                DEFINABLE NON-DIVISIBLE HENSELIAN VALUATIONS
                                          JIZHAN HONG

         Abstract. On a Henselian valued field (K, V) where V is the valuation ring, if
         the value group contains a convex p-regular subgroup which is not p-divisible,
         then V is definable in the language of rings. A Henselian valuation ring with
         a regular non-divisible value group is always 0-definable. In particular, some
         results of Ax's and Konenigmann's are generalized.
  In the model theoretic study of valued fields, it is natural to ask whether
the valuation rings are definable in terms of addition and multiplication.
Among the first and simplest observations about definable valuations is that
Z_{p} is definable over Q_{p} in the language of rings.  It is this very obser-
vation that inspired our investigation about definable Henselian valuations
with non-divisible value groups in this article. While there are many examples
where Henselian valuation rings with non-divisible value groups are not defin-
able (see e.g. Corollary 4.3 and the proof of Theorem 4.4 in [3].), we provide
some sufficient conditions for them to be definable. This enables us to gen-
eralize Ax's [1] (definability and decidability) and Koenigsmann's Lemma 3.6
in [8] to a broader class of non-divisible Henselian valuations.
  We use the following convention: K always denotes a field; if V is a val-
uation ring on K, then the corresponding valuation map is denoted by v : K ↠
vK = vK^{×} ∪ {∞}, where the value group is denoted by vK^{×} ; the residue field
and the maximal ideal of V are denoted by Kv and M_{v} respectively. The lan-
guage of rings is denoted by ℒ_{r}. Unless mentioned otherwise, `definable' means
with parameters.  Finally, p > 1 denotes a fixed prime number and `ζ_{p} ∈ K'
means that K contains a primitive p-th root of unity.
  Recall that an ordered abelian group is n-regular if every infinite convex
subset has at least one n-divisible element, or equivalently, if every quo-
tient over a non-zero convex subgroup is n-divisible. See e.g. [10] and [2].
A valued field is p-Henselian (our major reference would be [7]) if it has
exactly one extension of the valuation to its maximal Galois p-extension.
\underline{Definition} 1. An ordered abelian group would be called \underline{p-rendible} if it is
p-regular but not p-divisible; it would be called \underline{rendible} if it is regular
(i.e. p-regular for all p) but not divisible.

2010 Mathematics Subject Classification. Primary 03C60, 03C40; Secondary 12L12, 12J20, 13F30;

This article originates from an independent part of the author's to-appear Ph.D. thesis.

Among many things, the author would like to thank Dr. S. Azgin for introducing the notion of regular groups on a bus, Dr. F. Delon for introducing Schmitt's work and her careful reading of an earlier version of this article, Dr. D. Haskell for the discussion about Ax's proof and her comments on the style of writings and catching a lot of typos, Dr. J. Koenigsmann for the email messages about definable valuations, Dr. P. Speissegger for discussions about earlier versions of this paper, and McMaster University for the financial support.

2                                                JIZHAN HONG
\underline{Proposition} 2. Suppose that the valued field (K, V) satisfies one of the fol-
lowing conditions (in which case we call (K, V) \underline{p-leasant}):
        • V is Henselian.

• V is p-Henselian, ζ_{p} ∈ K, char(Kv) ≠ p;

        • V is p-Henselian, char(K) = p.
  If γ ∈ vK^{×} is not p-divisible, then the following set is definable in K in

ℒ_{r},

(1)  Φ = {x ∈ K | pv(x) + γ > 0}.

\underline{Proof}. We first show it for the Henselian case.
  Note that Φ ≠ ∅. Pick ϵ ∈ K with v(ϵ) = γ. The following formula defines Φ:

(2)  ϕ(x) := “∃y (y^{p} − y^{p−1} = ϵx^{p}) ” .

