Proof. Suppose $Kv_K$ admits a non-trivial henselian valuation $v$ . Via 2.1 this gives rise to a proper refinement of $v_K$ . Hence $H_2(K)$ is non-empty, so $Kv_K$ is separably closed.

Proof. Suppose there is $L\succeq Kv_K$ such that $(L,v_L)$ has mixed characteristic. Restricting $v_L$ down to $Kv_K$ yields a henselian valuation (because $Kv_K$ is relatively algebraically closed in $L$ ) of mixed characteristic on $Kv_K$ , which in particular must be non-trivial. By Lemma 2.2, $Kv_K$ must be separably closed. It follows that $L$ is also separably closed, but this implies that $v_L$ is trivial.

Fact 2.6. Let $K$ be t-henselian and $\aleph _1$ -saturated. Then $K$ admits a non-trivial henselian valuation. In particular, if $K$ is t-henselian and $K \preceq L$ is an $\aleph _1$ -saturated elementary extension, then $L$ admits a non-trivial henselian valuation.

Fact 3.2 [Reference Jahnke and Simon11, Lemma 3.1].

Let $(K,v)$ be a tame field of positive residue characteristic. Then the value group $vK$ is purely stably embedded as an ordered abelian group, i.e., for every $D \subseteq vK^n$ which is $\mathcal {L}_{{K,\Gamma ,k}}$ -definable (possibly with parameters), there is $\psi \in \mathcal {L}_{\mathrm {oag}}(vK)$ which defines $D$ .

Fact 3.3. Let $\Gamma $ be a divisible ordered abelian group. Then $\Gamma $ has no proper, $\mathcal {L}_{\mathrm {oag}}$ -definable (even with parameters) non-trivial convex subgroup.

Proof. Suppose $w$ is $\mathcal {L}_{\mathrm {val}}$ -definable (possibly with parameters); then the corresponding proper non-trivial convex subgroup

$$\begin{align*}\Delta = \{x \in vK \mid \exists y(v(y) = x \land w(y) = 0)\}, \end{align*}$$

is $\mathcal {L}_{{K,\Gamma ,k}}$ -definable (possibly with parameters) as a subset of $vK$ . By Fact 3.2, $\Delta $ is $\mathcal {L}_{\mathrm {oag}}(vK)$ -definable. This is a contradiction, as divisible ordered abelian groups have no definable proper non-trivial convex subgroups.

Proof. We assume

$$\begin{align*}\neg 1 \land \neg 2 \land \neg 3 \land \neg 4 \land \neg 5 \land \neg 6,\end{align*}$$

and want to show that $K$ then cannot admit a non-trivial definable henselian valuation. Note that $\neg 1$ means that $H_2(K) = \varnothing $ . Thus, all henselian valuations on $K$ are linearly ordered and coarser than $v_K$ .

Now, let ${(M,v) \succeq (K,v_K)}$ be an $\aleph _1$ -saturated extension in $\mathcal {L}_{\mathrm {val}}$ . Then, ${Mv \succeq Kv_K}$ is an $\aleph _1$ -saturated elementary extension of a t-henselian field (by $\neg 2$ ), hence it is henselian by Fact 2.6. Since ${L := Mv\equiv Kv_K}$ is henselian and not separably closed (by $\neg 1$ ), $v_L$ is non-trivial and in fact, by Remark 2.1, we have the following diagram (where the arrows are intended as places):

from which we can deduce that $v_M = v_{L} \circ v$ is a proper refinement of $v$ . Now, since $v_LL$ is divisible (because of $\neg 3$ ), so is $v_MM$ . Indeed,

$$\begin{align*}vM = {v_MM}/{v_LL}, \end{align*}$$

and we know $vM$ to be divisible, as $v_KK$ is (because of $\neg 4$ ).

Now, by assumption $Kv_K$ is a perfect field of positive characteristic and thus so is $L=Mv\succeq Kv_K$ . It follows that also the residue field $Lv_L=Mv_M$ of the valuation $v_L$ on $L$ is perfect. By assumptions $\neg 5$ and $\neg 6$ , $v$ and $v_L$ are defectless. Thus $v_M$ is defectless (as the composition of defectless valuations is defectless, see [Reference Anscombe and Jahnke2, Lemma 2.9]), and $(M,v_M)$ is a tame valued field.

