VALUATIONS AND ORDERINGS ON THE REAL WEYL ALGEBRA

. The ﬁrst Weyl algebra A 1 ( 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 mechanics. In this article, we classify all valuations on the real Weyl algebra A 1 ( R ) whose residue ﬁeld is R . We then use a noncommutative version of the Baer-Krull theorem to classify all orderings on A 1 ( R ). As a byproduct of our studies, we settle two open problems in real algebraic geometry. First, we show that not all orderings on A 1 ( 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 [15], 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.


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 [6] or Kuhlmann [11].Oscar Schilling wrote the first major work on valuations on (noncommutative) division rings in 1945 [21].
It follows that v is a homomorphism from D * to Γ.The set O v := {x ∈ D | v(x) ≥ 0} is called the valuation ring associated to v, and M v := {x ∈ M | v(x) > 0} is its maximal ideal.The division ring D := O v /M v is called the associated residue division ring.Since v is a group homomorphism, the subgroup O * v is normal in D * .Several alternative approaches to noncommutative valuations, where v does not define a group homomorphism, were introduced and studied recently by Nicolai Ivanovich Dubrovin in [4] and [5] (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 v deg valuation, defined by 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 [10].
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 commutative case.Additional difficulties arise from the fact that [f, g] = 0 does not hold for all f, g ∈ R[x; σ, δ].Granja, Martínez, and Rodríguez have shown in [7] 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 A 1 (R) or, equivalently, its quotient division ring D 1 (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 α i,j x i y j , α i,j ∈ R.
Because of this, our approach to constructing valuations on A 1 (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 many fewer valuations than in the commutative case.
As we will show, the valuations on A 1 (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 A 1 (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 A 1 (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 A 1 (R) in [15], Murray Marshall and Yufei Zhang introduced the Ore extensions R[y; δ] and R[y; δ], with As is often done in real algebraic geometry, all orderings are described 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 A 1 (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 A 1 (R) with Marshall and Zhang's description of valuations on R[y; δ] to characterize the valuations on A 1 (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 R[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 extensions 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 [14], 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 A 1 (R) for every valuation v on A 1 (R) constructed in Section 3 using a noncommutative version of the Baer-Krull theorem as given in [2] (see also [24], [1], [3] and [9] for modern treatments and extensions).We also characterize the v-compatible orders on A 1 (R) that extend to an order on R[y; δ] compatible with v's extension to R[y; δ].

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 A 1 (R) and some of its ring extensions.First, we define a property of valuations on division rings.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 than if a valuation v on a division ring D is strongly abelian, then the associated value group and residue division ring are commutative.Secondly, we are particularly interested in order-compatible valuations on A 1 (R); minimal such have residue field R, as it was shown in [15].It follows from Theorem 2.5 of [14] 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
and on the other, In the last equation, we used that (aba 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.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}: Proof.To prove (1), first observe ) is proved as for the first case.Since the value group is abelian, v[x m , y] ≥ v(x m y) holds, so we can observe 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 −1 x −1 [x, y]) = 0 and Proposition 2.2.So now we have 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 where the last equation follows from v[a 2 , y] > v(a 2 y), or, by Lemma 2.5 equivalently, 2 , y] = v(a 2 y) and v[a, y] = v(ay) follows from Lemma 2.5.Now we show v[a 4 , y] > v(a 4 y).On one hand, we can write 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.
follows from the 2-divisibility of the value group and Proposition 2.6.Let c := x m 1 y n 1 and d := x m 2 y n 2 .Then on the one hand, since v(a k 1 ) and v(c) are rationally dependent, v(b k 2 ) and v(d) are rationally dependent and v(b On the other hand, Here, the last equality follows from v[x, y] > v(xy) and Lemma 2.5.Thus we have The proof of the following proposition is the same as the proof of Proposition 2.7.
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 where k is odd.Then v is strongly abelian.
We will later use this result to show that all valuations v on A 1 (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 x 1 , . . ., x n ∈ D such that v(x 1 ), . . ., v(x n ) are rationally independent with v[x i , x j ] > v(x i x j ) 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 x 1 , . . ., for some k, m 1 , . . ., m n ∈ Z with k odd.Then v is strongly abelian.

