1 Introduction
In recent years, the study of definable Henselian valuations has received a considerable amount of attention and several definability conditions have been established. Early results of this study were related to decidability questions (cf., e.g., [Reference Ax2, Reference Robinson26]), whereas a more recent motivation is due to the Shelah–Hasson Conjecture on the classification of NIP fields (cf., e.g., [Reference Dupont, Hasson and Kuhlmann5, Reference Halevi and Hasson10, Reference Halevi and Hasson11, Reference Halevi, Hasson and Jahnke12, Reference Jahnke, Simon and Walsberg17, Reference Johnson18, Reference Krapp, Kuhlmann and Lehéricy22, Reference Shelah27]). We refer the reader to [Reference Fehm, Jahnke, Broglia, Delon, Dickmann, GondardCozette and Powers9] for a more detailed survey on the definability of Henselian valuations. While some of these results deal with the existence of nontrivial definable Henselian valuations and others consider the quantifier complexity of defining formulas (cf. [Reference Fehm, Jahnke, Broglia, Delon, Dickmann, GondardCozette and Powers9, Sections 2 and 3]), in this paper we are interested in the definability of a given Henselian valuation with a certain value group.Footnote ^{1}
The strongest currently known definability results which only pose conditions on the value group of a given Henselian valuation are exhibited in [Reference Hong14, Reference Krapp, Kuhlmann and Lehéricy21]. In the following, we state the results under consideration for this paper and outline how we strengthen these.
Fact 1.1 [Reference Hong14, Corollary 2]
Let $(K,v)$ be a Henselian valued field such that the value group $vK$ is discretely ordered. Then v is $\mathcal {L}_{\mathrm {r}}$ definable with one parameter from K.
For densely ordered value groups, the following is established.
Fact 1.2 [Reference Hong14, Theorem 3]
Let $(K,v)$ be a Henselian valued field such that $vK$ is densely ordered. Assume that $vK$ contains a convex subgroup that is pregular but not pdivisible, for some prime $p\in \mathbb {N}$ . Then v is $\mathcal {L}_{\mathrm {r}}$ definable with one parameter from K.
While in [Reference Hong14] definability is treated in the language of rings $\mathcal {L}_{\mathrm {r}}$ , the definability results in [Reference Krapp, Kuhlmann and Lehéricy21] are stated for ordered fields in the language of ordered rings $\mathcal {L}_{\mathrm {or}}$ .
Fact 1.3 [Reference Krapp, Kuhlmann and Lehéricy21, Theorem 5.3(2)]
Let $(K,<,v)$ be an ordered Henselian valued field such that $vK$ is not closed in its divisible hull. Then v is $\mathcal {L}_{\mathrm {or}}$ definable with one parameter from K.
After introducing preliminary notions in Section 2, we show in Section 3 that the hypothesis on $vK$ in Fact 1.2 is strictly stronger than that in Fact 1.3 (see Proposition 3.3 and Example 3.4). Thus, we obtain the following strengthening of both Facts 1.2 and 1.3:Footnote ^{2}
Theorem A. Let $(K,v)$ be a Henselian valued field such that $vK$ is not closed in its divisible hull. Then v is $\mathcal {L}_{\mathrm {r}}$ definable with one parameter from K.
In Section 4, we examine definability without parameters of the valuations. To this end, in Construction 4.5—the technical heart of this paper—we refine [Reference Prestel and Ziegler25, p. 338] to obtain a tHenselian nonHenselian field (see Definition 4.1) with prescribed value groups. More precisely, given an arbitrary family of nontrivial ordered abelian groups $\{A_{\gamma }\mid \gamma \in {{}\hspace{1pt}\omega} \}$ , Construction 4.5 produces a tHenselian nonHenselian ordered field $(K,<)$ as well as, for each $n\in \mathbb {N}$ , a convex valuation $u_n$ on K whose value group is the Hahn sum (see page 3) over $\{A_{\gamma }\mid \gamma \leq {{}\hspace{1pt}n}\}$ .
By applying Construction 4.5 to the specific case where all $A_{\gamma }$ are identical, we obtain the following:Footnote ^{3}
Theorem B. Let $A\neq \{0\}$ be an ordered abelian group. Then there exists an ordered Henselian valued field $(L,<,v)$ such that $vL$ is the Hahn sum $\coprod _{\omega }A$ and the only parameterfree $\mathcal {L}_{\mathrm {or}}$ definable Henselian valuation on $(L,<)$ is the trivial valuation. In particular, v is not $\mathcal {L}_{\mathrm {or}}$ definable without parameters.
Finally, by choosing appropriately the group in Theorem B, we can show that Facts 1.1–1.3 and Theorem A are optimal in the sense that, in general, one cannot obtain definability of the valuations without parameters (see Examples 4.9 and 4.10).
We conclude this work in Section 5 by collecting open questions motivated by our results.
2 Preliminaries
More details on the model and valuation theoretic notions we use can be found in [Reference Engler and Prestel6, Reference Kuhlmann23, Reference Marker24].
We denote by $\mathbb {N}$ the set of natural numbers without $0$ and by $\omega $ the set of natural numbers with $0$ . All topological properties are considered with respect to the order topology. Whenever an algebraic structure carries a standard linear ordering, we refer to this specific ordering if not explicitly stated otherwise.
Let K be a field and let v be a valuation on K. We denote the valuation ring of v (i.e., the subring $\{a\in K\mid v(a)\geq 0\}$ of K) by $\mathcal {O}_v$ and its valuation ideal (i.e., its maximal ideal $\{a\in \mathcal {O}_v\mid v(a)>0\}$ ) by $\mathcal {M}_v$ . Moreover, we denote the value group of v by $vK$ and the residue field $\mathcal {O}_v/\mathcal {M}_v$ by $Kv$ . An element $a+\mathcal {M}_v \in Kv$ with $a\in \mathcal {O}_v$ is also denoted by $av$ . Suppose that K admits an ordering $<$ . Then we say that v is convex (with respect to $<$ ) if $\mathcal {O}_v$ is a convex subset of K (with respect to the ordering $<$ ). By [Reference Knebusch and Wright19, Lemma 2.1], any Henselian valuation on an ordered field is convex. Given a valued field K we denote by ${K}^{\mathrm {h}}$ its henselisation.
Let $\mathcal {L}_{\mathrm {og}}=\{+,,0,<\}$ be the language of ordered groups, $\mathcal {L}_{\mathrm {r}}=\{+,,\cdot ,0, 1\}$ the language of rings, and $\mathcal {L}_{\mathrm {or}}=\mathcal {L}_{\mathrm {r}}\cup \{<\}$ the language of ordered rings. If no confusion is likely to arise, we denote an $\mathcal {L}_{\mathrm {og}}$ structure $(G,+,,0,<)$ simply by G, an $\mathcal {L}_{\mathrm {r}}$ structure $(R,+,,\cdot ,0,1)$ by R, and an $\mathcal {L}_{\mathrm {or}}$ structure $(R,+,,\cdot ,0,1, <)$ by $(R,<)$ . We say that a valuation v on a field K (an ordered field $(K,<)$ ) is $\mathcal {L}_{\mathrm {r}}$ definable ( $\mathcal {L}_{\mathrm {or}}$ definable) if $\mathcal {O}_v$ is an $\mathcal {L}_{\mathrm {r}}$ definable ( $\mathcal {L}_{\mathrm {or}}$ definable) subset of K. Here, definability is meant with parameters. We say that v is $0$  $\mathcal {L}_{\mathrm {r}}$ definable ( $0$  $\mathcal {L}_{\mathrm {or}}$ definable) if there is a defining formula without parameters from K.
