zk)i when expaned are of the form aℓ1k z
ℓ2
k . . . a
ℓi−1
k z
ℓi
k
with ℓ1, . . . , ℓi ≥ 0 and ℓ1 + · · ·+ ℓi = i. Since v(ak) = r1, v(zk) = rk+1 and
v[ak, zk] = v[
k∑
j=1
γjx
−rj , y −
k∑
j=1
γjx
−rj ] = v(
k∑
j=1
γj [x
−rj , y]) = 1 + r1,
it follows that the v-value of each term we get when we change the order of multiplication
is at least v(p(0)i (x)) + (i − 1)r1 + 1 for each 1 ≤ i ≤ n, which is, as is immediate from
(4.1), strictly greater than v(p(K)0 (x)). We thus conclude v(f) = v(p
(K)
0 (x)) ∈ Q. It
also follows that v(f) = vk(f) for all k ≥ K ′ with vk as defined in the beginning of this
section. As every element of D, the quotient ring for R[y; δ], can be written as fg−1 with
f, g ∈ R[y; δ], it follows that v is uniquely determined on D. We see that the value group
for v is equal to Q. We conclude from Corollary 2.4 that v is strongly abelian.
It remains to show that the residue field for v is equal to R. Suppose v(f) = 0 for some
f ∈ D. Then vk(f) = 0 for all k ≥ K for some K ∈ N. We can write
f = (
m∑
i=0
p
(K)
i z
i
K)(
n∑
j=0
q
(K)
j z
j
K)
−1
with p(K)i , q
(K)
j ∈ R and
v(
m∑
i=0
p
(K)
i z
i
K) = v(p
(K)
0 ) = v(
n∑
j=0
q
(K)
j z
j
K) = v(q
(K)
0 ) = q ∈ Q.
It follows that v(xq
∑m
i=0 p
(K)
i z
i
K) = v(x
q
∑n
j=0 q
(K)
j z
j
K) = 0 and α := xq
∑m
i=0 p
(K)
i z
i
K
= xqp
(K)
0 ∈ R = R, β := xq
∑n
j=0 q
(K)
j z
j
K = x
qq
(K)
0 ∈ R = R. We conclude
f = αβ−1 ∈ R. So the residue field for v is indeed equal to R.
L. Vukšić: Valuations and orderings on the real Weyl algebra 27
4.3 Extensions of valuations from A1(R) to R[x; δ]
In this section, we characterize the valuations on A1(R) with residue field R that have an
extension to R[y; δ] with the same residue field.
Since x
mi
ni ∈ R[y; δ], it follows that any valuation v′ that extends v to a valuation on
R[y; δ] with residue field R must satisfy v′(x
mi
ni ωi−1) = 0 and γ̃i := x
mi
ni ωi−1 ∈ R. In
the next proposition, we show the necessary condition for a valuation v on A1(R) to have
an extension v′ to R[y; δ] with the same residue field.
Proposition 4.1. Let v be a valuation on A1(R) with residue field R associated to a se-
quence (ωi)i≥−1 with v(ωi−1) = mini for i ≥ 1 and x
miωnii−1 = βi ∈ R. Let αi,j ∈ R
be as in Section 3. Let 2hi be the greatest power of two dividing ni for all i ≥ 0. Then
v can be extended to a valuation on R[y; δ] with the same residue field only if it fulfils the
following conditions:
(1) For each i such that ni is even, βi > 0 must hold, and
(2) for each i, j, ℓ with hi < hj ≤ hℓ, αi,jαi,ℓ > 0 must hold.
Proof. Since
γ̃i
ni = (x
mi
ni ωi−1)ni = xmiω
ni
i−1 = βi
due to Proposition 2.2, it is obvious that γ̃i must be equal to an ni-th root of βi. If ni is
odd, γ̃i ∈ R is uniquely determined regardless of sgn(βi), while if ni is even, γ̃i ∈ R only
if βi > 0. It is thus obvious that βi > 0 must hold for all i where ni is even if v can be
extended from a valuation on A1(R) to a valuation on R[y; δ] with the same residue field.
This proves the necessity of the first condition.
To prove the necessity of the second condition, we first observe that
αi,j = ω
Ki,j
i−1 ω
−Kj,i
j−1 = γ̃i
Ki,j γ̃j
−Kj,i ,
holds for all i, j ≥ 0. If hi < hj , Ki,j is odd while Kj,i is even, so sgn(γi) = sgn(αi,j)
must hold. We can therefore see that
αi,jαi,ℓ = γ̃i
Ki,j+Ki,ℓ γ̃j
−Kj,i γ̃ℓ
−Kℓ,i
for all i, j, ℓ ≥ 0. If hi < hj ≤ hℓ, both Ki,j and Ki,ℓ are odd while Kj,i and Kℓ,i are
even. It follows that if v can be extended to a valuation on R[y; δ] with residue field R,
αi,jαi,ℓ > 0 must hold for all i, j, ℓ ≥ 0 with hi < hj ≤ hℓ.
In this section, we show that the conditions (1) and (2) of Proposition 4.1 are also
sufficient for v to have an extension to R[y; δ] with residue field R. Let v be any valuation
on A1(R) satisfying the conditions described in Proposition 4.1. We will first determine
γ̃i ∈ R for all i ≥ 0. If ni is odd, there is a unique choice of γ̃i ∈ R. Suppose then ni
is even and βi > 0 for some i ≥ 0. If hi < hj for some j, then sgn(γ̃i) = sgn(αi,j) =
sgn(γ̃i
Ki,j γ̃j
−Kj,i) since Ki,j is even while Kj,i is odd.