Valuations on A 1 (R)
We now describe the construction of all valuations on A 1 (R) with residue field R that was sketched in [22] over the ground field of C. Since every f ∈ A 1 (R) can be written as m,n≥0 α m,n x m y n , 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 [10] or [13]), but with some additional constraints arising from the fact that the generators x, y ∈ A 1 (R) satisfy [y, x] = 1.We first note that it follows from Theorem 5.3 of [15] that the value group of any valuation on A 1 (R) is commutative.Also, since every valuation v on A 1 (R) can be uniquely extended to its quotient division ring D 1 (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 [15], 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 and as before compare v(ω 1 ) to v(x) in terms of rational dependence.
1 − β 2 also has value greater than zero.We continue this procedure.If we additionally define ω −1 = x and ω 0 = y, we thus get a sequence By the end of this section, we will prove a necessary and sufficient condition for the possibility to extend v from (ω i ) i≥−1 to a valuation on A 1 (R) with residue field R. Every such extension from (ω i ) i≥−1 to A 1 (R) will be uniquely determined.We will also show that every valuation on A 1 (R) with residue field R is strongly abelian.
3.1.Properties of the sequence (ω i ) i≥−1 associated to a valuation on A 1 (R).Thorough this subsection let v be a valuation on A 1 (R).
Proof.We prove the lemma by induction on i.
Now suppose that the equality holds for [ω i , x].Then we have We can then proceed by the induction hypothesis.
Before proving the next lemma, we define an equivalence relation between nonzero elements of A 1 (R) that have the same v-value, but their difference does not.For any . This is also a congurence relation, as ac ∼ bc and ca ∼ cb holds for all a, b, c If any and hence both sides of the equivalence hold, then for all i < j ≤ k.
Proof.Suppose v is a valuation on A 1 (R) and (ω i ) i≥−1 is a sequence as described in the lemma.So v(ω i ) ∈ Q either for all i ≥ 0 or for all 0 ≤ i < n for some n ≥ 0 and ) for all i, j < n since v(ω i ) and v(ω j ) are rationally dependent.We shall use this fact to evaluate v[ω i , ω j ] for all i, j ≤ k ≤ n.It follows from Lemma 3.1 that [ω k , x] is a sum of products P , all equal to , ω i ] and using both Lemma 3.1 and induction on k, we see that the first sum has v-value equal to ) must be strictly less than 0 for all k ≤ n in case v(ω n ) ∈ Q for some n, and for all k ≥ 0 if v(ω i ) ∈ 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 (ω i ) i≥−1 to a valuation on A 1 (R) with residue field R.
To ensure that the residue field is R, it is obviously necessary that ω are integer multiples of the pair (K i,j , −K j,i ) with n j = 0, we can write for some n ∈ Z, where we used Proposition 2.2 in the second equality.So for every We immediately see that α j,i = α −1 i,j and hence α i,i = 1 for all i, j ≥ 0. As for all i, j ≥ 0, α i,j is one of the possible d i,j -th roots for β If v is a valuation on D 1 (R) with residue field R, α i,j must be real for all i, j ≥ 0. For every i, j ≥ 0 with even d i,j , this means that β m j i β −m i j > 0 must hold.In the next lemma, we present a necessary condition on the sequence (β i ) i≥1 so that α i,j ∈ R can be chosen for all i, j.We also prove that if n i is odd, α i,j is uniquely determined for all j ≥ 0. Lemma 3.3.Let v be a valuation on D 1 (R) as in Lemma 3.2.Then the following holds: (1) If n i is odd, there is a unique possible choice for α i,j ∈ R for all j ≥ 0.
(2) Only if sgn(β i ) is constant on the set of all i ≥ 0 for which n i is even can we choose α i,j ∈ R for all i, j ≥ 0.
Proof.Suppose n i is odd.Then for any j ≥ 0, let di,j be the highest odd number dividing we can evaluate and since di,j is odd, α i,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 n i and n j are both even.As a consequence, both m i and m j are odd while d i,j is even.So, provided α i,j ∈ R, we compute 1 = sgn(α which proves the second part of the lemma. For even d i,j we have seemingly two choices for α i,j ∈ R -a positive and a negative one.We will show that in most cases, we cannot choose sgn(α i,j ) for all i, j ≥ 0 independently of each other.