If we expand $\mathcal {L}_{\mathrm {r}}$ by a unary predicate $\mathcal {O}_v$ , we obtain the language of valued fields $\mathcal {L}_{\mathrm {vf}}$ . Similarly, $\mathcal {L}_{\mathrm {ovf}}=\mathcal {L}_{\mathrm {or}}\cup \{\mathcal {O}_v\}$ denotes the language of ordered valued fields. Let $(K,\mathcal {O}_v)$ be a valued field. An atomic formula of the form $v(t_1)\geq v(t_2)$ , where $t_1$ and $t_2$ are $\mathcal {L}_{\mathrm {r}}$ terms, is equivalent to $t_1=t_2=0\vee (t_2\neq 0\wedge \tfrac {t_1}{t_2}\in \mathcal {O}_v)$ . Hence, by abuse of notation, we can also write $(K,v)$ for the $\mathcal {L}_{\mathrm {vf}}$ structure of a valued field. Likewise the $\mathcal {L}_{\mathrm {ovf}}$ structure of an ordered valued field is denoted by $(K,<,v)$ .
Let $(\Gamma ,<)$ be a linearly ordered set, and for each $\gamma \in \Gamma $ , let $A_\gamma \neq \{0\}$ be an abelian group (expressed additively). The corresponding Hahn product over $\left \{A_\gamma \mid \gamma \in \Gamma \right \}$ is given by
where $\prod _{\gamma \in \Gamma }A_\gamma $ denotes the group product over $(A_\gamma )_{\gamma \in \Gamma }$ and $\mathrm {supp}(s)$ denotes the support of s given by the set $\{\gamma \in \Gamma \mid s(\gamma )\neq 0\}$ . The Hahn product is an abelian group under pointwise addition. It carries a valuation with value set $\Gamma $ given by $v_{\min }\colon s\mapsto \min \mathrm {supp}(s)$ for $s\neq 0$ . Any $s\in \mathbf {H}_{\gamma \in \Gamma }A_\gamma $ can be expressed as a sum , where $s_\gamma = s(\gamma )$ and is the characteristic function mapping $\gamma $ to some fixed $1_\gamma \in A_\gamma $ (which in our application is always clear from the context) and everything else to $0$ . The valued subgroup
is called the Hahn sum over $(\Gamma ,(A_\gamma )_{\gamma \in \Gamma })$ . If $\Gamma $ is a convex subset of $\mathbb {Z}$ with $\min \Gamma =a$ and $\max \Gamma =b$ (where we set $a={{}\hspace{1pt}\infty}$ if $\Gamma $ is not bounded from below, and $b=\infty $ if $\Gamma $ is not bounded from above), then we express $\coprod _{\gamma \in \Gamma } A_\gamma $ as $\coprod _{\gamma =a}^b A_\gamma $ if $a\neq {{}\hspace{1pt}\infty} $ , and as $\coprod _{\gamma>  \infty }^b A_\gamma $ if $a = {{}\hspace{1pt}\infty}$ . For an ordinal $\alpha $ we have that $\coprod _{\gamma \in \alpha }A_\gamma = \mathbf {H}_{\gamma \in \alpha }A_\gamma $ . If there is an ordered abelian group A such that for each $\gamma \in \Gamma $ we have $A_\gamma =A$ , we simply write $\mathbf {H}_\Gamma A$ ( $\coprod _{\Gamma } A$ ) for the Hahn product (Hahn sum) above. If $A_\gamma $ is an ordered abelian group for each $\gamma \in \Gamma $ , then also $\mathbf {H}_{\gamma \in \Gamma }A_\gamma $ is an ordered abelian group with the ordering given as follows: for any nonzero we have $s> 0$ if and only if $s_{\gamma _0}> 0$ , where $\gamma _0 = v_{\min } \left (s\right )$ .
Let k be a field and let G be an ordered abelian group. The Hahn field (also called power series field) $k\!\left (\!\left ( G \right )\!\right )$ has as underlying additive group the Hahn product $\mathbf {H}_{G}k$ . Multiplication on $k\!\left (\!\left ( G \right )\!\right )$ is defined by $(rs)_h=\sum _{g\in G}r_g s_{hg}$ for any $h\in G$ . To distinguish between elements of Hahn products and Hahn fields, we write an element $s\in k\!\left (\!\left ( G \right )\!\right )$ as sum $s=\sum _{g\in G}s_gt^g$ , where $t^g$ is the characteristic function with $t^g(g)=1$ and $t^g(h)=0$ for any $h\neq g$ . The valuation $v_{\min }$ on $k\!\left (\!\left ( G \right )\!\right )$ is Henselian. Again, if k is ordered, then also $k\!\left (\!\left ( G \right )\!\right )$ carries a natural ordering as described above making it an ordered field. We denote by $k[G]$ the group ring with coefficient field k and value group G. It consists of all elements of $k\!\left (\!\left ( G \right )\!\right )$ with finite support. Its quotient field, denoted by $k(G)$ , is generated in $k\!\left (\!\left ( G \right )\!\right )$ by the set of all monic monomials, i.e., $k(G)=k(t^g\mid g\in G)$ .
Let G be an ordered abelian group. We say that G is discretely ordered if there exists some $1\in G$ with $0<1$ and there is no element in G strictly between $0$ and $1$ . If G is not discretely ordered, then it is densely ordered. Let $n\in \mathbb {N}$ . Then G is called nregular if every infinite convex subset of G contains at least one ndivisible element. If G is nregular for every $n\in \mathbb {N}$ , then we say that it is regular. We denote the divisible hull of G by ${G}^{\mathrm {div}}$ .
3 Definable valuations
We start by stating and proving the main result of this section. Subsequently, we will discuss how this generalises the known definability results from Facts 1.2 and 1.3.
Theorem 3.1. Let $(K,v)$ be a Henselian valued field. Suppose that $vK$ is not closed in ${vK}^{\mathrm {div}}$ . Then v is $\mathcal {L}_{\mathrm {r}}$ definable with one parameter.
Theorem 3.1 can essentially be proven by the methods in [Reference Hong14, p. 15] and [Reference Hong13, p. 19].Footnote ^{4} However, due to the more general context there are certain adjustments to be made, which is why we present the proof in detail.
Proof. We set $G=vK$ . Let $\varepsilon \in K$ and let $n\in \mathbb {N}$ such that $v(\varepsilon )$ is not ndivisible in G but $\tfrac {v(\varepsilon )}{n}$ is a limit point of G in ${G}^{\mathrm {div}}$ . Since $\tfrac {v(\varepsilon )}{n}$ is a limit point of G, also $\tfrac {v(\varepsilon )}{n}$ is a limit point of G. Thus, by replacing $\varepsilon $ by $\varepsilon ^{1}$ if necessary, we may assume that $\tfrac {v(\varepsilon )}{n}$ is a leftsided limit point of G, i.e., for any positive $h\in {G}^{\mathrm {div}}$ the halfopen interval $\left (\tfrac {v(\varepsilon )}{n}h,\tfrac {v(\varepsilon )}{n}\right ]$ in ${G}^{\mathrm {div}}$ contains some element from G.
Set $\Phi _\varepsilon =\{x\in K\mid v(\varepsilon x^n)>0\}$ . We show that the formula $\varphi (x,\varepsilon )$ given by
defines $\Phi _\varepsilon $ . Let $x\in \Phi _\varepsilon $ . Then $\varepsilon x^n\in \mathcal {M}_v$ and by Hensel’s Lemma there exists $y\in K$ such that $y^ny^{n1}=\varepsilon x^n$ , i.e., $K \models \varphi (x,\varepsilon )$ . Conversely, let $x\in K$ such that $K \models \varphi (x,\varepsilon )$ , i.e., there exists $y\in K$ such that $y^{n1}(y1)=\varepsilon x^n$ . By applying v, we obtain
and distinguish three cases.
Case 1: $v(y)>0$ . Then $v(y1)=0$ and thus $0<(n1)v(y)=v(\varepsilon )+nv(x)=v(\varepsilon x^n)$ , as required.
Case 2: $v(y)=0$ . Then $v(y1)=v(\varepsilon )+nv(x)$ . If $v(y1)>0$ , then we are done immediately. Otherwise, $v(y1)=0$ and we obtain $\tfrac {v(\varepsilon )}{n} = {{}\hspace{1pt}v}(x)$ , a contradiction to the fact that $v(\varepsilon )$ is not ndivisible in G.
Case 3: $v(y)<0$ . Then we obtain $v(\varepsilon ) = n (v(y)v(x))$ , again a contradiction, as $v(\varepsilon )$ is not ndivisible.