Assume now that there is a non-trivial definable henselian valuation $u$ on $K$ , with valuation ring defined by the $\mathcal {L}_{\mathrm {ring}}(K)$ -formula $\psi (x)$ . Because of $\neg 1$ , $u$ is a (not necessarily proper) coarsening of $v_K$ . Now, via the elementary embedding, $\psi (x)$ defines a valuation $u^*$ on $M$ , which is a coarsening of $v$ and hence a proper coarsening of $v_M$ . This is a contradiction to Lemma 3.4.

Proof. Let $L\equiv Kv_K$ be such that $v_L L$ is non-divisible. First, we notice that $Kv_K$ is not separably closed, and hence by Lemma 2.2 it does not admit any non-trivial henselian valuation. Indeed, if $Kv_K$ were separably closed, $L$ would also be separably closed, but separably closed valued fields always have divisible value group, and $v_LL$ is not divisible by assumption.

We now claim that $v_L \in H_1(L)$ . Assume for a contradiction that it is not. Then by Theorem 2.8, there is an $\varnothing $ -definable non-trivial henselian valuation on $L$ . Thus there is also an $\varnothing $ -definable non-trivial henselian valuation $u$ on $Kv_K$ by elementary equivalence, contradicting the considerations in the previous paragraph.

By the Keisler–Shelah isomorphism theorem [Reference Marker17, Theorem 2.5.36], there are an index set $I$ and an ultrafilter $\mathcal U$ on $I$ such that $L^{\mathcal U} \cong (Kv_K)^{\mathcal U}$ . Take the ultrapower $(L^{\mathcal U}, v_L^{\mathcal U}) := (L,v_L)^{\mathcal U}$ as an $\mathcal {L}_{\mathrm {val}}$ -structure. Then, $v_L^{\mathcal U}L^{\mathcal U} \equiv v_LL$ is not divisible. Since $v_L \in H_1(L)$ , then $v_L^{\mathcal U} \in H_1(L^{\mathcal U})$ and thus $v_L^{\mathcal U}$ coarsens the canonical henselian valuation $v_{L^{\mathcal U}}$ on $L^{\mathcal U}$ . In particular, also $v_{L^{\mathcal U}}L^{\mathcal U}$ cannot be divisible, say not $q$ -divisible for some prime $q$ . Moreover, if we let $(K^{\mathcal U}, v_K^{\mathcal U}) := (K,v_K)^{\mathcal U}$ as an $\mathcal {L}_{\mathrm {val}}$ -structure, then $K^{\mathcal U}v_K^{\mathcal U} = L^{\mathcal U}$ and $v_K^{\mathcal U}K^{\mathcal U} \equiv v_KK$ is divisible. Now:

• • $K^{\mathcal U}$ is henselian and $v_{K^{\mathcal U}} = v_{L^{\mathcal U}}\circ v_K^{\mathcal U} $ (Remark 2.1),

• • $v_{L^{\mathcal U}}L^{\mathcal U} \subseteq v_{K^{\mathcal U}}K^{\mathcal U}$ is a convex subgroup and thus also $v_{K^{\mathcal U}}K^{\mathcal U}$ is not $q$ -divisible, and

• • $v_{K^{\mathcal U}}K^{\mathcal U}/v_{L^{\mathcal U}}L^{\mathcal U} \cong v_K^{\mathcal U} K^{\mathcal U}$ is divisible, hence $v_{K^{\mathcal U}}K^{\mathcal U}$ is not $q$ -antiregular.

We can then apply Theorem 2.8 to $(K^{\mathcal U}, v_{K^{\mathcal U}})$ (note the different valuation!) and find an $\varnothing $ -definable non-trivial henselian valuation on $K^{\mathcal U}$ . Since ${K \preceq K^{\mathcal U}}$ , the same $\mathcal {L}_{\mathrm {ring}}$ -formula defines a non-trivial henselian valuation ring on $K$ .