Before that, we observe that for any i, j ≥ 0, at most one of K i,j and K j,i is even.In fact, if at most one of n i and n j is odd, K i,j is odd if and only if n i is divisible by the greatest power of two that divides n j .For each i ≥ 1, let 2 h i be the biggest power of two that divides n i .Define also m 0 = 1 and n 0 = −1.
Proof.To prove the necessity of the condition, suppose a, b, c ≥ 0 are such that with On the other hand, we see by analogous computations that sgn( We have chosen k a , k b odd and k c even.In this case, the greatest power of two that divides k c is 2 hc−ha+1 .On the other hand, the greatest power of two that divides K c,a and K c,b is 2 hc−ha .We can thus conclude from ℓK We first suppose 0 ∈ supp K.We will deal with the case 0 ∈ supp K at the end of our proof. If for some i, j > 0, and its value in the residue field is a power of α i,j .
We now need to show that in this way, Π r i=1 ω k i i−1 is uniquely determined, that is, if we choose another a ′ , b ′ ∈ supp K instead of a, b, we get the same value for Π r i=1 ω k i i−1 ∈ R. We will show this by choosing c ∈ supp K \ {a, b} and proving that the evaluated value of Π r i=1 ω k i i−1 is the same whether we factor a power of α a,b as above, or α a,c or α b,c instead.By transitivity of the equality relation, this will imply that the obtained value of Π r i=1 ω k i i−1 is independent of the choice of a, b ∈ supp K. Suppose without loss of generality that h a ≤ h b ≤ h c and that K a,b , K a,c and K b,c are odd.Above, we have evaluated We proceed by evaluating, in the same way as before, with ℓ i,1 and p i,1 for all 0 ≤ i ≤ r as above.In particular, we see that for i = a, b, c, and on the other hand, we see from 3.3 that We need to show that in both equations, we get the same value.We first see that for all we can see p c,1 K a,c = p c,2 K a,b holds as well, and thus we conclude , since both sides of the equation are equal to and the signs of α a,b , α a,c and α b,c were chosen so that the signs of both sides of the equality match.We conclude that the value of Π r i=1 ω is the same in both 3.4 and 3.5.As N is odd (since K a,b , K a,c and K b,c are all odd), we conclude that Π r i=1 ω k i i−1 is the same whether we factor a power of α a,b or α a,c .If we factored a power of α b,c , we would, as similar computations as above would show, get the same value for Π r i=1 ω k i i−1 .We have now shown that if the condition of the proposition is fulfilled, Π r i=1 ω . Now suppose n j is even for some j ∈ supp K. Then m j must be odd since gcd(m j , n We evaluate (Π r i=1 ω This concludes the proof of our proposition. In Lemma 3.6, we suppose that v is a valuation on D 1 (R) extended from (ω i ) i≥−1 to D 1 (R) and compute the value of certain elements of D 1 (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 value of the sum is equal to zero, proving the statement for positive integers n.For negative n ∈ Z, the statement follows from Proof.Let n be the least common multiple of n i 1 , . . ., n ir and c i j = n n i j for each i j .Since , by Proposition 2.2, we can compute .
For each j such that k i j > 0, In case k i j < 0, we see that We conclude, using Lemma 3.5, With the help of Lemma 3.6, we will evaluate v(x k 0 ω ir−1 ) = 0 in general.As in Lemma 3.6, we assume k 0 ∈ Z, k i j ∈ Z \ {0} for all 1 ≤ j ≤ r.This will be helpful when we will later construct a valuation v associated to a sequence (ω i ) i≥−1 .Let us assume for now that i 1 < i 2 < • • • < i r ; at the end of the calculation we will see that the order of i j does not affect the v-value.