We can conclude that if for every power of two 2h there is an i ≥ 0 (or, equivalently,
if the v-value group is 2-divisible), γ̃i ∈ R is uniquely determined for all i ≥ 0. If on the
other hand, the value group is non-2-divisible, there is an i ≥ 0 such that 2hi is maximal
28 Ars Math. Contemp. 24 (2024) #P2.04
for all i ≥ 0. We then have two choices for γ̃i - a positive or a negative one. The sign of γ̃j
is then uniquely determined for all j ≥ 0 since
αi,j = γ̃i
Ki,j γ̃j
−Kj,i ,
where Ki,j is even and Kj,i is odd. We will now take an arbitrary valuation v on A1(R)
satisfying the conditions of Proposition 4.1, constructed by a sequence of (ωi)i≥−1 as
shown in Section 3. We also pick γ̃i ∈ R for all i. Then we will describe v’s extension to
R[y; δ] with a sequence of (zi)i≥0 like in the beginning of Section 4.
Suppose the valuation v on A1(R) is given by a sequence (ωi)i≥−1 with ω−1 =
x, ω0 = y and ωi = xmiωnii−1 − βi, βi ∈ R and gcd(mi, ni) = 1 for all i. Suppose
also that v satisfies conditions (1) and (2) of Proposition 4.1. We will show that there is
exactly one extension of v from A1(R) to R[y; δ] for each appropriate choice of (γ̃i)i≥1.
Lemma 4.2. For each k ≥ ℓ ≥ 0, ωk can be written in the following form:
ωk = (Π
k
i=ℓ+1x
mi
ni )ωℓ(Π
k
i=ℓ+1Bi)−
k∑
i=ℓ+1
(Πkj=i+1x
mj
nj )γ̃i(Π
k
j=iBj)
+
k∑
i=ℓ+1
(Πkj=i+1x
mj
nj )Ai(Π
k
j=i+1Bi),
with
Ai =
ni−1∑
j=1
(x
mi
ni ωi−1)
ni−j−1x
mi
ni [x
j
mi
ni , ωi−1]ω
j
i−1
Bi =
nj∑
j=1
(x
mi
ni ωi−1)
ni−j γ̃i
j−1
for each 1 ≤ i ≤ k.
Proof. We prove the lemma by induction on k ≥ ℓ. For k = ℓ, it is trivially true since we
get ωk = ωk.
Suppose now the equation holds for some k ≥ ℓ. Then
ωk+1 = x
mk+1ω
nk+1
k − βk+1 = (x
mk+1
nk+1 ωk − γ̃k+1)Bk+1 +Ak+1
= x
mk+1
nk+1 ((Πki=ℓ+1x
mi
ni )ωℓ(Π
k
i=ℓ+1Bi)−
k∑
i=ℓ+1
(Πkj=i+1x
mj
nj )γ̃i(Π
k
j=iBj)
+
k∑
i=ℓ+1
(Πkj=i+1x
mj
nj )Ai(Π
k
j=i+1Bi))Bk+1 − γ̃k+1Bk+1 +Ak+1
= (Πk+1i=ℓ+1x
mi
ni )ωℓ(Π
k+1
i=ℓ+1Bi)−
k+1∑
i=ℓ+1
(Πk+1j=i+1x
mj
nj )γ̃i(Π
k+1
j=i Bj)
+
k+1∑
i=ℓ+1
(Πk+1j=i+1x
mj
nj )Ai(Π
k+1
j=i+1Bi),
L. Vukšić: Valuations and orderings on the real Weyl algebra 29
where we used the induction hypothesis, which is
ωk = (Π
k
i=ℓ+1x
mi
ni )ωℓ(Π
k
i=ℓ+1Bi)−
k∑
i=ℓ+1
(Πkj=i+1x
mj
nj )γ̃i(Π
k
j=iBj)
+
k∑
i=ℓ+1
(Πkj=i+1x
mj
nj )Ai(Π
k
j=i+1Bi)
in the second equation.
Since v(ωi−1) = mini and x
mi
ni ωi−1 = γ̃i, v(Bi) = 0 and Bi = niγ̃ini−1 must hold for
all i ≥ 1. From
[xmi , ωi−1] =
ni−1∑
j=0
x
mi
ni
·j
[x
mi
ni , ωi−1]x
mi
ni
·(ni−j),
we can see that v(Ai) = v[x
mi
ni , ωi−1] = v[x, ωi−1]+1− mini = −v(xyω1 . . . ωi−1). Since∑k
i≥−1 v(ωi) < 0 for every k by Lemma 3.2,
v(
k∑
i=ℓ+1
(Πj=i+1x
mj
nj )Ai(Π
k
j=iBi)) > v(ωk)
will hold for every k, which is why we can ignore the terms containing Ai during our
evaluations of v(zi) where for each i, zi ∈ R[y; δ] is as in the beginning of this section.
Lemma 4.3. Suppose v is a valuation on A1(R) constructed from a sequence (ωi)i≥−1
that extends to a valuation on R[y; δ] and thus D, its quotient division ring. For a given
i ≥ 1, define a sequence (Si,j)j≥1 by:
(a) Si,1 := (x
mi
ni ωi−1 − γ̃i)−1(Bi −Bi),
(b) Si,j+1 := (x
mi
ni ωi−1 − γ̃i)−1(Si,j − Si,j) for j ≥ 1.
Then for each j ≥ 1:
1 Si,j =
∑ni−j
k=1 Nk,j(x
mi
ni ωi−1)
ni−j−kγ̃i
k−1 with Nk,1 = k and Nk,j+1 =∑ni−j
ℓ=k Nℓ,j , and
2 v(Si,j) = 0, Si,j = N1,j+1γ̃ini−j−1
for all 1 ≤ j ≤ ni − 1.
Proof. We prove the first statement of the lemma by induction on j ≥ 1. To show the basis
of induction, we evaluate
Bi −Bi = (x
mi
ni ωi−1 − γ̃i)(
ni∑
k=1
((x
mi
ni ωi−1)
ni−k−1+
(x
mi
ni ωi−1)
ni−k−2γ̃i + · · ·+ γ̃ini−k−1)γ̃ik−1)
= (x
mi
ni ωi−1 − γ̃i)
ni−1∑
k=1
k(x
mi
ni ωi−1)
ni−k−1γ̃i
k−1.