Proof. We will use Beth’s definability theorem as explained in Remark 4.1. We work in $\mathcal {L}_{\mathrm {val}}$ : let $L$ and $\mathcal {O}_1, \mathcal {O}_2 \subseteq L$ be such that ${(L,\mathcal {O}_1), (L,\mathcal {O}_2) \equiv (K,\mathcal {O}_H)}$ . In particular, both $\mathcal {O}_1$ and $\mathcal {O}_2$ are henselian valuation rings on $L$ , corresponding to valuations $v_1$ and $v_2$ . We want to show that $\mathcal {O}_1 = \mathcal {O}_2$ . Note that, since $v_H \in H_1(K)$ , the same holds for $v_1$ and $v_2$ , and thus $\mathcal {O}_1$ and $\mathcal {O}_2$ are comparable. Without loss of generality, assume $\mathcal {O}_1 \subseteq \mathcal {O}_2$ . This implies that $\mathcal {O}_1^\times \subseteq \mathcal {O}_2^\times $ . We note that, if $D = \psi (K)$ for some $\mathcal {L}_{\mathrm {ring}}$ -formula $\psi $ (possibly with parameters), then

$$\begin{align*}(K,\mathcal{O}_H) \vDash \forall x \left(x \notin \mathcal{O}_H \iff \exists u \exists y \left(u \in \mathcal{O}_H^\times \land \psi(y) \land \left(x = u\frac{1}{y}\right)\right)\right). \end{align*}$$

The same $\mathcal {L}_{\mathrm {val}}$ -sentence is then true in both $(L,\mathcal {O}_1)$ and $(L,\mathcal {O}_2)$ .

Now, towards proving $\mathcal {O}_1 = \mathcal {O}_2$ , assume for a contradiction that there is $x \in \mathcal {O}_2$ but $x \notin \mathcal {O}_1$ : then, there exist $u \in \mathcal {O}_1^\times \subseteq \mathcal {O}_2^\times $ and $d \in \psi (L)$ such that $x = u\frac {1}{d}$ . In particular, then, by the $\mathcal {L}_{\mathrm {val}}$ -sentence above, $x \notin \mathcal {O}_2$ , a contradiction. It follows that $\mathcal {O}_1 = \mathcal {O}_2$ .

We can apply Beth’s definability theorem and get that $\mathcal {O}_H = \varphi (K)$ for some $\mathcal {L}_{\mathrm {ring}}$ -formula $\varphi (x)$ .

Fact 4.5 [Reference Kuhlmann and Rzepka16, Theorems 3.4 and 3.5].

$\Sigma _\sigma $ does not depend on the choice of $\sigma $ , and is a final segment of ${vK}_{\infty }$ . We denote it by $\Sigma _L$ .

Proof. Assume that $v = \bar {v} \circ w$ , with $w$ defectless and $(Kw, \bar {v})$ a perfect independent defect field. If $(K,v)$ has mixed characteristic $(0,p)$ we assume without loss of generality that $K$ contains a primitive $p$ th root of unity.

Let $(L,v)$ be a Galois defect extension of degree $p$ of $(K,v)$ . In particular, this extension is immediate, i.e., $Lv=Kv$ and $vL=vK$ . We want to show that $(L,v)\supseteq (K,v)$ has independent defect.

We first argue that $(Kw,\bar {v}) \subseteq (Lw,\bar {v})$ is a defect extension. We prove that it is immediate of degree $p$ . As $(K,w)$ is defectless, we have the fundamental equality

$$\begin{align*}p=[L:K]=[Lw:Kw](wL:wK).\end{align*}$$

We argue that we must have $[Lw:Kw] = p$ and $wL = wK$ . Suppose not, i.e., that $Lw=Kw$ : then $\bar {v}(Lw)=\bar {v}(Kw)$ , and thus

contradicting $(wL:wK)=p$ .