To start, we introduce some abbreviations to make the written equations easier to read.Let n and c i j for all j be as in the proof of Lemma 3.6, Since B 0 is in R, we can write ).To evaluate the right-hand side of (3.6), we first proceed as we have done in the proof of Lemma 3.6, so If k i j > 0, we proceed by For i j > 0, k i j < 0, we can on the other hand write We now define, if k j > 0, and, if k j < 0, for each 1 ≤ j ≤ r.Further, we define 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 ω i ℓ (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 min 1≤j≤r {v(ω i j )} is achieved at more than one j, we take the sum of all ω i j C j that have the minimal v-value, i.e., v(A 1,j ) is minimal A 1,j , then factor ω i 1 , so the sum now looks like and, since v(ω −1 i 1 ω i j C j ) = 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(A 0 − B 0 ) = min 1≤j≤r {v(A 1,j )}.Otherwise write For every j, we write (in the same way we did with A 0 − B 0 ).We sum all of the newly obtained terms, as well as the terms in v(A 1,j ) is not minimal A 1,j , and relabel them as A 2,j where j goes from 1 to the number of all terms.As A 0 − B 0 can be written in the form where we use D as the label of the product of all terms of the form . ., k ′ ir ∈ Z and B = A, that we factor out when we evaluate ω −1 i 1 ω i j C j − ω −1 i 1 ω i j C j for each j.All terms ( i A k−i B i ) −1 have v-value equal to zero and their image in the residue field, which is of the form (mB m−1 ) −1 ∈ R for some m ∈ N, is easy to determine.
We repeat the described procedure, writing A 0 − B 0 = ( j A k,j )D for increasing k.We stop when for some k, j,v(A k,j ) is minimal A k,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(A 0 − B 0 ) is equal to v(A k,j ) for any term of the sum j,v(A k,j ) is minimal A k,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 as a sum of terms with strictly positive v-value, the value of each of these terms is v(ω i ℓ ) for some ℓ = 1, . . ., r.It follows that v(A 0 − A 0 ) is a sum of v(ω ℓ ) for some ℓ ≥ 1.
If v(ω N ) is irrational for some N ≥ 0, the process either stops beforehand or, after k ≥ N − i r steps we get a unique term A k,j that has v-value equal to v(ω ir ω ir+1 • • • ω N ).This is the term we get when we take the last term of A 0 , written as a sum of terms A 1,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 A k,j is written as a sum of terms with higher v-value.
If on the other hand, v(ω k ) ∈ Q for an infinite sequence (ω k ) k≥−1 , then lim k→∞ v(ω k ) = 0 since by Lemma 3.2, k i≥−1 v(ω k ) < 0 for all k ≥ −1 and v(ω i ) > 0 for i ≥ 1.Then for some N ≥ 1, v(ω N ) < v(ω i ) for all 1 ≤ i < N. The evaluation of v(A 0 − B 0 ) again either stops beforehand or we get a unique term A k,j that has v-value equal to v(ω ir ω ir+1 • • • ω N ).As in the first case, this term is the one we get when we take the last term of A written as a sum of terms A 1,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 A k,j is written as a sum of terms with higher v-value.
In both cases, the value of this term, v(ω ir ω ir+1 • • • ω N ) 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 (ω i ) i≥−1 , v(A 0 − B 0 ) is the same as it would be if all elements of the sequence (ω i ) i≥−1 commuted.This follows from the fact that when we change the order of factors ω i and ω j in some A k,j 1 , the term we obtain has v-value greater by ) is for any j 1 , j 2 equal to v(ω i ℓ +1 ) for some i ℓ such that A k,j 2 contains a power of ω i ℓ .Since, as we have shown in the proof of Lemma 3.2, N i=−1 v(ω i ) < 0, it follows that the terms we obtain by changing the order of the factors have v-value greater than v(A 0 − B 0 ).And ) is higher than the v-value of any terms we get when we change the order of factors, the order of ω i 1 , . . ., ω ir does not matter.

3.2.
Extending v from the sequence (ω i ) i≥−1 to A 1 (R).We can now prove that every v associated to either a finite or an infinite sequence (ω i ) i≥−1 can be extended to a valuation on D 1 (R).Lemma 3.7.For every r ≥ 0, there exists a finite number M of elements of the form n j+1 ∈ Q for j ≥ 0, the problem translates to finding general classes of solutions to the diophantine equation j=1 n i for all 0 ≤ i ≤ k.Theorem 3.8.Let v and (ω i ) i≥−1 be as described in the beginning of the section, i.e., Suposse that sgn β i is constant on the set of all i for which n i is even.Then v can be extended to a valuation on D 1 (R) with residue field R. The valuation is unique for every choice of {α i,j } i,j≥0 where α i,j = Proof.The following construction of the valuation v associated to the sequence (ω i ) 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 D 1 (R), we define it for some specific elements of D 1 (R).
(1) Since we have defined v(ω i ) for all which must be strictly greater than v(ω i ω j ) for all i < j.