Now set
Again, $\Psi _\varepsilon $ is $\mathcal {L}_{\mathrm {r}}$ definable only using the parameter $\varepsilon $ . Finally, set
We show that $\Omega _\varepsilon =\mathcal {M}_v$ , as then $\mathcal {O}_v$ is defined by $\forall (u\in \Omega _\varepsilon ) \ xu\in \Omega _\varepsilon $ .
First let $a\in \Omega _\varepsilon $ and fix $x\in K$ such that $x^nx^{n1} = a$ . Then by definition of $\Omega _\varepsilon $ there are $y\in K$ and $z\in \Psi _\varepsilon $ such that $z(y^ny^{n1})=a$ . Let $r\in K$ with $z=\varepsilon r^n\in \mathcal {M}_v$ . Assume, for a contradiction, that $v(a)\leq 0$ . If $v(a)<0$ , then
Since
we obtain
contradicting the choice of $\varepsilon $ . If $v(a)=0$ , then $0=(n1)v(x)+v(x1)$ and thus $v(x)=v(x1)=0$ . Moreover, $0=v(z)+(n1)v(y)+v(y1)$ and thus $0>{{}\hspace{1pt}v}(z)=(n1)v(y)+v(y1)=nv(y)$ . This implies
which again contradicts the fact that $v(\varepsilon )$ is not ndivisible. This shows that $v(a)>0$ , i.e., $a\in \mathcal {M}_v$ .
For the converse, let $a\in \mathcal {M}_v$ . By Hensel’s Lemma there exists $x\in K$ such that $x^nx^{n1}=a$ . Since $\tfrac {v(\varepsilon )}{n}$ is a leftsided limit point of G, we obtain some $b \in K^\times $ with
Hence, $v(\varepsilon )< {{}\hspace{1pt}nv}(b)<v(a)v(\varepsilon )$ . We set $z=\varepsilon b^{n}$ and obtain that $0<v(z)<v(a)$ , whence $z\in \Psi _\varepsilon $ and $v\!\left ( \frac {a}{z} \right )>0$ . Again by applying Hensel’s Lemma, we find some $y\in K$ such that $y^ny^{n1}=\frac {a}{z}$ . This yields $a\in \Omega _\varepsilon $ and therefore $\Omega _\varepsilon = \mathcal {M}_v$ , as required. ⊣
Remark 3.2. Note that the class of nontrivial ordered abelian groups that are closed in their divisible hull is elementary in the language of ordered groups. Indeed, it is axiomatised by the axiom scheme
one for each $n\in \mathbb {N}$ .⊣
Our next aim is to show that Theorem 3.1 implies Fact 1.2. Note that in both of these statements the value group is densely ordered, in Fact 1.2 by assumption and in Theorem 3.1 since any discretely ordered abelian group is already closed in its divisible hull.
Proposition 3.3. Let G be a densely ordered abelian group. Suppose that there exists some $n\in \mathbb {N}$ such that G contains a convex subgroup C that is nregular but not ndivisible. Then G is not closed in ${G}^{\mathrm {div}}$ .
Proof. Note that since G contains a convex subgroup that is not ndivisible, we have that G is a nondivisible ordered abelian group. Assume, for a contradiction, that G is closed in ${G}^{\mathrm {div}}$ . Since C is not ndivisible, there exists $a \in C$ such that $\tfrac {a}{n} \in {G}^{\mathrm {div}}\setminus G$ . Moreover, since G is closed in ${G}^{\mathrm {div}}$ , there exists a positive $\delta \in {G}^{\mathrm {div}}$ such that
As $0$ is a limit point of G in ${G}^{\mathrm {div}}$ (cf. [Reference Krapp, Kuhlmann and Lehéricy21, Lemma 3.6]), there exists $b\in G$ with $0<b<\delta $ and hence
By convexity of C, we may choose $b\in C^{>0}$ . Further, since G is densely ordered and thus C is densely ordered, the interval $(ab,a+b)$ is an infinite convex subset of C. By nregularity of C, there is some $z\in C$ such that $ab \leq nz \leq a+b$ . This is equivalent to
a contradiction to (1). Hence, G is not closed in ${G}^{\mathrm {div}}$ .⊣
Proposition 3.3 shows that the condition on the value group in Fact 1.2 is covered by the condition in Theorem 3.1. The converse of Proposition 3.3 does not hold, as verified by the following example.
Example 3.4. We consider the ordered abelian group
Note that G consists of all elements
for which the following holds: there exist $m\in \mathbb {Z}$ and $n_0\in \omega $ such that for any $n \geq n_0$ we have $s_n = 2m$ . We now show that G is not closed in ${G}^{\mathrm {div}}$ but for any $n\in \mathbb {N}$ and any convex subgroup C of G, if C is nregular, then it is already ndivisible.
By [Reference Krapp, Kuhlmann and Lehéricy21, Lemma 3.6], we have that G is densely ordered, as its value set $v_{\min } G=\omega $ does not have a maximum. We first note that the divisible hull of G is given by
Let . By the description of the elements of G above, one immediately sees that $s \notin G$ . We now show that s is a limit point of G in ${G}^{\mathrm {div}}$ , whence G is not closed in ${G}^{\mathrm {div}}$ . Let $b\in {G}^{\mathrm {div}}$ be positive and set $\ell = v_{\min } (b)$ . Further, let and note that $g<s$ . As $b_{\ell }> 0$ , we obtain that . We have thus found an element $g\in G$ with $sb<g<s$ , showing that s is a limit point of G in ${G}^{\mathrm {div}}$ .
Let $n\in \mathbb {N}$ with $n\geq 2$ . We show that G contains no nontrivial nregular convex subgroup. Thus, in particular, the only nregular convex subgroup of G is the trivial group, which is ndivisible.
Let C be a nontrivial convex subgroup of G. Then there exists $\ell \in \omega $ such that C is of the form
(cf. [Reference Kuhlmann23, p. 50 f.]). Now let $c,d \in C$ with
and
, where ${q}\in \mathbb {N}$ is some prime with ${q}>n$ . Then $c<d$ and $(c,d)$ is an infinite convex subset of the densely ordered group C. Since for any $z \in (c,d)$ we have that $z_{\ell } = q$ and $n \nmid q$ in $\mathbb {Z}$ , there is no ndivisible element in $(c,d)$ . Hence, C is not nregular. ⊣
Example 3.4 shows that the condition in Theorem 3.1 is indeed more general than the condition in Fact 1.2. More precisely, let k be any field and let G be the ordered abelian group constructed in Example 3.4; then Theorem 3.1 implies that the Henselian valuation $v_{\min }$ on $k\!\left (\!\left ( G \right )\!\right )$ is $\mathcal {L}_{\mathrm {r}}$ definable (with one parameter), whereas Fact 1.2 cannot be applied directly to obtain the $\mathcal {L}_{\mathrm {r}}$ definability of $v_{\min }$ .
In conclusion, Theorem 3.1 is a strict generalisation of both Facts 1.2 and 1.3 for densely ordered value groups: Theorem 3.1 clearly generalises Fact 1.3, as it has the same condition on the value group but does not rely on the existence of an ordering on the field. Moreover, Proposition 3.3 shows that the value group condition of Theorem 3.1 follows from that of Fact 1.2 but Example 3.4 illustrates that the converse does not hold.
4 Main construction and parameterfree definability
The main aim of this section is to show that the conclusions of Fact 1.1 and Theorem 3.1 can, in general, not be strengthened to $0$  $\mathcal {L}_{\mathrm {r}}$ definability. More precisely, we construct a class of ordered Henselian valued fields that do not admit any nontrivial $0$  $\mathcal {L}_{\mathrm {or}}$ definable Henselian valuation (see Theorem 4.7). This class contains ordered Henselian valued fields $(K,<,v)$ with discretely ordered value group $vK$ (see Example 4.9) as well as such with a value group $vK$ not closed in ${vK}^{\mathrm {div}}$ (see Example 4.10).