Using $wL = wK$ and $vL=vK$ , we get

and thus $\bar {v}(Lw)=\bar {v}(Kw)$ . Moreover, $(Lw)\bar {v}=Lv=Kv=(Kw)\bar {v}$ . This shows that $ (Kw,\bar {v}) \subseteq (Lw,\bar {v}) $ is an immediate extension of degree $p$ , in particular a defect extension. Note that as $Kw$ is perfect, the extension is separable; moreover, since $K \subseteq L$ is normal, the same holds for $Kw \subseteq Lw$ (for example, by [Reference Engler and Prestel4, Proposition 3.2.16]), and thus the latter is Galois. Since $(Kw,\bar {v})$ is an independent defect field, the extension $ (Kw,\bar {v}) \subseteq (Lw,\bar {v}) $ has independent defect. By definition, there is a convex subgroup $H$ of $\bar {v}(Kw)$ such that has no minimum positive element, and further

where $\tau \in {\operatorname {Gal}({Lw} | {Kw})}$ is a generator, i.e., ${\langle \tau \rangle ={\operatorname {Gal}({Lw} | {Kw})}}$ . Since $\bar {v}(Kw)\subseteq vK$ is a convex subgroup, $H$ is also a convex subgroup of the bigger group $vK$ . Then, the embedding $\bar {v}(Kw) \subseteq vK$ gives rise to an embedding as convex subgroup, thus also has no minimum positive element. We will now show that

where $\sigma \in {\operatorname {Gal}({L} | {K})}$ is such that $\langle \sigma \rangle ={\operatorname {Gal}({L} | {K})}$ . This will prove that the defect in the extension $(L,v)\supseteq (K,v)$ is independent.

As $\Sigma _L$ is a final segment in ${vK}_{\infty }$ (by Fact 4.5), it is enough to show that $\Sigma _{L}\cap {\bar {v}(Kw)}_{\infty } = \Sigma _{Lw}$ .

Note that since $(K,w)$ is henselian, $\mathcal {O}_{(L,w)}$ is fixed setwise by the action of ${\operatorname {Gal}({L} | {K})}$ . Thus by [Reference Engler and Prestel4, Proposition 3.2.16(3)], $\phi $ induces an automorphism $\bar {\phi }\in {\operatorname {Gal}({Lw} | {Kw})}$ given by

We argue that the map

$$\begin{align*}\left\lbrace \begin{array}{rcl} {\operatorname{Gal}({L} | {K})} &\to&{\operatorname{Gal}({Lw} | {Kw})}\\ \phi&\mapsto &\bar{\phi} \end{array}\right. \end{align*}$$

is a surjective group homomorphism (cf. the proof of [Reference Engler and Prestel4, Lemma 5.2.6]): indeed, we write $Lw = Kw(\operatorname {\mathrm {res}}_w(a))$ , and we take any $\rho \in {\operatorname {Gal}({Lw} | {Kw})}$ . We let $a_1, \dots , a_p$ be the ${\operatorname {Gal}({L} | {K})}$ -conjugates of $a$ .

For some $i \leq p$ , ${\operatorname {\mathrm {res}}_w(a_i) = \rho (\operatorname {\mathrm {res}}_w(a))}$ , so there is $\sigma \in {\operatorname {Gal}({L} | {K})}$ such that

$$\begin{align*}{\operatorname{\mathrm{res}}_w(\sigma(a)) = \bar{\sigma}(\operatorname{\mathrm{res}}_w(a)) = \rho(\operatorname{\mathrm{res}}_w(a))}. \end{align*}$$

Thus $\bar {\sigma }$ and $\rho $ coincide on $Lw$ .

Now, since $(\phi \mapsto \bar {\phi })$ is surjective between finite groups with the same order, it is an isomorphism. It follows that it maps generators of ${\operatorname {Gal}({L} | {K})}$ to generators of ${\operatorname {Gal}({Lw} | {Kw})}$ .

We can now compare the sets $\Sigma _L$ and $\Sigma _{Lw}$ .

Step 1: $\Sigma _{Lw}\subseteq \Sigma _{L}\cap {\bar {v}(Kw)}_{\infty }$ .