(3) We also see that is uniquely determined by {α i,j } i,j≥0 as in shown in Proposition 3.4.In this case, v(ω ) 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 (ω i ) i≥−1 if we want v to be a valuation.
To determine v(F ) for any F ∈ A 1 (R), we first note that F can be written as a finite sum Let F 1 be the sum of all terms α ℓ x i ℓ y j ℓ such that i ℓ v(x) + j ℓ v(y) is equal to u := min ℓ {i ℓ v(x) + j ℓ v(y)}.
If F 1 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 x i 1 y j 1 with the smallest power of x and get where In this case, v(F 1 ) = u and since all terms in F − F 1 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 the abelian group, generated by {v If on the other hand f (β 1 ) = 0, write f (t) = g 1 (t)(t − β 1 ) k 1 with g 1 (β 1 ) = 0 and we have We set v(F 1 ) = u + k 1 v(ω 1 ) and add all terms we get from exchanging the order of factors, whose v-value can be lower than the newly set v(F 1 ), although still strictly higher than ) is in the subgroup of Γ, generated by v(x), v(y) and v(ω 1 ) and that if v(F 1 ) = 0, F 1 ∈ R. It is important to note that in both cases, we consider F 1 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, F 1 was just a single term in the beginning and we get v(F ) = v(F 1 )).
We now consider the values of the terms in F − F 1 .If all of them have v-value strictly greater than that of F 1 , we conclude v(F ) = v(F 1 ).Otherwise, we take all terms of (y)} and then as before define As above, we write and add all the terms we get when we change the order of factors to We continue this process, defining Afterwards, we sum together all those F i for 1 ≤ i ≤ k for which v(F i ) = u 1 := min{v(F 1 ), . . ., v(F k )}.If the minimum is achieved at exactly one such F i , we set v(F ) = v(F i ).This is always the case whenever v(ω 1 ) ∈ Q. Otherwise we can relabel the terms so the minimum is achieved at F 1 , . . ., F r for some r ≤ k.As we have shown, each F i can be written as We sum the terms together, factor out x i 1 y j 1 ω k 1 1 , the term that has, written as a polynomial in x and y, the lowest power of x, and label the new sum F 1,1 .
To evaluate v(F 1,1 ), we follow a procedure similar to the one evaluating v(F 1 ).After factoring x i 1 y j 1 ω k 1 1 , we are left with Each term in the sum has v-value zero.Let a 1 , a 2 , • • • , a ℓ be the terms such that each product of the form x i y j ω k 1 , i, j, k ∈ Z that fulfills the condition v(x i y j ω k 1 ) = 0 is a product of positive integer powers of some of a i up to the order of factors x, y and ω 1 .
The existence of a 1 , . . ., a ℓ is assured by Lemma 3.7.We can then write with γ J ∈ R, m I,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 ω 1 in a product to F − F 1,1 since the value of its terms is strictly greater than u 1 .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 a i ∈ R is for each 1 ≤ i ≤ ℓ, v(g(a 1 , . . ., a ℓ )) will have to be greater than or equal to zero, we can define g(a 1 , . . ., a ℓ ) = g(a 1 , . . ., a ℓ ).
If g(a 1 , . . ., a ℓ ) = 0, then we set v(g(a 1 , . . ., a ℓ )) = 0 and v(F . ., t n ] and h i (a 1 , . . ., a ℓ ) = 0 for all i.We factor out the Π j (t j − a j ) m i,j for those i for which j v(a j − a j ) m i,j is minimal.Then If g(a 1 , . . ., a k ) = 0, we set If on the other hand, g(a 1 , . . ., a k ) = 0, we do the same thing as we did with g.The process cannot go on indefinetly since g is a polynomial and hence of finite degree.All terms we get when we exchange the order of x, y and ω 1 are added to F − F 1,1 .Their v-value must be strictly greater than u 1 .It follows from the construction that v(F 1,1 ) must be in the group generated by {v(ω i )} i≥−1 since this holds for v(a i − a i ) for all i and ) is, as we have shown in Lemma 3.6 and the discussion following it, a sum of v(ω j ) and thus v(Π j ω −1 j (a i − a i )) = 0, we can write F 1,1 as one term of the form Π n i=−1 ω k i i g(a 1 , a 2 , . . ., a ℓ ) with n ∈ N and g ∈ R[t 1 , . . ., t ℓ ] and g(a 1 , a 2 , . . ., a ℓ ) = 0.After v(F 1,1 ) is set, we compare it to both v(F i ) for all F i that are not part of F 1,1 and the terms of . If all of these terms have v-value strictly greater than v(F 1,1 ), then we can set v(F ) = v(F 1,1 ).Otherwise, we collect all terms with minimal v-value in a sum which we label F 1,2 .We determine v(F 1,2 ) in the same way we determined F 1,1 and then sum all of the remaining terms that have v-value less or equal to min{v(F 1,1 ), v(F 1,2 )} to a sum labeled F 1,3 .