The method for constructing an ordered Henselian valued field $(K,<,v)$ that does not admit a nontrivial $0$  $\mathcal {L}_{\mathrm {or}}$ definable Henselian valuation relies on the construction of a tHenselian nonHenselian ordered field.Footnote ^{5} We start by clarifying the notion of tHenselianity, which was first introduced in [Reference Prestel and Ziegler25, Section 7].Footnote ^{6}
Definition 4.1. We call a field Henselian if it admits a nontrivial Henselian valuation. A field K is tHenselian if there exists a Henselian field L such that $K\equiv L$ (in the language $\mathcal {L}_{\mathrm {r}}$ ).
We now establish a sufficient condition for tHenselianity based on $n_\leq $ Henselian valuations: Let $n\in \mathbb {N}$ . Following the terminology of [Reference Fehm and Jahnke8, Definition 6.1], we say that a valued field $(K,v)$ is $n_\leq $ Henselian if Hensel’s Lemma holds in $(K,v)$ for all polynomials of degree at most n. Note that $(K,v)$ is Henselian if and only if it is $n_\leq $ Henselian for any $n\in \mathbb {N}$ . Arguing similarly as in the proof of [Reference Engler and Prestel6, Theorem 4.1.3], one can show that $(K,v)$ is $n_\leq $ Henselian if and only if every polynomial $X^\ell +X^{\ell 1}+a_{\ell 2}X^{\ell 2}+\cdots +a_0 \in \mathcal {O}_v[X]$ with $\ell \leq n$ and $a_{\ell 2},\ldots ,a_0\in \mathcal {M}_v$ has a zero in K. Moreover, by the arguments in the proof of [Reference Knebusch and Wright19, Lemma 2.1], for $n\geq 2$ any $n_\leq $ Henselian valuation on a real field K is convex with respect to any ordering on K.
Lemma 4.2. Let K be a field. Suppose that for any $n\in \mathbb {N}$ , there exists a nontrivial $n_\leq $ Henselian valuation on K. Then K is tHenselian.
Proof. Let $h_n$ be the $\mathcal {L}_{\mathrm {vf}}$ sentence stating that v is a nontrivial valuation for which Hensel’s Lemma holds for polynomials of degree at most n. Then any finite subset of the $\mathcal {L}_{\mathrm {vf}}$ theory $\operatorname {\mathrm {Th}}(K)\cup \{h_n\mid n\in \mathbb {N}\}$ is satisfied by $(K,v_m)$ for some $m\in \mathbb {N}$ , where $v_m$ is $m_\leq $ Henselian. Hence, by the Compactness Theorem, there exists a Henselian field L with $L\models \operatorname {\mathrm {Th}}(K)$ , i.e., $L\equiv K$ .⊣
Our aim is to refine the construction of a tHenselian nonHenselian field in [Reference Prestel and Ziegler25, p. 338] by gaining influence on the resulting value group. More precisely, starting with a family of nontrivial ordered abelian groups $\{A_{\gamma }\mid \gamma \in \omega \}$ , we construct a nonHenselian ordered field $(K,<)$ such that for any $n\in \mathbb {N}$ , there is an $n_\leq $ Henselian valuation $v_n$ on K with $v_nK = \coprod _{\gamma>\infty }^{n} A_{\gamma }$ .
In the following, for an ordered abelian group G we consider ${G}^{\mathrm {div}}$ also as a $\mathbb {Q}$ vector space with the usual scalar multiplication. For an algebraic field extension $K\subseteq L$ and an element $\alpha \in L$ , we denote by $\deg _K(\alpha )$ the degree of the minimal polynomial of $\alpha $ over K.
Lemmas 4.3 and 4.4 will be used in Construction 4.5. Note that the statement of Lemma 4.4 can be adjusted for an arbitrary field k of characteristic $0$ (instead of an ordered field).
Lemma 4.3. Let k be a field of characteristic $0$ and let G be an ordered abelian group. Moreover, let $a\in k[G]$ such that for some $n\geq 2$ we have $\mathrm {supp}(a) = \{g_1,\dots ,g_n,0\}$ and $g_1,\dots ,g_n$ are linearly independent in the $\mathbb {Q}$ vector space ${G}^{\mathrm {div}}$ . Then a is prime in $k[G]$ .
Proof. First note that the group of units of $k[G]$ is given by the multiplicative group of monomials of $k\!\left (\!\left ( G \right )\!\right )$ , i.e., $k[G]^\times =\{ct^g\mid c\in k^\times , g\in G\}$ . By [Reference Biljakovic, Kochetov, Kuhlmann, Enayat, Kalantari and Moniri3, Proposition 4.1], every irreducible element of $k[G]$ is prime. It thus suffices to show that a is irreducible. Let $r,s\in k[G]$ with $a=rs$ . We obtain from [Reference Biljakovic, Kochetov, Kuhlmann, Enayat, Kalantari and Moniri3, Proposition 4.8] that a is prime in $k[{G}^{\mathrm {div}}]$ and hence irreducible in $k[{G}^{\mathrm {div}}]$ (as $k[{G}^{\mathrm {div}}]$ is an integral domain). We may thus assume that r is a unit in $k[{G}^{\mathrm {div}}]$ . Hence, $r\in k[G]\cap \{ct^g\mid c\in k^\times , g\in {G}^{\mathrm {div}}\}=k[G]^\times $ . This shows that r is a unit in $k[G]$ , as required.⊣
For two ordered abelian groups A and B, we denote by $A\amalg B$ the abelian group $A\times B$ ordered lexicographically.
Lemma 4.4. Let $(k,<)$ be an ordered field and let $A_1, A_2\neq \{0\}$ be ordered abelian groups. Moreover, set $K=k(A_1\amalg A_2)$ and let $({K}^{\mathrm {h}},<)$ be a henselisation of K in $(k\!\left (\!\left ( A_1\amalg A_2 \right )\!\right ),<,v_{\min })$ . Then for any $n\in \mathbb {N}$ , there exists an intermediate field extension $K\subseteq L \subsetneq {K}^{\mathrm {h}}$ such that $(L,v_{\min })$ is $n_\leq $ Henselian but not Henselian.
Proof. Let $n\in \mathbb {N}$ and let $p\in \mathbb {N}$ be prime with $p>n$ . Since $p_{\leq }$ Henselianity implies $n_\leq $ Henselianity, it suffices to construct an ordered valued field $(L, <,v_{\min })$ with $(K,<,v_{\min })\subseteq (L, <,v_{\min }) \subsetneq ({K}^{\mathrm {h}},<,v_{\min })$ such that for any polynomial of the form
where $m\leq p$ , and any $\alpha \in {K}^{\mathrm {h}}$ with $f(\alpha )=0$ , we have that $\alpha \in L$ . Let M be the set of all natural numbers m such that for any prime number $q\in \mathbb {N}$ dividing m we have $q\leq p$ . We set $L=K\!\left (\alpha \in {K}^{\mathrm {h}}\mid \deg _K(\alpha )\in M\right )$ , i.e., L is generated by all elements in ${K}^{\mathrm {h}}$ whose degree over K only has prime factors less than or equal to p. Note that since M is closed under multiplication and divisors, $\deg _K(x)\in M$ for any $x\in L$ . Let $m\leq p$ and let $f(T)=T^m + a_{m1}T^{m1}+ \cdots + a_0 \in L[T]$ . By construction of L, we have $\deg _K(a_i)\in M$ for any $i\in \{0,\ldots ,m1\}$ . Moreover, since $[K(a_0,\ldots ,a_{m1}):K]$ divides the product $[K(a_0):K]\ldots [K(a_{m1}):K]=\deg _K(a_0)\ldots \deg _K(a_{m1})$ , we obtain $[K(a_0,\ldots ,a_{m1}):K] \in M$ . Suppose that $\alpha \in {K}^{\mathrm {h}}$ is a zero of f. Since also $[K(a_0,\ldots ,a_{m1},\alpha ):K(a_0,\ldots ,a_{m1})]\leq m\leq p$ , we obtain
Finally, since $\deg _K(\alpha )=[K(\alpha ):K]$ divides $[K(a_0,\ldots ,a_{m1},\alpha ):K]$ , we obtain $\deg _K(\alpha )\in M$ and thus $\alpha \in L$ .