30 Ars Math. Contemp. 24 (2024) #P2.04
We can thus see Si,1 =
∑ni−1
k=1 k(x
mi
ni ωi−1)
ni−k−1γ̃i
k−1 and since x
mi
ni ωi−1 = γ̃i, we
see that the lemma holds in case j = 1. Now we suppose that the statement is true for
some j ≥ 1, i.e., Si,j =
∑ni−j
k=1 Nk,j(x
mi
ni ωi−1)
ni−j−kγ̃i
k−1 and v(Si,j) = 0, Si,j =
N1,j+1γ̃i
ni−j−1. We then write
Si,j − Si,j =
ni−j∑
k=1
Nk,j(x
mi
ni ωi−1)
ni−j−kγ̃i
k−1 −N1,j+1γ̃ini−j−1
=
ni−j∑
k=1
Nk,j(x
mi
ni ωi−1)
ni−j−kγ̃i
k−1 −
ni−j∑
k=1
Nk,j γ̃i
ni−j−1
=
ni−j−1∑
k=1
Nk,j((x
mi
ni ωi−1)
ni−j−k − γ̃ini−j−1)γ̃ik−1
= (x
mi
ni ωi−1 − γ̃i)
ni−j−1∑
k=1
Nk,j((x
mi
ni ωi−1)
ni−j−1−k +· · ·+ γ̃ini−j−1−k)γ̃ik−1
= (x
mi
ni ωi−1 − γ̃i)
ni−j−1∑
k=1
(Nk,j +Nk+1,j + · · ·+Nni−j−1,j)
(x
mi
ni ωi−1)
ni−j−kγ̃i
k−1
= (x
mi
ni ωi−1 − γ̃i)
ni−j−1∑
k=1
Nk,j+1(x
mi
ni ωi−1)
ni−j−kγ̃i
k−1
proving the first statement of the lemma. The second statement immediately follows from
the first since x
mi
ni ωi−1 = γ̃i and thus
v(Si,j) = v(
ni−j∑
k=1
Nk,j(x
mi
ni ωi−1)
ni−j−kγ̃i
k−1) = 0,
Si,j =
ni−j∑
k=1
Nk,j(x
mi
ni ωi−1)ni−j−kγ̃i
k−1 = N1,j+1γ̃i
ni−j−1.
Lemma 4.4. Supose v is as in Lemma 4.3. For a given i ≥ 1, define a sequence (Di,j)j≥1
by:
(a) Di,1 = B1B2 · · ·Bi −B1B2 · · ·Bi,
(b) Di,j+1 = xv(Di,j)Di,j − xv(Di,j)Di,j .
Then for each j ≥ 1:
1. Di,j is a R-linear sum of terms which are products of elements from the set
{ωi}i ∪ {Bi}i ∪ {B−1i }i ∪ {Si,j}i,j , (4.2)
where parts of the product are conjugated by a rational power of x.
L. Vukšić: Valuations and orderings on the real Weyl algebra 31
2. v(Di,j) is a sum of v(ωℓ) for finitely many ωℓ.
3. xv(Di,j)Di,j is sum of products of γ̃k for various k.
Proof. Since
B1B2 · · ·Bi −B1B2 · · ·Bi =
i∑
j=1
B1 · · ·Bj−1(Bj −Bj)Bj+1 · · ·Bi
=
i∑
j=1
B1 · · ·Bj−1ωjB−1j Sj,1Bi+1 · · ·Bi,
and
x
mj+1
nj+1 B1 · · ·Bj−1ωjB−1j Sj,1Bi+1 · · ·Bi
= (x
mj+1
nj+1 B1 · · ·Bj−1x
−
mj+1
nj+1 )x
mj+1
nj+1 ωjB
−1
j Sj,1Bi+1 · · ·Bi,
for each 1 ≤ j ≤ k, the first two statements of the lemma follow from:
1. x
mi
ni ωi−1 − γ̃i = ωiB−1i ,
2. Bi −Bi = (x
mi
ni ωi−1 − γ̃)Si,1 = ωiB−1i Si,1,
3. Si,j − Si,j = (x
mi
ni ωi−1 − γ̃Si,j+1 = ωiB−1i Si,j+1,
4. B−1i −Bi
−1
= −Bi−1(Bi −Bi)Bi
−1
= −Bi−1ωiB−1i Si,1Bi
−1
,
where we ignore the terms we get when we change the order of multiplication. We can do
that that since these terms are procucts of Ai as defined in Lemma 4.2 and terms with zero
v-value. As we have already mentioned, these terms will not influence the construction
of the extension of a valuation on A1(R) to R[y; δ]. Indeed - we can see by induction on
j ≥ 1 that each term of the sum is a product of factors equal to, modulo conjugation by a
rational power of x, one of the elements of the set (4.2); that is,
1. either equal to ωi, or
2. equal to a power of Bj or Sj,i for some i, j.
Since the latter have v-value equal to zero and since both Bi − Bi and Si,j − Si,j are
products of ωi, a power of B−1i and Si,j for some i, j, v(Di,j) is a sum of v(ωi) for some
i. The last statement of the lemma follows from the fact that for all i, j, Bi and Si,j are of
the form Nγ̃i for N ∈ N.
If v is a valuation on R[y; δ] with residue field R, then, as we have presented in Sec-
tion 4, v can be constructed from a sequence (zi)i≥0 ⊂ R[y; δ]. In the next proposition, we
make the first comparison between this construction and the construction of a valuation on
A1(R) from a sequence (ωi)i≥−1 described in Section 3.