We repeat the process until for some k, min{v(F 1,1 ), • • • , v(F 1,k )} is strictly smaller than the v-value of any of the remaining terms.

If min{v(F
Otherwise we sum all the terms with the minimal v-value and label the sum F 2,1 .We evaluate v(F 2,1 ) in the same way we evaluated v(F 1,1 ).We repeat the process, defining F i,j and determining its v-value in the same way as above.We point out that after v(F i,j ) is defined, we regard F i,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(F i,j ) is strictly smaller than the v-value of all other terms.This holds because each time we define F i,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 ∈ A 1 (R).What we essentially did was that we wrote where F is written as a single term, v( F ) is computed as if x and y commuted and the v-value of each term of F1 is strictly greater than v( F ).For another G ∈ A 1 (R), we can write It follows from the construction that for each i, j, v(F i,j ) is a linear combination of {v(ω i )} i≥−1 and that in case v(F i,j ) = 0, F i,j ∈ R. Theorem 3.9.Let v be a valuation on A 1 (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(ω N ) ∈ Q for some N by our construction.But as we have shown in Lemma Since the value group is generated by {v(ω i )} i≥−1 , it follows from Proposition 2.7 that v is strongly abelian.

Valuations on R[y; δ]
In this section, we explain a construction of valuations on the ring R[y; δ] with and δ(p(x)) = p ′ (x).This construction, which was first introduced in [15], 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 A 1 (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 [15].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, z 0 := y and consider v(y).
) is greater than r 2 .We repeat this process to construct a sequence (z i ) i≥0 . If The value group is then group-isomorphic to v is stongly abelian by Proposition 2.7.Otherwise, the sequence We take note of the fact that v(z i+1 ) > v(z i ) and since [z i , x] = [y, x] = 1 for all i, v(z i ) < 1 for all i.We define r := lim i→∞ r i ≤ 1.
4.1.Case r < 1.If r < 1, it has been shown in [15] that v can be extended to a valuation on R[y; δ] with residue field R. We first extend in a natural way, i.e., by defining v( q∈A a q x −q ) = min A for each q∈A a q x −q ∈ R[y; δ].Then for every 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 2r k > 1 + r 1 , which exists since r = 1, and On the other, , contradicting the assumption that v is a valuation, as shown in [15].
Of course, even in case r < 1, there may also exist a k such that 2r k > 1 + r 1 .But the important difference between the two cases is that if r < 1, there is always an ℓ ∈ N such that 1 + r ℓ > 2r k for all k ∈ N, which does not hold in case r = 1.Then, since is a real algebra automorphism of R[y; δ], we can translate the sequence by replacing y with z ℓ .
We see that since the associated value group is Q, v is a strongly abelian valuation by Corollary 2.4.

4.2.