It remains to show that $(L,v_{\min })$ is not Henselian, i.e., that there exists $y\in {K}^{\mathrm {h}}$ such that $y\notin L$ . Let $q \in \mathbb {N}$ be a prime number greater than p. We set
where $b_1\in A_1$ and $b_2\in A_2$ are fixed positive elements, and observe that a is prime in $k[A_1\amalg A_2]$ by Lemma 4.3. Further, let
Note that $v_{\min }(a)=0$ and $v_{\min }(a t^{(b_1,0)})=(b_1,0)>0$ . Hence, by applying the residue map, we obtain the polynomial $gv_{\min } (T)=T^q+T^{q1}\in k[T]$ . Since $gv_{\min }(T)$ has a simple zero in k, we obtain that g has a zero $y\in {K}^{\mathrm {h}}$ . Moreover, by Eisenstein’s Criterion we obtain that g is irreducible over K and thus $q = [K(y) : K]$ . Since q does not divide any element in M, we obtain by construction of L that $y\notin L$ . Hence, $L\subsetneq {K}^{\mathrm {h}}$ as required.⊣
We now proceed with our main construction.
Construction 4.5 (Main Construction)
Let $\{A_\gamma \mid \gamma \in \omega \}$ be a family of nontrivial ordered abelian groups and set $G:= \coprod _{\gamma \in \omega }A_{\gamma }$ . We construct an ordered field $(K,<)$ such that for any $n\in \mathbb {N}$ there exists a convex $n_\leq $ Henselian valuation $ u_n$ on K with value group $u_nK = \coprod _{\gamma> \infty }^{n} A_ {\gamma }$ and K does not admit a nontrivial Henselian valuation. In particular, K is tHenselian (by Lemma 4.2) but not Henselian.
By induction on $m\in \omega $ we construct a chain of ordered fields $(K_m, <)$ each endowed with a valuation $v_{\mathrm {min}}^m$ . From this the ordered field $(K,<)$ will be obtained by an inverse limit construction, as well as a chain of convex valuations $v_m$ on K for $m\in \omega $ . Ultimately, for each $n\in \omega $ we set $u_{2n}$ to be $v_{n}$ and $u_{2n+1}$ to be a certain coarsening of $u_{2n}$ .
To initiate the induction, we set $(K_{0},<)=(\mathbb {R},<)$ . Now let $m\in \mathbb {N}$ and suppose that the ordered field $(K_{m1},<)$ has already been defined. We denote the valuation given by $s\mapsto \min \mathrm {supp} (s)$ on $K_{m1}\!\left (\!\left ( A_{2m1}\amalg A_{2(m1)} \right )\!\right )$ by $v_{\min }^m$ . By Lemma 4.4, there is an ordered valued field $K_m$ with
where the ordering is inherited from $K_{m1}\!\left (\!\left ( A_{2m1}\amalg A_{2(m1)} \right )\!\right )$ and the valuation is $v_{\min }^m$ , such that any zero in ${K_{m1}\!\left (A_{2m1}\amalg A_{2(m1)}\right )}^{\mathrm {h}}$ of a polynomial over $K_m$ of degree at most $ 2m+1$ already lies in $K_m$ . Since the extensions above are immediate, we have $K_mv_{\min }^m =K_{m1}$ . Let $\mathcal {O}_m'$ and $\mathcal {M}_m'$ be the valuation ring and valuation ideal of $v_{\min }^m$ in $K_m$ and let $\pi _m\colon \mathcal {O}_m' \to K_{m1}$ denote the corresponding residue map $a \mapsto av_{\min }^m$ .
For $i,j\in \omega $ with $j\leq i$ we define the map $ \psi _{i,j}\colon K_i \cup \{\infty \} \to K_j \cup \{\infty \}$ as follows: We set
For $0\leq j<i1$ we set $\psi _{i,j} = \psi _{j+1,j} \circ \dots \circ \psi _{i,i1}$ . Now we define I to be the inverse limit of $\left (K_m\cup \{\infty \}\right )_m$ , i.e., I is the following set of sequences:
Note that if for a given sequence $(x_m) \in I$ , we have $x_{\ell }=0$ (respectively $x_{\ell }=\infty $ ) for some $\ell \in \omega $ , then $x_k=0$ (respectively $x_k=\infty $ ) for any $k\leq \ell $ . On the other hand, if $x_{\ell } \notin \{0,\infty \}$ for some $\ell \in \omega $ , then
i.e., $v_{\min }^k(x_k)=0$ , for all $k>\ell $ . Therefore, if $x_{\ell } \notin \{0,\infty \}$ and $x_{\ell 1} \in \{0,\infty \}$ for some $\ell \in \mathbb {N}$ , then $v_{\min }^\ell (x_{\ell }) \neq 0. $
Claim 1. The set $K:= I\setminus \{(\infty )\}$ can be made an ordered field.
Proof. Proof of Claim 1
Addition, multiplication, and an ordering on K can be defined by a standard procedureFootnote ^{7} making K an ordered field: Let $(x_m),(y_m) \in K$ and let $\ell \in \omega $ be least such that $x_{\ell }\neq \infty $ and $y_{\ell }\neq \infty $ . For any $i\geq \ell $ set $z_i=x_i+y_i$ . For any $i<\ell $ set $z_i=\psi _{\ell ,i}(x_{\ell }+y_{\ell })$ . Then $(x_m) + (y_m)$ is given by $(z_m) \in K$ . Likewise, multiplication can be defined via the multiplication operations on the fields $K_m$ . Finally, a nonzero element $(x_m)$ of K is positive if and only if $x_i$ is nonnegative for any $i\in \omega $ with $x_i\neq \infty $ .⊣
For any $n\in \omega $ we define the projection map $\psi _n \colon K \to K_n\cup \{\infty \}, (x_m) \mapsto x_n$ . This gives rise to the following commutative diagram and equation for $j\leq i$ :
One can easily verify that $\psi _n$ is a place, and therefore $\mathcal {O}_n = \psi _n^{1}(K_n)$ defines a valuation ring on K with maximal ideal $\mathcal {M}_n=\psi _n^{1}(\{0\})$ and residue field $K_n$ . By (3) we compute for any $a\in \mathcal {M}_{n+1}$ :
Therefore we have $a\in \mathcal {M}_n$ , and thus $\mathcal {O}_n \subseteq \mathcal {O}_{n+1}$ .
We denote the valuation on K with valuation ring $\mathcal {O}_n$ by $v_n$ and fix $n\in \mathbb {N}$ .
Claim 2. $v_n$ is a $( 2n+1)_\leq $ Henselian valuation on K.
We observe that, if Claim 2 is established, then we also have that $v_n$ is $( 2n)_\leq $ Henselian and convex (as $ 2n+1\geq 2$ ).
Proof. Proof of Claim 2
Let
where $a^{( 2n1)},\dots ,a^{(0)}\in \mathcal {M}_n$ . We need to find a zero $(y_m)\in K$ of f. For any $i\in \omega $ , set $a_i^{( 2n1)} = \psi _i\big (a^{( 2n1)}\big ),\dots ,a_i^{(0)}=\psi _i\big (a^{(0)}\big ) \in K_i$ and
First let $i\leq n$ . Then $a_i^{( 2n2)}= \dots = a_i^{(0)}=0$ and therefore $f_i(T) = T^{ 2n+1} + T^{ 2n}$ . Set $y_i={}1$ , which is a simple zero of $f_i$ . Now let $i>n$ . We construct $y_i$ iteratively. To this end, assume that $y_{i1}\in K_{i1}$ is a simple zero of $f_{i1}$ with $\psi _{i1,j}(y_{i1})=y_j$ for $0\leq j\leq i1$ . Applying $\pi _i$ to the coefficients of $f_i(T)\in K_i[T]$ we obtain the residue polynomial
Thus, $f_i$ has a simple zero $y_i$ in the henselisation of $K_i$ with $\pi _i(y_i)=y_{i1}$ . By construction of $K_i$ (as $i>n$ ), we obtain $y_i\in K_i$ , as required. Finally, one can verify by direct computation that the element $(y_m)\in K$ we obtained from this construction is a zero of f.