32 Ars Math. Contemp. 24 (2024) #P2.04
Lemma 4.5. Suppose v is as in Lemma 4.3. Suppose that for some k, ℓ ≥ 0, we can write
ωk = (Π
k
i=1x
mi
ni )zℓ(Π
k
i=1Bi) + C,
where C is a R-linear sum of elements of the form Di,j for some i, j ≥ 0. Then:
1. If v(ωk) > v(Πki=1x
mi
ni zℓ) ∈ Q, then
ωk = (Π
k
i=1x
mi
ni )zℓ+1(Π
k
i=1Bi) + C1.
2. If v(ωk) = v(Πki=1x
mi
ni zℓ) ∈ Q, then
ωk+1 = (Π
k+1
i=1 x
mi
ni )zℓ+1(Π
k+1
i=1Bi) + C2.
3. If v(ωk) < v(Πki=1x
mi
ni zℓ), then, if v(ωk) ∈ Q,
ωk+1 = (Π
k+1
i=1 x
mi
ni )zℓ(Π
k+1
i=1Bi) + C3.
Here, C1, C2 and C3 are other R-linear sums of elements of the form Di,j as in Lemma 4.5
for some i, j ≥ 0.
Proof. Suppose first v(ωk) > v(Πki=1x
mi
ni zℓ) = v(C). Since v(C) = rℓ+1 −
∑k
i=1
mi
ni
∈
Q, then rℓ+1 := v(zℓ) ∈ Q as well. So, for zℓ+1 = zℓ − x−rℓ+1γℓ+1, we can write
ωk = (Π
k
i=1x
mi
ni )zℓ+1(Π
k
i=1Bi) + γℓ+1(Π
k
i=1x
mi
ni )x−rℓ+1(Πki=1Bi) + C
= (Πki=1x
mi
ni )zℓ+1(Π
k
i=1Bi) + x
−v(C)(γℓ+1(Π
k
i=1Bi) + x
v(C)C)
Since v(zℓ+1) > v(zℓ), we see that v(γℓ+1Πki=1Bi + x
v(C)C) > 0. It follows that γi+1 =
−Πki=1Bi
−1
x−v(C)C, hence
C1 := γℓ+1Π
k
i=1Bi + x
−v(C)C = γℓ+1Π
k
i=1Bi + x
−v(C)C − x−v(C)C + x−v(C)C
= γℓ+1(Π
k
i=1Bi −Πki=1Bi) + (x−v(C)C − x−v(C)C)
We see that since C is an R-linear sum of Di,j for various i, j, so is, by Lemma 4.4,
x−v(C)C − x−v(C)C. Hence, C1 is an R-linear sum of Di,j .
Now consider the case v(ωk) = v(Πki=1x
mi
ni zℓ) ≤ v(C). Since v(ωk) = mk+1nk+1 ∈ Q,
we can evaluate
˜γk+1 = x
mk+1
nk+1 ωk = (Π
k+1
i=1 x
mi
ni )zℓ(Πki=1Bi) + x
mk+1
nk+1 C
= γℓ+1Π
k
i=1Bi + x
mk+1
nk+1 C,
L. Vukšić: Valuations and orderings on the real Weyl algebra 33
and then deduce
ωk+1 = (x
mk+1
nk+1 ωk − ˜γk+1)Bk+1
= (Πk+1i=1 x
mi
ni )zℓ(Π
k+1
i=1Bi) + x
mk+1
nk+1 CBk+1 − ˜γk+1Bk+1
= (Πk+1i=1 x
mi
ni )zℓ+1(Π
k+1
i=1Bi) + γℓ+1(Π
k+1
i=1Bi) + x
mk+1
nk+1 CBk+1 − ˜γk+1Bk+1
= (Πk+1i=1 x
mi
ni )zℓ+1(Π
k+1
i=1Bi) + γℓ+1(Π
k
i=1Bi −Πki=1Bi)Bk+1+
(x
mk+1
nk+1 C − x
mk+1
nk+1 C)Bk+1
= (Πk+1i=1 x
mi
ni )zℓ+1(Π
k+1
i=1Bi) + C2
where C2 is, again, an R-linear sum of Di,j for some i, j.
Lastly, we consider the case v(Πki=1x
mi
ni zℓ) > v(ωk) = v(C) ∈ Q. In this case,
˜γk+1 = x
mk+1
nk+1 ωk = x
mk+1
nk+1 C, so we can write
ωk+1 = (x
mk+1
nk+1 ωk − ˜γk+1)Bk+1
= (Πk+1i=1 x
mi
ni )zℓ(Π
k+1
i=1Bi) + (x
mk+1
nk+1 C − x
mk+1
nk+1 C)Bk+1
and the statement again follows.
Theorem 4.6. Suppose v is a valuation on A1(R), constructed from an either finite or
infinite sequence (ωi)i≥−1 with v(ωi) =
mi+1
ni+1
. Suppose also that v satisfies the following
conditions:
1. For each i such that ni is even, βi > 0 must hold, and
2. for each i, j, ℓ ≥ 0 with hi < hj ≤ hℓ, αi,jαi,ℓ > 0 must hold.
Then v has a unique extension to a valuation on R[y; δ] with residue field R for each choice
of (γ̃i)i≥1.
Proof. As we know from the beginning of Section 4, each valuation on R[y; δ] and D
with residue field R can be constructed by either a finite or an infinite sequence (zi)i≥0.
For every sequence (ωi)i≥−1, we will use the lemmas proved in this section to find the
unique sequence (zi)i≥0 which, as we have shown in the beginning of this section, uniquely
determines a valuation on R[y; δ]. Our calculations will then show that the valuation on D
defined by the sequence (zi)i≥0 is the extension of the valuation on A1(R) associated to
the sequence (ωi)i≥−1.