  For the other two cases, replace y^{p} − y^{p−1} in ϕ(x) above, by y^{p} − 1 and y^{p} − y
respectively. Note that these polynomials split in K in each case and then
one can use the “p-Hensel's Lemma” (see e.g. [7]).  □
\underline{Corollary} 3. Suppose that the valued field (K, V) is p-leasant. Then vK^{×} be-
ing discrete implies that V is definable in ℒ_{r}.
\underline{Proof}. Let v(ϵ) be the smallest positive element in vK^{×} , where ϵ ∈ K.  Then
v(ϵ) is not p-divisible. So Φ = {x ∈ K | v(ϵx^{p}) > 0} is definable in ℒ_{r}. Clearly,
V ⊆ Φ. On the other hand if v(x) < 0 then v(ϵx^{p}) < 0 as well since v(ϵ) is the
smallest positive element of vK^{×} .  □
\underline{Theorem} 4. Suppose that (K, V) is p-leasant.  If vK^{×} contains a p-rendible
convex subgroup C, then V is definable in ℒ_{r}.
\underline{Proof}. We assume that vK^{×} is dense (Corollary 3 proves the discrete case).
  Let ϵ ∈ K be such that v(ϵ) ∈ C is not p-divisible. Now by Proposition 2,

Ψ := {ϵx^{p} | v(ϵx^{p}) > 0} is definable in ℒ_{r}.

  We show it for the case where (K, V) is Henselian, as proofs for the other
two cases are similar. We define

(3)  Ω := {x^{p} − x^{p−1} ∈ K | (∃y ∈ K) (∃z ∈ Ψ) [z(y^{p} − y^{p−1} ) = x^{p} − x^{p−1} ] } ,

Then Ω ⊆ M_{v} because v(z) is not p-divisible and is positive for all z ∈ Ψ;
M_{v} ⊆ Ω because of Hensel's Lemma and the fact that Ψ contains arbitrarily
small positively valued elements of K thanks to the p-regularity of C. □
  The above results about definable valuations all involve one parameter whose
value is not p-divisible. We can say more in the case where the value group
is rendible; in particular, Koenigsmann's Lemma 3.6 in [8] follows from the
following.
\underline{Theorem} 5. Suppose that (K, V) is a Henselian valued field with vK^{×} rendible.
Then V is 0-definable in ℒ_{r}.
\underline{Proof}. If vK^{×} is discrete, then the minimal element of vK^{×} is not l-divisible
for all prime l. Choose one l not equal to char(Kv), then one applies Corol-
lary 3 and Ax's trick in [1], to show that if vK^{×} is l-rendible with v being
Henselian and l ≠ char(Kv), then V is 0-definable. The proof is essentially
the same as Ax's.

If vK^{×} is dense, then for ϵ ∈ K^{×} , let Ψ_{ϵ} = {ϵx^{p} | ∃y(y^{p}−y^{p−1} ) = ϵx^{p})}; define

Ω_{ϵ} := {x^{p} − x^{p−1} ∈ K | (∃y ∈ K) (∃z ∈ Ψ_{ϵ}) [z(y^{p} − y^{p−1} ) = x^{p} − x^{p−1} ] } .