Fix $\sigma \in {\operatorname {Gal}({L} | {K})}$ such that $\langle \sigma \rangle ={\operatorname {Gal}({L} | {K})}$ . Then, since the set $\Sigma _{Lw}$ does not depend on the choice of the generator, we can write

$$ \begin{align*} \Sigma_{Lw}&=\left\{ \bar{v}\left( \frac{f-\bar{\sigma}(f)}{f} \right) \mathrel{\Big|} f\in(Lw)^\times \right\}\\ &=\left\{ \bar{v}\left( \frac{\operatorname{\mathrm{res}}_w(f)-\bar{\sigma}(\operatorname{\mathrm{res}}_w(f))}{\operatorname{\mathrm{res}}_w(f)} \right) \mathrel{\Big|} f\in \mathcal{O}_{(L,w)}^\times \right\}\\ &=\left\{ \bar{v}\left(\operatorname{\mathrm{res}}_w\left( \frac{f-\sigma(f)}{f} \right)\right) \mathrel{\Big|} f\in \mathcal{O}_{(L,w)}^\times \right\}\\ &= \left\{ v\left( \frac{f-\sigma(f)}{f} \right) \mathrel{\Big|} f\in \mathcal{O}_{(L,w)}^\times \right\}\cap {\bar{v}(Kw)}_{\infty} \subseteq\Sigma_{L}\cap {\bar{v}(Kw)}_{\infty}, \end{align*} $$

since $\bar {v}(\operatorname {\mathrm {res}}_w(x))=v(x)$ for $x\in L$ such that $v(x)\in \bar {v}(Kw)$ .

Step 2: $\Sigma _{Lw} \supseteq \Sigma _L \cap {\bar {v}(Kw)}_{\infty }$ .

Let $v\Big (\frac {f-\sigma (f)}{f}\Big )\in \Sigma _L\cap \bar {v}(Kw)$ , i.e., $f \in L^\times $ , then because $wL=wK$ , there is $g\in K^\times $ with $w(f)=w(g)$ and so for we have

$$\begin{align*}\frac{f-\sigma(f)}{f}=\frac{h-\sigma(h)}{h}, \end{align*}$$

so

$$\begin{align*}v\left(\frac{f-\sigma(f)}{f}\right)=v\left(\frac{h-\sigma(h)}{h}\right)\in\Sigma_{Lw}.\\[-42pt] \end{align*}$$

Question 4.9. Let $(K,v)$ be a henselian valued field with $v=\bar {v}\circ w$ , such that both $(K,w)$ and $(Kw,\bar {v})$ are independent defect fields. Must then $(K,v)$ be an independent defect field?

Proof. Note that we can assume that $v \in H_1(K)$ , by Remark 4.10. Let $K \subseteq L$ be the degree $p$ Galois extension with independent defect and take $\theta $ such that $L=K(\theta )$ . Let $\sigma $ be a generator of the (cyclic) Galois group ${\operatorname {Gal}({L} | {K})}$ , and let

As the extension has independent defect, there is a (possibly trivial) convex subgroup $H \subsetneq vK$ such that

$$\begin{align*}\Sigma_{L} = v(D) = \{\alpha \in {vK}_{\infty} \mid \alpha> H\}.\end{align*}$$

Recall that $L=K(\theta )$ can be interpreted in $K$ via the $K$ -linear isomorphism

$$\begin{align*}f\colon\left\{ \begin{array}{rcl} K^p & \to & L=K(\theta)\\ (a_0,\ldots,a_{p-1})&\mapsto &\sum_{i=0}^{p-1} a_i\theta^i. \end{array}\right. \end{align*}$$

Note that the coefficients of the minimal polynomial of the generator $\theta $ over $K$ are needed as parameters to describe the multiplication. Now, the action of $\sigma $ is definable in $K$ (via the interpretation) using the coefficients of the change of basis matrix. If $\operatorname {\mathrm {char}}(K)=p$ then the matrix has integer coefficients, so no additional parameter is needed. If $\operatorname {\mathrm {char}}(K) = 0$ , then we need to use $\zeta _p$ . We will denote the parameter tuple needed to define $D$ in $K$ by $\overline {c}$ (i.e., the coefficients of the minimal polynomial of the generator $\theta $ over $K$ , and $\zeta _p$ if necessary).