We determine the finite or infinite sequence (zi)i≥0 associated to v’s extension to
R[y; δ]. In the first step, we consider v(y). If v(y) ̸∈ Q, then v’s extension to R[y; δ]
is clearly uniquely determined, namely the one defined by
v(
n∑
i=0
pi(x)y
i) = min
0≤i≤n
{v(pi(x)) + iv(y)}
for every
∑n
i=0 pi(x)y
i ∈ R[y; δ].
34 Ars Math. Contemp. 24 (2024) #P2.04
So suppose v(y) = m1n1 ∈ Q. Then, in the second step of our evaluation, we write
ω1 = x
m1yn1 − β1 = (x
m1
n1 y − γ̃1)B1 = x
m1
n1 (y − x−
m1
n1 γ̃1)B1
and since v(ω1) > 0, v(y − x−
m1
n1 γ̃1) >
m1
n1
= v(y). We deduce z1 = y − x−
m1
n1 γ1 with
γ1 = γ̃1. Obviously, v(z1) = v(ω1) + m1n1 ∈ Q if and only if v(ω1) ∈ Q. If either and
hence both values are irrational, we get a unique extension of v to R[y; δ]. Otherwise, if
v(ω1) =
m2
n2
∈ Q and hence r2 = v(z2) = m1n1 +
m2
n2
, we continue with the third step of
our evaluation by writing
ω2 = (x
m2
n2 ω1 − γ̃2)B2 = (x
m2
n2 x
m1
n1 z1B1 − γ̃2)B2.
Since v(ω2) > 0, we conclude that γ2 = x
m2
n2 x
m1
n1 z1 must be equal to γ̃2B1
−1
. To evaluate
the v-value of z2 = z1 − γ2x−r2 , we write
ω2 = (x
m2
n2 x
m1
n1 z2B1 + γ2B1 − γ̃2)B2 = x
m2
n2 x
m1
n1 z2B1B2 + (γ2B1 − γ̃2)B2.
We note that ω2 is here written as a sum of Π2i=1x
mi
ni z2Π
2
i=1Bi + C where C is as in
Lemma 4.5. To determine v(z2), we compare v(ω2) and v(γ2B1 − γ̃2), the latter being
equal to v(ω1) since γ2B1 − γ̃2 = γ2(B1 −B1). There are three possible cases:
1. If v(ω2) < v(ω1), then v(x
m2
n2 x
m1
n1 z2) must be equal to v(ω2), so r3 = v(z2) is
determined by v(z2) = v(ω2) − m1n1 −
m2
n2
. If v(ω2) = m3n3 ∈ Q, then v(z2) =
r3 ∈ Q and γ3 = γ̃3(B1B2)−1 must hold. It follows that v(z2) ∈ Q if and only
if v(ω2) ∈ Q. In this case, the v-value of z3 = z2 − γ3x−r3 will be determined in
the subsequent steps, i.e., by considering (v(ωi))i≥3 and (γ̃i)i≥3. By Lemma 4.5,
ω3 = Π
3
i=1x
mi
ni z3Π
3
i=1Bi + C1, C1 being an R-linear sum of Di,j .
2. If v(ω2) = v(ω1), then v(z2) depends on γ̃3:
(a) If γ̃3 = x
m2
n2 (γ2B1 − γ̃2)B2, then v(x
m2
n2 x
m1
n1 z2) must be greater than v(ω1)
since in this case, ω2 ∼ (γ2B1 − γ̃2)B2 and will be determined in the sub-
sequent steps, i.e., by considering (v(ωi))i≥3 and (γ̃i)i≥3. By Lemma 4.5,
ω3 = Π
3
i=1x
mi
ni z2Π
3
i=1Bi + C2, C2 being an R-linear sum of Di,j .
(b) Otherwise, v(x
m2
n2 x
m1
n1 z2) = v(ω2). In this case, we get
γ3 = γ̃3 − x
m2
n2 (γ2B1 − γ̃2)B2.
By Lemma 4.5, ω3 = Π3i=1x
mi
ni z3Π
3
i=1Bi + C3, with C3 an R-linear sum of
Di,j .
3. If v(ω2) > v(ω1), then v(x
m2
n2 x
m1
n1 z2) = v(ω1) and γ3 = −x
m2
n2 (γ2B1 − γ̃2)B−11 .
By Lemma 4.5, ω2 = Π2i=1x
mi
ni z3Π
2
i=1Bi + C4, with C4 an R-linear sum of Di,j .
L. Vukšić: Valuations and orderings on the real Weyl algebra 35
The general step of the evaluation is similar to the first three. Suppose that in the
previous steps, we have evaluated v(z1) = r2, . . . , v(zℓ−1) = rℓ ∈ Q. In the last step, we
have, by considering (ωi)ki=−1 for some k, begun to evaluate v(zℓ) and we are, with
ωk = (Π
k
i=1x
mi
ni )zℓ+1(Π
k
i=1Bi) + C,
where C is as in Lemma 4.5, in one of the five situations:
1. If v(ωk) < v(C), then v(zℓ) = v(ωk) −
∑k
i=1
mi
ni
and γℓ+1 = ˜γℓ+1Πki=1Bi
−1
.
In case rℓ+1 = v(zℓ) ∈ Q, our next step is to evaluate the v-value of zℓ+1 =
zℓ − rℓ+1γℓ+1 by writing
ωk+1 = (Π
k+1
i=1 x
mi
ni )zℓ(Π
k+1
i=1Bi) + C1.
2. If v(ωk) > v(C), then v(zℓ) = v(C) and γℓ+1 = CΠki=1Bi
−1
. In case rℓ+1 =
v(zℓ) ∈ Q, our next step is to evaluate the v-value of zℓ+1 = zℓ − rℓ+1γℓ+1 by
writing
ωk = (Π
k
i=1x
mi
ni )zℓ+1(Π
k
i=1Bi) + C2.