Case r = 1.The question whether in case r = 1, v can be extended from a sequence (z i ) i≥0 to a valuation on R[y; δ] was left open in [15].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 a strictly increasing sequence with r = lim i→∞ r i = 1.Then for each n ≥ 0, there is a valuation We now present the first-order theory that the valuation associated to the infinite sequence 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 where F is the set of all constants c a for each a ∈ D, +, • and < are binary function symbols, − and −1 are unary function symbols, O is an unary relation symbol and c z i 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: We add all sentences C that will give proper meaning to the constants c z i for all i ≥ −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 v n described in the beginning of this subsection, so does, by compactness, the whole theory.Since the theory contains F , the set of all constants c a for each a ∈ D, the models are valued division rings which all contain D. We pick a model of the theory, a pair (D 1 , v), where D 1 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 For the time being, we ignore the terms we get when we change the order of multiplication.At the end of this subsection, we will see that they do not influence v(f ).For each k ≥ 1, we define for each 0 ≤ j ≤ n and k ≥ 1.For each 0 ≤ j ≤ n, g j (t) := is the algebraic closure of the quotient field of R, as shown in for example [16].Since ∞ i=1 γ i x −r i is not in the quotient field of C, it is not a root of g j (t) for any 0 ≤ j ≤ n.We conclude that for some K ∈ N, v(p ) for all k ≥ K and all 0 ≤ j ≤ n.We can then write ) for all k ≥ K.We now show that from some . Since (r i ) i≥1 is an increasing sequence with lim i→∞ r i = 1, there exists K ′ ≥ K such that min i=0,...,n {v(p is achieved at exactly one 0 ≤ i ≤ n.We conclude v(p 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 with q j ∈ Q for all 1 ≤ j ≤ k i and 1 ≤ i ≤ n.Since for all 0 ≤ i ≤ n, p i (x) ∈ R and (r i ) i≥1 is an increasing sequence with lim i→∞ r i = 1, there exists some k ≥ 1 such that the term of the sum with v-value min i=1,...,n {v(p is the only term in the sum with its v-value.We conclude v(p for all 1 ≤ i ≤ n.On the other hand, we can write For all 0 ≤ i ≤ n, all terms of (a k + z k ) i when expaned are of the form a ℓ 1 k z ℓ 2 k . . .a it follows that the v-value of each term we get when we change the order of multiplication is at least v(p i (x)) + (i − 1)r 1 + 1 for each 1 ≤ i ≤ n, which is, as is immediate from (4.1), strictly greater than v(p for all k ≥ K ′ with v k as defined in the beginning of this section.As every element of D, the quotient ring for R[y; δ], can be written as f g −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 v k (f ) = 0 for all k ≥ K for some K ∈ N. We can write We conclude f = αβ −1 ∈ R.So the residue field for v is indeed equal to R.

Extensions of valuations from
Let α i,j ∈ R be as in Section 3. Let 2 h i be the greatest power of two dividing n i 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 n i is even, β i > 0 must hold, and (2) for each i, j, ℓ with it is obvious that γi must be equal to an n i -th root of β i .If n i is odd, γi ∈ R is uniquely determined regardless of sgn(β i ), while if n i is even, γi ∈ R only if β i > 0. It is thus obvious that β i > 0 must hold for all i where n i is even if v can be extended from a valuation on A 1 (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 holds for all i, j ≥ 0. If h i < h j , K i,j is odd while K j,i is even, so sgn(γ i ) = sgn(α i,j ) must hold.We can therefore see that for all i, j, ℓ ≥ 0. If h i < h j ≤ h ℓ , both K i,j and K i,ℓ are odd while K j,i and K ℓ,i are even.It follows that if v can be extended to a valuation on R[y; δ] with residue field R, α i,j α i,ℓ > 0 must hold for all i, j, ℓ ≥ 0 with h i < h j ≤ h ℓ .
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 A 1 (R) satisfying the conditions described in Proposition 4.1.We will first determine γi ∈ R for all i ≥ 0. If n i is odd, there is a unique choice of γi ∈ R. Suppose then n i (1) D i,j is a R-linear sum of terms which are products of elements from the set where parts of the product are conjugated by a rational power of x.
(3) x v(D i,j ) D i,j is sum of products of γk for various k.
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 A i 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 A 1 (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 ω i , or (2) equal to a power of B j or S j,i for some i, j.
Since the latter have v-value equal to zero and since both B i − B i and S i,j − S i,j are products of ω i , a power of B −1 i and S i,j for some i, j, v(D i,j ) is a sum of v(ω i ) for some i.The last statement of the lemma follows from the fact that for all i, j, B i and S i,j are of the form N γi for N ∈ N.
If v is a valuation on R[y; δ] with residue field R, then, as we have presented in Section 4, v can be constructed from a sequence (z i ) i≥0 ⊂ R[y; δ].In the next proposition, we make the first comparison between this construction and the construction of a valuation on A 1 (R) from a sequence (ω i ) i≥−1 described in Section 3. Lemma 4.5.Suppose v is as in Lemma 4.3.Suppose that for some k, ℓ ≥ 0, we can write where C is a R-linear sum of elements of the form D i,j for some i, j ≥ 0. Then: ( ( Here, C 1 , C 2 and C 3 are other R-linear sums of elements of the form D i,j as in Lemma 4.5 for some i, j ≥ 0.