Then the proof of Theorem 4 implies that M_{v} = ⋂ _{ϵ≠0} Ω_{ϵ}, because for all ϵ ≠ 0,
Ψ_{ϵ} contains arbitrarily small positively valued elements.  □
  The conditions in Theorem 4 are not necessary for a non-divisible Henselian
valuation ring to be definable in ℒ_{r}. To see this, we recall that for a field
K which is not p-closed, there is a so-called “canonical p-Henselian valua-
tion” on it, denoted by O_{p}, whose valuation ring is the smallest (with respect
to “⊆”) p-Henselian valuation ring if there is no p-Henselian valuation on K
with a p-closed residue field, or the biggest p-Henselian valuation on K with
a p-colsed residue field. See e.g. [7] for details.
  We have the following fact about the canonical p-Henselian valuation O_{p}.
\underline{Fact} 6 ( [7]). Assume that K, not p-closed, satisfies that ζ_{p} ∈ K or char(K) =
p. Then O_{p} is 0-definable in ℒ_{r}, unless p = 2 and the residue field of O_{p} is
euclidean.
\underline{Remark} 7. Given any ordered abelian group G, consider the ordered valued field
Q((t^{G})) with the natural valuation V. By Lemma 4.3.6 of [4], every 2-Henselian
valuation ring on K is convex, hence comparable to V.  But V must be the
O_{2} which is 0-definable by Fact 6. So the non-necessity of the conditions of
Theorem 4 follows; e.g.  when G is the lexicographical product of Q and Z
with elements in Z bigger.
  For an ordered abelian group G, recall that in Schmitt's [9], for g ∈ G,
B(g) is the smallest convex subgroup containing g; for n > 1 an integer, A_{n}(g)
is the smallest convex subgroup C not containing g in G such that B(g)/C is
n-regular; A_{n}(0) := ∅. Then A_{n}(g) is definable using g in G in the language of
ordered abelian groups, for all n. Some cases of Theorem 4 follow from Fact
6:
\underline{Corollary} 8. Suppose that K satisfies the hypothesis of Fact 6.  If V is a
p-Henselian valuation on K with vK^{×} containing a p-rendible convex subgroup
C, then O_{p} ⊆ V and V is definable in ℒ_{r}.
\underline{Proof}. If a field K′ is p-closed, and K′ contains ζ_{p} or char(K′) = p, then
wK^{′×} is p-divisible for all valuations w on K′. If V ⊊ O_{p}, then because C
is p-rendible (so any convex subgroup of vK^{×} is not p-divisible), the residue
field Ko_{p} (by our convention) of O_{p} is not p-closed, because ζ_{p} ∈ Ko_{p} or
char(Ko_{p}) = p. This contradicts the definition of O_{p}.
  Therefore, O_{p} ⊆ V. Let γ ∈ o_{p}K^{×} be such that γ + Δ ∈ C is not p-divisible,
where Δ is the convex subgroup of o_{p}K^{×} corresponding to V. Then A_{p}(g) = Δ
in o_{p}K^{×} . So V is definable using O_{p}.  □
  One can combine Corollary 8 and Theorem 4 to get that if K, being not p-
closed, with V p-Henselian and either ζ_{p} ∈ K or char(K) = p, then V is defin-
able if vK^{×} contains a p-rendible convex subgroup.
\underline{Proposition} 9. Given an ordered abelian group G, not p-divisible, with a non-
dense order type (henceforth called the \underline{principal p-regular rank}) of the col-
lection of A_{p}(g) in G for g ≥ 0, there is some convex subgroup C of G such
that G/C contains a convex p-rendible subgroup.

\underline{Proof}. It is enough to show that B(g)/A_{p}(g) is not p-divisible for some g. By
assumption, there are some g_{1} < g_{2} such that A_{p}(g_{1}) ⊊ A_{p}(g_{2}) and for all g_{1} <

g < A_{p}(g_{2}), A_{p}(g) = A_{p}(g_{1}). We may assume that g_{1} ≠ 0. Because B(g_{2})/A_{p}(g_{1}) is

not p-regular,

B(g_{2})/A_{p}(g_{2}) = \frac{B(g_{2})/A_{p}(g_{1})}{A_{p}(g_{2})/A_{p}(g_{1})}

implies that either B(g_{2})/A_{p}(g_{2}) is not p-divisible, or B(g)/A_{p}(g) = B(g)/A_{p}(g_{1})
is not p-divisible for some g_{1} < g ∈ A_{p}(g_{2}).  □
Without the assumption of the non-denseness of the principal p-regular rank
of G, the conclusion of Proposition 9 could be false. For example, let I = Q
be the index set, and let H := ∏ ′_{i∈I} G_{i} be the subgroup of ∏ _{i∈I} G_{i} consisting
of those elements with well-ordered support, where G_{i} = Q. H is linearly
ordered lexicographically. For a, b ∈ I with a < b, let s(a, b) = {b−(b−a)/n}_{n=1}^{∞} .
Then the subgroup of H generated by ⨁ _{i∈I} G_{i} and the characteristic functions
of s(a, b) (for all a < b), does not have that property.
It is then clear that if V is p-leasant on K with vK^{×} not p-divisible with
a non-dense principal p-regular rank, then some valuation ring containing V
is definable. This is a result similar to that about defining the valuation
up to equivalence of topologies in Koenigsmann [6].
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Department of Math. & Stats., McMaster University, Canada.
E-mail address: hongj37@math.mcmaster.ca