3. If v(ωk) = v(C) ̸∈ Q, then v(zℓ) = v(ωk − C) −
∑k
i=1
mi
ni
̸∈ Q since in case
v(ωk − C) > v(ωk), C ∼ ωk must hold, but since, given that C ̸= ωk and that C
is a sum of Di,j , v(ωk − C) must be in Q. This terminates our evaluation of the
sequence (zi)i≥0 associated to v’s extension to R[y; δ].
4. If v(ωk) = v(C) =
mk+1
nk+1
∈ Q and x
mk+1
nk+1 ωk = x
mk+1
nk+1 C, v((Πki=1x
mi
ni )zℓ) >
v(ωk). We continue with our evaluation by writing
ωk+1 = (Π
k+1
i=1 x
mi
ni )zℓ(Π
k+1
i=1Bi) + C3.
5. If v(ωk) = v(C) =
mk+1
nk+1
∈ Q and x
mk+1
nk+1 ωk ̸= x
mk+1
nk+1 C, then v((Πki=1x
mi
ni )zℓ) =
v(ωk) and γℓ+1 = (x
mk+1
nk+1 ωk − x
mk+1
nk+1 C)Πki=1Bi
−1
. We write
ωk+1 = (Π
k+1
i=1 x
mi
ni )zℓ+1(Π
k+1
i=1Bi) + C4.
For each 1 ≤ i ≤ 4, Ci is, as is C, an R-linear sum of Di,j for various i, j. This is assured
by Lemma 4.5.
We point out that for each ℓ, v(zℓ) is determined in a finite number of steps. In case v is
determined by a finite sequence (ωi)Ni≥−1 for some N ≥ 0, this is immediate. In the infinite
case, it follows from Lemma 3.2 that limk→∞ v(ωk) = 0, so, given that by Lemma 4.4,
v(C) is a sum of v(ωi), v(ωk) < v(C) must hold for some k. And since, in the step where
v(zℓ) is determined, we get v(Πki=1x
mi
ni zℓ) = v(zℓ) −
∑k
i=1 v(ωi−1) ≤ v(ωk) for each
ℓ ≥ 0, we get v(zℓ) ≤
∑k
i=0 v(ωi) < 1 by Lemma 3.2. Since for each ℓ ≥ 0, zℓ+1 =
zℓ − x−rℓ+1γℓ+1, we did indeed find a unique sequence (zi)i≥0 that uniquely determines a
valuation v on R[y; δ]. This valuation is v’s extension from A1(R) to R[y; δ].
The construction introduced in the proof of Theorem 4.6 can be reversed. Given a val-
uation v on R[y; δ], we could use the reverse construction to find the sequence (ωi)i≥−1 ⊆
A1(R) associated to v’s restriction to A1(R).
36 Ars Math. Contemp. 24 (2024) #P2.04
5 Valuations on R̃[y; δ]
The ring R̃[y; δ] is an extension of R[y, δ] where R̃ is defined as R((xQ)), the generalized
power ring of sums
∑
q∈Q αqx
−q with well-ordered support. We first show that every
valuation on R[y; δ] can be easily extended to R̃[y; δ].
Lemma 5.1. Every valuation on R[y; δ] with residue field R can be extended to a valuation
on R̃[y; δ] with the same residue field.
Proof. Suppose first v is defined on R[y; δ] by a finite sequence (zi)ki=0 with v(zk) ̸∈ Q.
Then, as we can write every f ∈ R̃[y; δ] as
∑n
i=0 pi(x)z
i
k with pi(x) ∈ R̃, we define
v(f) = min1≤i≤n{v(pi(x)) + iv(zk)}. This gives us a well-defined valuation on R̃[y; δ]
which clearly extends the one we defined on R[y; δ] in the previous section.
Now suppose v is defined by an infinite sequence (zi)i≥0 with r = limi→∞ v(zi) ≤ 1.
Define z := y−
∑∞
i=1 αix
−ri ∈ R̃[y; δ]. In the same way as before, write every f ∈ R̃[y; δ]
as
∑n
i=0 pi(x)z
i with pi(x) ∈ R̃ and define v(f) = min1≤i≤n{v(pi(x))y + i(r − µ)}
where µ is a positive infinitesimal. Thus we once more get v’s extension to R̃[y; δ].
In case r = 1, this is the only possible extension up to isomorphism of the value group,
for v(z) must be 1 − µ. This is because on the one hand, since v(z) > v(zk) = rk+1
for all k ≥ 0, v(z) is greater than any rational number q < 1. On the other hand, since
v[y− z, x] = v[y, x] = 0 and the value group is commutative, v(z) ≤ 1. Thus if v(z) ∈ R,
v(z) = 1. But if we restricted v to the quotient ring of the R-algebra, generated by x and
z, we would get a valuation on a division ring, isomorphic to D1(R), with a rational value
group, residue field R and v[z, x] = v(zx), which contradicts Corollary 2.4.
Proposition 5.2. Let v be a valuation on R̃[y; δ] with residue field R. Then the value group
is not of rational rank one.
Proof. The only case in the proof of Lemma 5.1 where it does not immediately follow
that the value group is not Q is when v’s restriction to R[y; δ] is constructed by an infinite
sequence (zi)i≥0 with r := limi→∞ v(y − zi) < 1. In this case, we can set v(z) = r ∈ R
and if r ∈ Q, define y(1) := z and restart the construction of v. We may get another
infinite sequence (z(1)i )i≥0 with r
(1) := limi→∞ v(z
(1)
i ) < 1. If r
(1) ∈ Q, we start
over with y(2) := z(1). Though we may have to repeat the process infinitely many times,
the set {z(j)k }j≥1,k≥0 is countable and A := {v(z
(j)
k )}j≥1,k≥0 is a well-ordered set of
rational numbers smaller than one. At one point, v(z) will have to be irrational for some
z =
∑
q∈A αqx
−q since we would otherwise get z ∈ R̃ such that v(z) = 1 which, as we
have shown, contradicts the fact that the value group is rational.