The argument above shows that $v_n$ satisfies the condition for $( 2n+1)_\leq $ Henselianity only for polynomials of degree equal to $ 2n+1$ . However, since the coarsening of a $( 2n+1)_\leq $ Henselian valuation is also $( 2n+1)_\leq $ Henselian (see the proof of [Reference Krapp20, Lemma A.4.24]), we immediately obtain $( 2n+1)_\leq $ Henselianity of $v_n$ .⊣
It remains to show that $v_n$ is not Henselian.
Claim 3. $v_n$ is not Henselian.
Proof. Proof of Claim 3
Assume for a contradiction that $v_n$ is Henselian. Let $m\in \mathbb {N}$ and let
for some $a_{m2},\dots ,a_0\in \mathcal {M}^{\prime }_{n+1}$ . Let $a^{(m2)},\dots ,a^{(0)}\in K$ such that $\psi _{n+1}\!\big (a^{(i)}\big )=a_i$ for $0\leq i \leq m2$ . Then $a^{(m2)},\dots ,a^{(0)}\in \psi _n^{1}(\{0\})=\mathcal {M}_n$ . As $v_n$ is Henselian, there exists $x\in \mathcal {O}_n$ such that $x^m + x^{m1} + a^{(m2)}x^{m2} + \dots + a^{(0)} = 0$ . A direct computation shows that $\psi _{n+1}(x)$ is a zero of the polynomial $g(X)$ . Thus, we have that $\left (K_{n+1},v_{\min }^{n+1}\right )$ is Henselian, which is a contradiction to the construction of $K_{n+1}$ .⊣
So far we have seen that for any $n\in \mathbb {N}$ there exists a nonHenselian but $( 2n+1)_\leq $ Henselian valuation $v_n$ on K. It remains to show that K does not admit a nontrivial Henselian valuation.
Claim 4. K does not admit a nontrivial Henselian valuation.
Proof. Proof of Claim 4
Let u be any nontrivial convex valuation on K and pick $a=(a_m)\in K$ with $u(a)<0$ , i.e., $a\notin \mathcal {O}_u$ . Let $\ell \in \mathbb {N}$ be least such that $a_{\ell }\neq \infty $ . Then $\psi _{\ell }(a)=a_{\ell }\in K_{\ell }$ , whence $a\in \mathcal {O}_{\ell }$ . Since $v_{\ell }$ is convex, we obtain that $v_{\ell }$ is a proper coarsening of u. Therefore, since $v_{\ell }$ is not Henselian, also u is not Henselian. Finally, as any Henselian valuation on an ordered field is convex (cf. [Reference Knebusch and Wright19, Lemma 2.1]), this suffices to show that K does not admit a nontrivial Henselian valuation.⊣
Finally, we show inductively that $v_nK = \coprod _{\gamma>\infty }^{2n}A_{\gamma }$ . Since $v_n$ is a coarsening of $v_0$ for all $n \in \mathbb {N}$ , we have that $v_nK$ is the quotient of $v_0K$ by $v_0(\mathcal {O}^\times _n)$ . Therefore our next task is to compute $v_0K$ and $v_0(\mathcal {O}^\times _n)$ . In order to do so, let us first recall the construction of the fields $K_m$ .
We have the inclusions as ordered fields
where the last inclusion is obtained by embedding $A_1 \amalg A_0$ into G as its convex subgroup $\coprod _{\gamma = 1}^0 A_{\gamma }$ . In the next step of our iterative field construction we obtained the ordered field inclusions
The last inclusion is given via the isomorphismFootnote ^{8}
and the embedding of $(A_3\amalg A_2)\amalg (A_1\amalg A_0)$ into G as its convex subgroup $\coprod _{\gamma =3}^0 A_{\gamma }$ . Continuing this process, we have that
We denote by $v_{\min }$ the standard valuation $s\mapsto \min \mathrm {supp} (s)$ on $\mathbb {R}\!\left (\!\left ( G \right )\!\right )$ . By the inclusion described above, $v_{\min }$ restricted to $K_m$ defines a valuation with value group $v_{\min } K_m=\coprod _{\gamma =(2m1)}^0 A_{\gamma }$ . For any $m\in \omega $ we have that $v_{\min } K_m$ can be regarded as a convex subgroup of G. Note that the valuation $v_{\min }$ restricted to $K_m$ is a refinement of $v_{\min }^m$ : Both $v_{\min }$ and $v_{\min }^m$ are convex, $ v_{\min } K_m=\coprod _{\gamma =(2m1)}^0 A_{\gamma }$ and $v_{\min }^m K_m = \left (A_{2m1} \amalg A_{2(m1)}\right )$ . Thus, $v_{\min }^m$ is the coarsening of $v_{\min }$ induced by the convex subgroup $\coprod _{\gamma =(2m3)}^0 A_{\gamma }$ of $v_{\min } K_m$ .
We now want to define a map v on K such that v is a valuation on K and $\mathcal {O}_v = \mathcal {O}_0$ . We set $v(0) = \infty $ .
Claim 5. For any element $(x_m) \in K^\times $ and for any $x_i, x_j \notin \{0,\infty \}$ we have that $v_{\min }(x_i) = v_{\min }(x_j)$ .
Proof. Proof of Claim 5
Let $\ell \in \omega $ be maximal with $x_{\ell 1} \in \{0,\infty \}$ , if such $\ell $ exists, and $\ell =0$ otherwise. It suffices to show that $v_{\min }(x_k) = v_{\min }(x_{\ell })$ for any $k\geq \ell $ . We proceed by induction on k. For $k=\ell $ there is nothing to prove. Now assume that $v_{\min }(x_{\ell }) = v_{\min }(x_k)$ for some fixed $k \geq \ell $ . We first express $x_k \in K_k$ as
Now set $g_0 = v_{\min }(x_k)$ . Then $v_{\min }(x_k)= v_{\min }(x_{\ell }) \in \coprod _{\gamma = (2\ell 1)}^0 A_{\gamma } \subseteq G$ if $\ell \geq 1$ , and $g_0=v_{\min }(x_0)=0$ if $\ell =0$ . Further we set
where $h_{\gamma }\in A_{\gamma }$ for each $\gamma $ . By (2) we have that $v_{\min }^{k+1}(x_{k+1}) = 0$ and by construction $x_{k+1}v_{\min }^{k+1} = x_k$ . Therefore we can set
for some $\varepsilon \in \mathcal {M}^\prime _{k+1}$ . Considering $\varepsilon $ as an element of $\mathbb {R}\!\left (\!\left ( G \right )\!\right )$ we can write
Let $\delta = v_{\min }(\varepsilon )$ and write
Now since $\varepsilon \in \mathcal {M}^\prime _{k+1}$ , we have that
. Hence,
and therefore $\delta> g_0$ . Thus, $v_{\min }(x_{k+1}) = v_{\min }(x_k + \varepsilon ) = v_{\min }(x_k) = v_{\min }(x_{\ell })$ , as required.⊣
Let $(x_m)\in K^\times $ . We define
for any $x_i \notin \{0, \infty \}$ .
Claim 6. The map v defines a valuation on K with $vK = G$ .
Proof. Proof of Claim 6
Note that by Claim 5, v is welldefined. Let $(x_m),(y_m)\in K^\times $ and let i be large enough such that $x_i,y_i\notin \{0,\infty \}$ . We obtain
Similarly, we prove that $v((x_m) + (y_m)) \geq \min \{v((x_m)), v((y_m))\}.$
Moreover, $vK = G$ : For any $g \in G$ we can choose the least $\ell $ such that there is an element $a_{\ell }\in K_{\ell }$ with $v_{\min }(a_{\ell })=g$ . Now for any $(x_m) \in K$ with $x_{\ell }=a_{\ell }$ a direct calculation shows $v((x_m))=v_{\min }(a_{\ell })=g$ .⊣
We now show that $\mathcal {O}_v=\mathcal {O}_0$ , this will yield that $v_0K$ is (isomorphic to) G, as required.