Thus, $f^{-1}(D) \subseteq K^p$ is definable in $K$ , using $\overline {c}$ as parameters; choose an $\mathcal {L}_{\mathrm {ring}}(\overline {c})$ -formula $\psi (\overline {X},\overline {c})$ so that $f^{-1}(D) = \psi (K^p,\overline {c})$ . Denote by $\mathcal {O}_H \subseteq L$ the coarsening of (the unique extension of) $v$ corresponding to $H$ . We can see $\mathcal {O}_H$ as a subset $f^{-1}(\mathcal {O}_H)$ of $K^p$ .

We will use Beth’s definability theorem. We work in , where $P(\overline {X})$ is a $p$ -ary predicate. Under the bijection $f$ that establishes the interpretation, the predicate will represent the valuation subring $\mathcal {O}_H$ of $K(\theta )$ . We now let $(M,\overline {c}^{\prime },P_1),(M,\overline {c}^{\prime },P_2) \equiv _{\mathcal {L}} (K,\overline {c},f^{-1}(\mathcal {O}_H))$ . Note that the interpretation given by $f$ gives rise to an $\mathcal {L}_{\mathrm {ring}}$ -structure $N$ and a map $f'$ which is establishing an interpretation of $N$ in $M$ .

Furthermore, because $(M,\overline {c}^{\prime },P_1)$ and $(M,\overline {c}^{\prime },P_2)$ are elementarily equivalent, we have that:

• • $M\subseteq N$ are fields, and the polynomial with coefficients from $\overline {c}'$ is of degree $p$ and irreducible over $M$ ; thus, the field extension $M\subseteq N$ is algebraic of degree $p$ ;

• • $\mathcal {O}_1 = f^{\prime }(P_1)$ and $\mathcal {O}_2 = f'(P_2)$ are non-trivial henselian valuation rings on $N$ ; we denote the corresponding valuations by $v_1$ and $v_2$ , respectively;

• • there is $\ell \geq 2$ such that there is a separable polynomial of degree $\ell $ over $Lv_H$ with no root in $Lv_H$ ; thus, there are separable polynomials of degree $\ell $ over $Nv_1$ and $Nv_2$ with no root, and thus $v_1, v_2 \in H_1(N)$ ,

• • for $i =1,2$ , $x \notin \mathcal {O}_i$ if and only if there are $u \in \mathcal {O}_i^\times $ and $y \in \psi (M^p,\overline {c}')$ such that $x = u \frac {1}{f'(y)}$ .

Since $\mathcal {O}_1, \mathcal {O}_2 \in H_1(N)$ , they must be comparable. Assume, for example, that ${\mathcal {O}_1 \subseteq \mathcal {O}_2}$ . Towards a contradiction, assume now that there is $x \in \mathcal {O}_2$ such that $x \notin \mathcal {O}_1$ . In particular, this means that there are $u \in \mathcal {O}_1^\times $ and $y \in \psi (M^p,\overline {c}')$ such that ${x = u\frac {1}{f'(y)}}$ . But since $\mathcal {O}_1^\times \subseteq \mathcal {O}_2^\times $ , and thus $u \in \mathcal {O}_2^\times $ , we get that $x \notin \mathcal {O}_2$ , a contradiction. Then $\mathcal {O}_1 = \mathcal {O}_2$ , so $P_1 = P_2$ and we can apply Beth’s definability. This means that there is an $\mathcal {L}_{\mathrm {ring}}$ -formula $\psi (\overline {X},\overline {c})$ , such that ${f^{-1}(\mathcal {O}_H) = \psi (K^p,\overline {c})}$ . $\blacksquare _{\text {Claim}}$

Now, we can define the (non-trivial) restriction $\mathcal {O}_H \cap K$ as follows: given $z \in K$ , we have that

$$\begin{align*}z \in \mathcal{O}_H \cap K \iff K \vDash \psi(z, \underbrace{0, \dots, 0}_{(p-1)\text{-times}}, \overline{c}).\end{align*}$$

This exhibits a non-trivial definable henselian valuation ring on $K$ .