Recall that a pseudo-Cauchy sequence in a division ring D with a valuation v is a
sequence (aλ)λ∈Λ ⊆ D, where Λ is an ordinal such that there exists λ ∈ Λ for which
v(aσ − aρ) < v(aρ − aτ ) for all σ, ρ, τ ∈ Λ with λ ≤ σ < ρ < τ . Let D′ be an extension
of D and v′ and extension of v to D′. Then a ∈ D′ is a limit of the pseudo-Cauchy
sequence (aλ)λ∈Λ if v′(a − aσ) = v′(aσ+1 − aσ) for all σ ∈ Λ, λ ≤ σ. Recall also the
definition of a immediate extension of a valuation.
L. Vukšić: Valuations and orderings on the real Weyl algebra 37
Definition 5.3. Let v be a valuation on a division ring D, the division ring D′ an extension
of D and v′ a valuation on D′ that extends v. Let Γ and D and Γ′ and D′ be the value
groups and residue division rnigs associated to v and v′. Then we say v′ is an immediate
extension of v if [Γ′ : Γ] = 1 and [D′ : D].
As a byproduct of our investigations, we show that not every extension of a valued
division ring by limits of pseudo-Cauchy sequences is immediate. This differs from the
commutative case since, as Kaplansky proved in [8], every extension of a valued field by
limits of pseudo-Cauchy sequences is immediate.
Corollary 5.4. There exist division rings D ⊆ D′ and a valuation v on D which extends
to a valuation v′ on D′ such that D′ is an extension of D by limits of pseudo-Cauchy
sequences in D whereas v′ is not an immediate extension of v.
Proof. Let v be a valuation on R[y; δ] with residue field R and value group Q as described
in Section 4. Then v(x) = −1 and R̃[y; δ] is an extension of R[y; δ] by limits of pseudo-
Cauchy sequences. This holds because every
∑
i∈Q aix
−i ∈ R̃ is a limit of the pseudo-
Cauchy sequence (
∑k
i=1 aix
−qi)k≥1 in R[y; δ].
As we have shown in this section, v can be uniquely extended to R̃[y; δ]. By Proposi-
tion 5.2, this extension is not immediate. Let D be the quotient division ring of R[y, δ], to
which v uniquely extends, and D′ the quotient division ring of R̃[y; δ], to which v′ uniquely
extends, since both rings are Ore domains. D′ is not an immediate extension of the ring
D with valuation v, even though D′ is an extension of D by limits of pseudo-Cauchy
sequences.
6 Compatibility with orderings on A1(R) and R[y; δ]
In Section 3, we mentioned that every strongly abelian valuation on a division ring with
an ordered residue field is compatible with an ordering on the valued division ring. In
this section, we will use a noncommutative version of the Baer-Krull theorem to determine
all orderings on A1(R) compatible with one of the valuations v we have described in the
previous sections. We will then show which of these orderings on A1(R) can be extended
to an ordering on R[y; δ] compatible with a v’s extension to R[y; δ].
Recall that an order P on a division ring D is compatible with a valuation v on D if for
every a, b ∈ D∗ such that v(a) = v(b) < v(a− b), ab ∈ P holds.
Let v be a strongly abelian valuation on a division ring D with a formally real residue
field D. Let Γ be its value group. Let s : Γ → D∗ be a semisection of v, i.e., a map for
which
1. s(0) = 1,
2. v(s(g)) = g for all g ∈ Γ,
3. s(g1 + g2) = s(g1)s(g2)u2 for some u ∈ D∗ for all g1, g2 ∈ Γ.
Let χ : Γ/2Γ → {−1, 1} be a group homomorphism called a character and let P be an
ordering of D. Then, as it was shown in [24],
Pχ = {a ∈ D | a · s(v(a))−1χ(v(a) + 2Γ) ∈ P}
38 Ars Math. Contemp. 24 (2024) #P2.04
is an order of D compatible with v. Moreover, if X denotes all orders of the residue field,
Xv denotes all v-compatible orders on D and (Γ/2Γ)∗ denotes the set of all characters on
Γ/2Γ, then by Proposition 3 of [24], the map
f : X × (Γ/2Γ)∗ → Xv
f((P , χ)) = Pχ
is a bijection. The choice of a semisection s on Γ does not matter. Using f , we will now
describe all orders on A1(R) that are compatible with a valuation described in Section 3.
Suppose v is a valuation on A1(R) associated to an infinite sequence (ωi)i≥−1 with R
as a residue field. There is only one possible order of R, so the orders of A1(R) compatible
with v will only depend on the characters χ : Γ/2Γ → {−1, 1}. Then there are three
different options for the value group Γ ⊆ Q× Z.
1. Γ is a 2-divisible subgroup of Q,
2. Γ is a non-2-divisible subgroup of Q,
3. Γ is a direct sum of a non-2-divisible subgroup of Q with Z.
Since in each case the value group Γ is generated by {v(ωi)}i≥−1, the characters and thus
the v-compatible orders will be determined by the signs of the ωi.
In the first two cases, v is determined by an infinite sequence (ωi)i≥−1 with v(ωi) =
mi+1
ni+1
∈ Q for each i ≥ −1. In the third case, v is determined by a finite sequence (ωi)Ni=−1
with v(ωi) =
mi+1
ni+1
∈ Q for each −1 ≤ i ≤ N − 1 and v(ωN ) ̸∈ Q.
If the value group is a 2-divisible subgroup of Q, then for each ωi, there is a j ̸= i such
that Ki,j ∈ Z is odd while Kj,i ∈ Z is even. Since ω
Ki,j
i−1 ω
Kj,i
j−1 = αi,j , it follows that
ωi−1 > 0 if and only if αi,j > 0.
Conversely, if the value group is a non-2-divisible subgroup of Q, we choose one i
such that ni is divisible by the greatest power of two that divides nj for any j ≥ 0. After
choosing either ωi < 0 or ωi > 0, the order on A1(R) is defined.