We see that since C is an R-linear sum of D i,j for various i, j, so is, by Lemma 4.4, and then deduce where C 2 is, again, an R-linear sum of D i,j for some i, j.

Lastly, we consider the case
n k+1 C, so we can write and the statement again follows.
Theorem 4.6.Suppose v is a valuation on A 1 (R), constructed from an either finite or infinite sequence (ω i ) i≥−1 with v(ω i ) = m i+1 n i+1 .Suppose also that v satisfies the following conditions: (1) For each i such that n i is even, β i > 0 must hold, and (2) for each i, j, ℓ ≥ 0 with h i < h j ≤ h ℓ , α i,j α i,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 ( γi ) 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 (z i ) i≥0 .For every sequence (ω i ) i≥−1 , we will use the lemmas proved in this section to find the unique sequence (z i ) 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 (z i ) i≥0 is the extension of the valuation on A 1 (R) associated to the sequence (ω i ) i≥−1 .
We determine the finite or infinite sequence (z i ) i≥0 associated to We note that ω 2 is here written as a sum of Π 2 i=1 x m i n i z 2 Π 2 i=1 B i + C where C is as in Lemma 4.5.To determine v(z 2 ), we compare v(ω 2 ) and v(γ 2 B 1 − γ2 ), the latter being equal to v(ω 1 ) since γ 2 B 1 − γ2 = γ 2 (B 1 − B 1 ).There are three possible cases: (1) If v(ω 2 ) < v(ω 1 ), then v(x n i z 3 Π 2 i=1 B i + C 4 , with C 4 an R-linear sum of D i,j .The general step of the evaluation is similar to the first three.Suppose that in the previous steps, we have evaluated v(z 1 ) = r 2 , . . ., v(z ℓ−1 ) = r ℓ ∈ Q.In the last step, we have, by considering (ω i ) k i=−1 for some k, begun to evaluate v(z ℓ ) and we are, with where C is as in Lemma 4.5, in one of the five situations: (2) If v(ω k ) > v(C), then v(z ℓ ) = v(C) and γ ℓ+1 = CΠ k i=1 B i −1 .In case r ℓ+1 = v(z ℓ ) ∈ Q, our next step is to evaluate the v-value of z ℓ+1 = z ℓ − r ℓ+1 γ ℓ+1 by writing as n i=0 p i (x)z i with p i (x) ∈ R and define v(f ) = min 1≤i≤n {v(p i (x))y + i(r − µ)} where µ is a positive infinitesimal.Thus we once more get v's extension to R[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(z k ) = r k+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 D 1 (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 R[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 (z i ) i≥0 with r := lim i→∞ v(y − z i ) < 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 i ) i≥0 with r (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 α q x −q since we would otherwise get z ∈ R 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 (a λ ) λ∈Λ ⊆ D, where Λ is an ordinal such that there exists λ ∈ Λ for which v(a σ − a ρ ) < v(a ρ − a τ ) 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 (a λ ) λ∈Λ if v ′ (a − a σ ) = v ′ (a σ+1 − a σ ) for all σ ∈ Λ, λ ≤ σ.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.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 R[y; δ] is an extension of R[y; δ] by limits of pseudo-Cauchy sequences.This holds because every i∈Q a i x −i ∈ R is a limit of the pseudo-Cauchy sequence ( k i=1 a i x −q i ) k≥1 in R[y; δ].

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.
c and K b,c are all odd.So we see that sgn(ω ka a−1 ω k b b−1 ω kc c−1 ) = sgn(α a,b ) = sgn(α a,c α b,c ), which proves the necessity of the condition.Now suppose sgn(α a,b α a,c α b,c ) = 1 for all a, b, c ≥ 0 with

Proposition 4 . 1 .
A 1 (R) to R[x; δ].In this section, we characterize the valuations on A 1 (R) with residue field R that have an extension to R[y; δ] with the same residue field.Since x m i n i ∈ 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 m i n i ω i−1 ) = 0 and γi := x m i n i ω i−1 ∈ R. In the next proposition, we show the necessary condition for a valuation v on A 1 (R) to have an extension v ′ to R[y; δ] with the same residue field.Let v be a valuation on A 1 (R) with residue field R associated to a sequence

Corollary 5 . 3 .
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.