Claim 7. $\mathcal {O}_v = \mathcal {O}_0$ .
Proof. Proof of Claim 7
Let $(x_m) \in K$ . First suppose that $(x_m) \notin \mathcal {O}_0= \psi _0^{1}(K_0)$ . Then $x_0=\infty $ . Let i be maximal such that $x_i=\infty $ . Then $v((x_m))=v_{\min }(x_{i+1})<0$ . Hence, $(x_m) \notin \mathcal {O}_v$ . For the converse, suppose that $(x_m) \in \mathcal {O}_0\setminus \{0\}$ . Then $x_0\neq \infty $ . If $x_0=0$ , then let $\ell \in \omega $ be maximal with $x_{\ell 1} =0$ . We obtain $v((x_m))=v_{\min }(x_{\ell })>0$ and thus $(x_m) \in \mathcal {O}_v$ . If $x_0\neq 0$ , then $x_i\neq 0$ for every $i\in \omega $ . Hence, $v((x_m))=0$ , also showing $(x_m) \in \mathcal {O}_v$ .⊣
Claim 8. For any $n\in \omega $ we have $v_0(\mathcal {O}_n^\times ) = \coprod _{\gamma = 2n+1}^0 A_{\gamma }$ and therefore
Proof. Proof of Claim 8
Let $(x_m)\in \mathcal {O}_1^\times $ . Then $x_1\in K_1\setminus \{0\}$ . Hence, by the inclusion of ordered fields above, we have $x_1\in \mathbb {R}\!\left (\!\left ( \coprod _{\gamma =1}^0 A_{\gamma } \right )\!\right )$ and thus $v_0((x_m))=v_{\min }(x_1)\in \coprod _{\gamma =1}^0 A_{\gamma }$ . By iteration of this argument, we obtain (4) for any $n\in \omega $ .⊣
To complete our construction, set $u_{2n}=v_n$ for any $n\in \omega $ . Claims 2 and 8 yield that $u_{2n}$ is a $(2n+1)_{\leq }$ Henselian (and thus $(2n)_{\leq }$ Henselian) valuation with $u_{2n}K=\coprod _{\gamma>\infty }^{2n} A_{\gamma }$ . Now let $u_{2n+1}$ be the coarsening of $u_{2n}$ induced by the convex subgroup $A_{2n}$ of $u_{2n}K$ . Since $u_{2n+1}$ is the coarsening of a $(2n+1)_{\leq }$ Henselian valuation, it is itself also $(2n+1)_{\leq }$ Henselian. By definition of $u_{2n+1}$ , its value group is given by $u_{2n+1}K=\coprod _{\gamma>\infty }^{2n1} A_{\gamma }$ , as required.⊣
Remark 4.6. In Construction 4.5 we could have started with any ordered field $K_0$ (instead of $\mathbb {R}$ ) to obtain the required tHenselian nonHenselian ordered field. In fact, one can also start with any field $K_0$ of characteristic $0$ (so Lemma 4.3 is applicable) to obtain a tHenselian nonHenselian field (without an ordering) with the specified value groups.⊣
Theorem 4.7. Let $A\neq \{0\}$ be an ordered abelian group. Then there exists an ordered Henselian valued field $(L,<,v)$ such that $vL=\coprod _{\omega }A$ and $(L,<)$ admits no nontrivial $0$  $\mathcal {L}_{\mathrm {or}}$ definable Henselian valuation. In particular, v is not $0$  $\mathcal {L}_{\mathrm {or}}$ definable.
Proof. For any $\mathcal {L}_{\mathrm {og}}$ sentence $\sigma $ , let $\sigma ^*$ be the $\mathcal {L}_{\mathrm {vf}}$ sentence such that for any valued field $(F,w)$ we have $wF\models \sigma $ if and only if $(F,w)\models \sigma ^*$ .Footnote ^{9} Let $G=\coprod _{\omega }A$ and let $\operatorname {\mathrm {Th}}(G)$ be the complete $\mathcal {L}_{\mathrm {og}}$ theory of G. We set $\Sigma $ to be the $\mathcal {L}_{\mathrm {vf}}$ theory $\{\sigma ^*\mid \sigma \in \operatorname {\mathrm {Th}}(G)\}$ .
Let $(K,<)$ and $ u_n$ (for any $n\in \mathbb {N}$ ) be as in Construction 4.5 for the case $A_\gamma =A$ for any $\gamma \in \omega $ . For any $n\in \mathbb {N}$ we let $h_n$ be the $\mathcal {L}_{\mathrm {vf}}$ sentence stating that v is a nontrivial $n_\leq $ Henselian valuation. Finally, let $\operatorname {\mathrm {Th}}(K,<)$ be the complete $\mathcal {L}_{\mathrm {or}}$ theory of $(K,<)$ .
For any $n\in \mathbb {N}$ , the ordered valued field $(K,<, u_n)$ satisfies $\{h_i\mid i\leq n\}$ . Since $ u_nK=G$ , we also have $(K, u_n)\models \Sigma $ . Hence, any finite subset of the $\mathcal {L}_{\mathrm {ovf}}$ theory $\Sigma \cup \{h_n\mid n\in \mathbb {N}\}\cup \operatorname {\mathrm {Th}}(K,<)$ is satisfiable. By the Compactness Theorem, we obtain an ordered valued field $(F,<,w)$ such that $(F,<)\models \operatorname {\mathrm {Th}}(K,<)$ , $wF\models \operatorname {\mathrm {Th}}(G)$ , and w is nontrivial, $n_\leq $ Henselian for any $n\in \mathbb {N}$ , and thus Henselian. We set $L=Fw\!\left (\!\left ( G \right )\!\right )$ and $v=v_{\min }$ on L. Since $Fw=Lv$ and $wF\equiv G=vL$ , the Ax–Kochen–Ershov principle for ordered fields (cf. [Reference Farré7, Corollary 4.2]) implies $(L,<,v)\equiv (F,<,w)$ . As $(F,<)\equiv (K,<)$ , we also obtain $(L,<)\equiv (K,<)$ .
By elementary equivalence, any nontrivial $0$  $\mathcal {L}_{\mathrm {or}}$ definable Henselian valuation in $(L,<)$ corresponds to a nontrivial $0$  $\mathcal {L}_{\mathrm {or}}$ definable Henselian valuation in $(K, <)$ . However, since $(K,<)$ does not admit any nontrivial Henselian valuation, this shows the required conclusion.⊣
Remark 4.8.

1. Let $A\neq \{0\}$ be an ordered abelian group. In order to obtain an ordered Henselian valued field $(L,<,v)$ such that $vL = \coprod _{\omega } A$ and v is not $0$  $\mathcal {L}_{\mathrm {or}}$ definable, one can also proceed as follows:Footnote ^{10} Let $L = K\!\left (\!\left ( \coprod _{\mathbb {Z}} A \right )\!\right )$ for some ordered field $(K,<)$ . Now let v be the coarsening of $v_{\min }$ induced by the convex subgroup $\coprod _{\mathbb {N}} A$ of $v_{\min } L$ . Then $Lv = K\!\left (\!\left ( \coprod _{\mathbb {N}} A \right )\!\right )$ . Now let w be the strict coarsening of v induced by the convex subgroup $\coprod _{\{0\}} A$ of $vL$ . Then $wL = \coprod _{\mathbb {N}} A \cong vL$ and $Lw = K\!\left (\!\left ( \coprod _\omega A \right )\!\right ) \cong Lv$ . We fix $a \in \mathcal {O}_w \setminus \mathcal {O}_v$ . Assume, for a contradiction, that there is a parameterfree $\mathcal {L}_{\mathrm {or}}$ formula $\varphi (x)$ defining $\mathcal {O}_v$ . Then we have that
$$ \begin{align*}(L,<) \models \neg \varphi(a).\end{align*} $$By the Ax–Kochen–Ershov principle for ordered fields we have that $(L, <,v) \equiv (L,<,w)$ , whence $\varphi (x)$ also defines $\mathcal {O}_w$ . Since $a\in \mathcal {O}_w$ , we obtain$$ \begin{align*}(L,<) \models \varphi(a),\end{align*} $$a contradiction. 