In the last case, where the value group is a direct sum of a non-2-divisible subgroup of
Q and Z, the order is determined by choosing either ωi < 0 or ωi > 0 and, independently,
either ωN > 0 or ωN < 0 where i is as in the second case and v(ωN ) ̸∈ Q. All four
combinations define an ordering on A1(R).
We have thus proven the following proposition.
Proposition 6.1. Suppose v is a valuation on A1(R) with residue field R and value group
Γ. Then:
1. If Γ is a 2-divisible subgroup of Q, there is a unique v-compatible ordering on
A1(R).
2. If Γ is a non-2-divisible subgroup of Q, there are two v-compatible orderings on
A1(R).
3. If Γ is a direct sum of a non-2-divisible subgroup of Q with Z, there are four possible
v-compatible orderings on A1(R).
L. Vukšić: Valuations and orderings on the real Weyl algebra 39
6.1 Extensions of orderings on A1(R) to orders on R[y; δ]
In this section, we show which orders on A1(R) are extendable to an order on R[y; δ],
thereby answering the question posed by Marshall and Zhang in [15].
Every order on A1(R) is compatible with a unique finest valuation v on the same ring
with residue field R, as proved in [15]. Suppose v is a valuation on A1(R) associated to
an infinite sequence (ωi)i≥−1 with v(ωi−1) = mini and ωi = x
miωnii−1 − βi. In Section 4,
we showed that provided both conditions of Theorem 4.6 are fulfilled, v can be uniquely
extended to a valuation v′ on R[y; δ] with residue field R if the value group is 2-divisible
and that it has two extensions to R[y; δ] with the same residue field if the value group is
non-2-divisible. Here we show all v-compatible orders on A1(R) we have described in
the first part of this section can be extended to a v′-compatible order on R[y; δ] for some
extension v′ of v from A1(R) to R[y; δ].
Theorem 6.2. Let P be an ordering on A1(R) and v be the unique finest valuation on
A1(R) compatible with P . Then:
1. The order P can be extended to an ordering on R[y; δ] if and only if v can be extended
to a valuation on R[y; δ] with residue field R.
2. If the v-value group Γ is a 2-divisible subgroup of Q, then the extension of P is
unique. If on the other hand, Γ is a subgroup of Γ1 × Z, where Γ1 is a non-2-
divisible subgroup of Q, i.e., when Γ is not a 2-divisible subgroup of Q, there are
two extensions of P to R[y; δ]. Each of the two extensions of v to a valuation v′ on
R[y; δ] with residue field R uniquely determines one of the two of P ’s extensions to
R[y; δ].
Proof. The first statement of the theorem follows from the fact that every ordering on
R[y; δ] is compatible with a valuation on the same ring with residue field R.
To prove the second statement, suppose that v is the unique P -compatible valuation on
A1(R) with residue field R that extends to a valuation on R[y; δ] with the same residue
field.
If the value group Γ of v on A1(R) is either a 2-divisible or non-2-divisible subgroup
of Q, then the value group Γ′ of v’s extension to R[y; δ] is Q. In this case, there is exactly
one v′-compatible order of R[y; δ] for each of v’s extensions v′ to R[y; δ].
Suppose Γ is a 2-divisible subgroup of Q. Then there is a unique extension v′ of v to
R[y; δ]. It follows that in case Γ is a 2-divisible subgroup of Q, the only v-compatible order
on A1(R) extends to an order of R[y; δ] that is compatible with v′.
If, on the other hand, Γ is a non-2-divisible subgroup of Q, there are two extensions v′
of v to R[y; δ]. We will now show that for each of the v-compatible orderings on A1(R),
there is a unique extension v′ of v to R[y; δ] such that the ordering on A1(R) can be
extended to the unique v′-compatible ordering on R[y; δ].
In this case, a v-compatible ordering on A1(R) is, as we have shown in the beginning
of this section, uniquely determined by the sign of ωi−1 where i ≥ 1 is such that ni is
divisible by the greatest power of two that divides nj for any j ≥ 1. Furthermore, v′, the
extension of v to R[y; δ], is uniquely determined by choosing the sign of γ̃i for this i.
We first choose an extension v′ of v to R[y; δ]. We observe that x
m
n > 0 must hold for
every mn ∈ Q since all rational powers of x are in R[y; δ]. Since x
mi
ni ωi−1 = γ̃i for each
i ≥ 1, γ̃iωi−1 > 0 must hold for all i ≥ 0 for the order to be extendable to a v-compatible
40 Ars Math. Contemp. 24 (2024) #P2.04
order on R[y; δ]. This holds for exactly one of the two v-compatible orders on A1(R). It is
clear from the construction that for each ordering on A1(R), there is exactly one extension
v′ of v to R[y; δ] such that this ordering is extendable to the unique v′-compatible ordering
on R[y; δ].
In case Γ is a subgroup of Q× Z of rational rank two, Γ′ = Q× Z holds. In this case,
there are two v′-compatible orderings on R[y; δ] for every extension v′ of v to R[y; δ]. We
will now show that for each of the four v-compatible orders on A1(R), there is a unique
extension v′ of v to R[y; δ] and a unique v′-compatible ordering P ′ on R[y; δ] such that P ′
is an extension of P . The ordering P compatible to a valuation v on A1(R) is determined
by the signs of ωi−1 and ωN where i ≥ 1 is such that ni is divisible by the greatest power of
two that divides nj for any j ≥ 1, and v(ωN ) ̸∈ Q. The extension v′ of v to R[y; δ] and the
v′-compatible ordering on R[y; δ] that extends P are the valuation v′ for which ωi−1γ̃i > 0
and the v′-compatible ordering that agrees with the signs of ωi−1 and ωN in P .
We have thus proved the second statement of the theorem.
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