Proof. Take a defect extension $K \subseteq L$ . In particular, $p^n \mid [L:K]$ for some $n \geq 1$ . We let $N$ be the normal hull of $L$ ; note that $p^n \mid [N:K]$ and $K \subseteq N$ is Galois. Consider now $H \subseteq {\operatorname {Gal}({N} | {K})}$ to be a $p$ -Sylow subgroup. Then is such that $L' \subseteq N$ is Galois and it is a tower $L' = L_0 \subseteq L_1 \subseteq \cdots \subseteq L_n = N$ of normal degree $p$ extensions. Moreover, since $K \subseteq L'$ has degree prime to $p$ , it is defectless, hence $L' \subseteq N$ is defect. In particular, there is some $\ell \leq n$ such that $L_\ell \subseteq L_{\ell +1}$ is a defect Galois extension of degree $p$ , so we take $K' = L_{\ell }$ .

Proof. We may assume that $v \in H_1(K)$ , by Remark 4.10.

In the case where $\operatorname {\mathrm {char}}{K} = 0$ , we make the following two observations:

• • Since $K$ always admits a definable valuation if $v$ is finitely ramified, we may assume that it is infinitely ramified.

• • The assumptions on $(K,v)$ imply that it is a roughly deeply ramified valued field in the sense of [Reference Kuhlmann and Rzepka16, p. 2696] (note that by [Reference Kuhlmann and Rzepka16, Lemma 4.1], we only need to check that $\mathcal {O}_{v}/p$ is semi-perfect, without looking at the completion of $K$ ). Thus, by [Reference Kuhlmann and Rzepka16, Theorem 1.8] any finite extension of it is also roughly deeply ramified.

If $\operatorname {\mathrm {char}}{K} = 0$ , let $K_0 = K(\zeta _p)$ , where $\zeta _p$ is some primitive $p$ th root of unity. Otherwise, take $K_0 = K$ . By Lemma 4.12, we have a finite extension $K_0 \subseteq K_1$ which admits a Galois defect extension of degree $p$ . Since $K\subseteq K_1$ is a finite separable field extension, we can find $b\in K_1$ such that $K_1=K(b)$ .

The field $K_1$ admits a Galois defect extension of degree $p$ , i.e there is $c \in K_1$ which has no Artin–Schreier root in $K$ if $\operatorname {\mathrm {char}}{K} = p$ (resp. no $p$ th root, if $\operatorname {\mathrm {char}}{K} = 0$ ), and the extension $(K_1,v) \subseteq (K_1(\theta ),v)$ , where $\theta ^p-\theta = c$ (resp. $\theta ^p = c$ ), is an immediate defect extension. Denote by $L = K_1(\theta )=K(b,\theta )$ . Then, we have to distinguish two cases:

• • If $\operatorname {\mathrm {char}}{K} = p$ , then $K_1$ is a perfect field.

• • If $\operatorname {\mathrm {char}}{K} = 0$ , then $K_1$ is a finite extension of a roughly deeply ramified field, and thus it is roughly deeply ramified itself.

In both cases, $K_1$ is an independent defect field (see [Reference Kuhlmann and Rzepka16, Theorem 1.10]), and thus the extension $(K_1,v) \subseteq (L,v)$ is a degree $p$ defect extension with independent defect. By Theorem 4.11, then, $K_1$ admits a non-trivial definable henselian valuation $v_H$ , with valuation ring defined by $\varphi (X,\overline {c})$ . As $K_1=K(b)$ can be interpreted in $K$ , by Remark 4.13 the restriction of $v_H$ to $K$ is then $\mathcal {L}_{\mathrm {ring}}$ -definable with parameters (namely, the coefficients of the minimal polynomial of $b$ , and the coordinates of $\overline {c}$ in a chosen $K$ -basis of $K_1$ ).

Proof. We may assume that $v \in H_1(K)$ , by Remark 4.10. Now, $v_K$ is also of mixed characteristic, thus Corollary 4.14 yields the existence of a definable non-trivial henselian valuation, and by [Reference Anscombe and Jahnke1, Theorem 1.1(B)] we can conclude that there is an $\varnothing $ -definable non-trivial henselian one.

Proof. We may assume that $v \in H_1(K)$ : otherwise, we can use [Reference Jahnke and Koenigsmann9, Theorem 3.10] to find an $\varnothing $ -definable non-trivial henselian valuation. Using Lemma 4.12, there is a finite extension $K(\zeta _p) \subseteq K_1$ such that $(K_1,v)$ admits a defect extension of degree $p$ , and we find $b\in K_1$ such that $K_1=K(b)$ .