2. Let $(L,<,v)$ be as in Theorem 4.7. Since any nontrivial coarsening w of v is also Henselian, also w is not $0$  $\mathcal {L}_{\mathrm {or}}$ definable. Moreover, for any convex subgroup $B\subseteq A$ , the corresponding coarsening w of v with value group $\left (\coprod _{\mathbb {N}}A\right )\amalg (A/B)$ is not $0$  $\mathcal {L}_{\mathrm {or}}$ definable.

3. Theorem 4.7 stands in contrast to [Reference Hong14, Theorem 4], which implies that any Henselian valuation v such that $vK$ is nondivisible and archimedean is already $0$  $\mathcal {L}_{\mathrm {r}}$ definable. In this regard, note that the rank $\omega $ of $\coprod _{\omega }A$ has no minimum, whence the valuation v in Theorem 4.7 has no coarsening with archimedean value group.⊣
We can now use Theorem 4.7 to show that the conclusion in Fact 1.1 cannot be strengthened to parameterfree definability.
Example 4.9. The ordered abelian group $\coprod _{\omega }\mathbb {Z}$ is discretely ordered (with as least positive element). By Theorem 4.7, there exists an ordered Henselian valued field $(L,<,v)$ with $vL=\coprod _{\omega }\mathbb {Z}$ such that v is not $0$  $\mathcal {L}_{\mathrm {or}}$ definable.⊣
Similarly, we show that Facts 1.2 and 1.3 and Theorem 3.1 cannot be strengthened to parameterfree definability.
Example 4.10. Consider the densely ordered abelian group
Then A is regular but not $3$ divisible, and is (isomorphic to) a convex subgroup of $G:=\coprod _{\omega }A$ . Proposition 3.3 yields that G is not closed in ${G}^{\mathrm {div}}$ . By Theorem 4.7, there exists an ordered field $(L,<)$ and a Henselian valuation v on L such that $vL=G$ and v is not $0$  $\mathcal {L}_{\mathrm {or}}$ definable.⊣
Remark 4.11. By [Reference Hong13, Lemma 2.3.7], the following holds:Footnote ^{11} Let $(K,v)$ be a Henselian valued field and let $p\in \mathbb {N}$ be prime. Suppose that $vK$ is densely ordered and pregular but not pdivisible. Then v is $0$  $\mathcal {L}_{\mathrm {r}}$ definable.
In contrast to this result, Example 4.10 shows that the mere existence of a convex pregular subgroup that is not pdivisible does, in general, not ensure definability without parameters.
5 Further work
We conclude this work by collecting open questions which may serve as a possible starting point for future work on topics related to this paper.
For a class $\mathcal {C}$ of nontrivial ordered abelian groups we consider the following:

(†)_{0} If $(K,v)$ is a Henselian valued field such that $vK \in \mathcal {C}$ , then v is $0$  $\mathcal {L}_{\mathrm {r}}$ definable.

(†) If $(K,v)$ is a Henselian valued field such that $vK \in \mathcal {C}$ , then v is $\mathcal {L}_{\mathrm {r}}$ definable.
The corresponding properties ( ) $_0$ and ( ) stand for the negations of the above; that is, a class $\mathcal {C}$ of nontrivial ordered abelian groups satisfies ( ) $_0$ (respectively ( )) if there exists a Henselian valued field $(K,v)$ such that $vK\in \mathcal {C}$ but v is not $0$  $\mathcal {L}_{\mathrm {r}}$ definable (respectively $\mathcal {L}_{\mathrm {r}}$ definable).
To summarise the results from the literature as well as this work, we distinguish between discretely and densely ordered abelian groups.
5.1 Discrete case
We consider the following two elementary classes of ordered abelian groups.
The class $\mathcal {C}^{\mathrm {discr}}_0$ consists of all ordered abelian groups that are elementarily equivalent to $\mathbb {Z}$ as $\mathcal {L}_{\mathrm {og}}$ structures. These are also called $\mathbb {Z}$ groups (cf., e.g., [Reference Krapp, Kuhlmann and Lehéricy21, Fact 3.4]). By [Reference Hong14, Theorem 4], $\mathcal {C}^{\mathrm {discr}}_0$ satisfies $(\dagger )_0$ .
The ordered abelian group $\mathbb {Z}\amalg \mathbb {Z}$ serves as an example of a nonregular discretely ordered abelian group, showing that $\mathcal {C}^{\mathrm {discr}}_0$ is strictly contained in $\mathcal {C}^{\mathrm {discr}}_1$ . By Fact 1.1 the class $\mathcal {C}^{\mathrm {discr}}_1$ satisfies ( $\dagger $ ), and due to Example 4.9 this class does not satisfy ( $\dagger $ ) $_0$ . We thus obtain the following picture:
5.2 Dense case
We consider the following four elementary classes of ordered abelian groups $\mathcal {C}^{\mathrm {dense}}_i$ consisting of all densely ordered abelian groups G satisfying the condition $c_i$ , where $c_i$ is given as follows:

c _{0} : for some prime p, the group G is pregular but not pdivisible;

c _{1} : for some prime p, some convex subgroup of G is pregular but not pdivisible;

c _{2} : G is not closed in ${G}^{\mathrm {div}}$ ;

c _{3} : for some prime p, the group G contains no nontrivial pdivisible convex subgroup.
While $\mathcal {C}^{\mathrm {dense}}_0$ satisfies $(\dagger )_0$ , the class $\mathcal {C}^{\mathrm {dense}}_1$ satisfies ( $\dagger $ ) but not $(\dagger )_0$ (see Example 4.10 and Remark 4.11). By Proposition 3.3 and Example 3.4, the class $\mathcal {C}^{\mathrm {dense}}_1$ is strictly contained in $\mathcal {C}^{\mathrm {dense}}_2$ . Moreover, Theorem 3.1 yields that $\mathcal {C}^{\mathrm {dense}}_2$ satisfies ( $\dagger $ ).
Further, $\mathcal {C}^{\mathrm {dense}}_2$ is strictly contained in $\mathcal {C}^{\mathrm {dense}}_3$ (cf. [Reference Krapp, Kuhlmann and Lehéricy21, Proposition 3.18 and Example 3.20]). The remarks in [Reference Delon and Farré4, p. 1147] show that for any densely ordered abelian group G not satisfying condition $c_3$ above there exists an (almost real closed) valued field $(K,v)$ with $vK=G$ such that v is not $\mathcal {L}_{\mathrm {r}}$ definable. Hence, any class of densely ordered abelian groups $\mathcal {D}$ that strictly contains $\mathcal {C}^{\mathrm {dense}}_3$ does not satisfy ( $\dagger $ ).
As a result of the discussion above, we obtain the following picture:
5.3 Open questions
While from the observations above it is clear that the class $\mathcal {G}_0$ of ordered abelian groups for which $(\dagger )_0$ holds is a proper subclass of the class $\mathcal {G}$ of ordered abelian groups satisfying $(\dagger )$ , none of the described dividing lines are yet proven to be sharp. Therefore, we pose the following questions:
Question 5.1.

(i) What is the largest class of ordered abelian groups satisfying $(\dagger )$ , respectively $(\dagger )_0$ ?

(ii) Is there a characterisation of $\mathcal {G}_0$ inside $\mathcal {G}$ ?
5.4 Concluding Remark
As noted in the introduction, in this paper we only consider conditions on the value group. There are also conditions for the definability of a given Henselian valuation only depending on its residue field (cf., e.g., [Reference Jahnke and Koenigsmann15, Proposition 3.1 and Corollaries 3.3 and 3.8] and [Reference Krapp, Kuhlmann and Lehéricy21, Theorem 5.3(3)]). In particular, the analogous of $(\dagger )$ and $(\dagger )_0$ can be formulated for residue fields. Such conditions will be the subject of a further publication.
Acknowledgments
The first author was partially supported by Werner und Erika Messmer Stiftung. During parts of this project, the second and third authors were supported by the German Excellence Strategy and the German Research Foundation.