We now show that $(K_1,v)$ is an independent defect field. For this we consider the decomposition of $(K_1,v)$ with respect to the coarsening $(K_1,w)$ and we denote the induced valuation on $K_1w$ by $\bar {v}$ :

Note that $(K_1,w)$ is defectless, because it is a finite extension of $(K,w)$ and $K_1w$ is perfect because it is a finite extension of $Kw$ . From Lemma 4.8 it follows that also $(K_1,v)$ is an independent defect field.

By Theorem 4.11, then, $K_1$ admits a non-trivial definable henselian valuation $v_H$ , with valuation ring defined by $\varphi (X,\overline {c})$ . As $K_1=K(b)$ can be interpreted in $K$ , by Remark 4.13 the restriction of $v_H$ to $K$ is then $\mathcal {L}_{\mathrm {ring}}$ -definable with parameters (namely, the coefficients of the minimal polynomial of $b$ , and the coordinates of $\overline {c}$ in a chosen $K$ -basis of $K_1$ ).

By our assumptions, $v_K$ is also of mixed characteristic, thus [Reference Anscombe and Jahnke1, Theorem 1.1(B)] yields the existence of an $\varnothing $ -definable non-trivial henselian valuation.

Proof. Choose $L$ such that $L\equiv Kv_K$ and $(L,v_L)$ is not defectless. By the Keisler–Shelah isomorphism theorem [Reference Marker17, Theorem 2.5.36], take an index set $I$ and an ultrafilter $\mathcal U$ on $I$ such that $(Kv_K)^{\mathcal U} \cong L^{\mathcal U}$ . Consider the ultrapowers and in $\mathcal {L}_{\mathrm {val}}$ . Since $v_L^{\mathcal U}$ is a valuation with defect, while $v_K^{\mathcal U}$ is defectless, the composition $v_1$ is a non-trivial henselian valuation on $K^{\mathcal U}$ with defect. The valuation $v_1$ is not defectless and of mixed characteristic, since $v_K^{\mathcal {U}}$ is, and its coarsening $v_K^{\mathcal {U}}$ is defectless and has perfect residue field $K^{\mathcal {U}}v_K^{\mathcal {U}}=L^{\mathcal {U}}$ . Thus by Lemma 4.17, there is an $\varnothing $ -definable valuation on $K^{\mathcal {U}}$ . By elementary equivalence, we can find an $\varnothing $ -definable henselian valuation on $K$ .

Fact 4.19 (cf. [Reference Jahnke8, Proposition 2.4]).

Let $(K,v)$ be a henselian valued field with non-separably closed nor real closed residue field. Then $v$ has an $\mathcal {L}_{\mathrm {ring}}$ -definable refinement.

Proof. Note that we may assume that $Kv_K$ is not separably closed; otherwise, Theorem 2.8 gives an $\varnothing $ -definable non-trivial henselian valuation.

By the Keisler–Shelah isomorphism theorem [Reference Marker17, Theorem 2.5.36], take an index set $I$ and an ultrafilter $\mathcal U$ on $I$ such that ${(Kv_K)^{\mathcal U} \cong L^{\mathcal U}}$ . Consider the ultrapowers and in $\mathcal{L}_{\mathrm{val}}$ . Since $v_L^{\mathcal U}$ is a valuation with defect, the composition $v_1=v_L^{\mathcal U}\circ v_K^{\mathcal U}$ is a non-trivial henselian valuation on $K^{\mathcal U}$ with defect. Thus, by Corollary 4.14, there is an $\mathcal {L}_{\mathrm {ring}}(K^{\mathcal U})$ -definable non-trivial henselian valuation $v_1^{\prime }$ coarsening $v_1$ , say given by $\varphi (K^{\mathcal U},\overline {\beta })$ for some parameters $\overline {\beta } \in (K^{\mathcal U})^\ell $ , $\ell \geq 1$ .

Since $Kv_K$ is not separably closed, Fact 4.19 yields an $\mathcal {L}_{\mathrm {ring}}(K)$ -definable valuation $w$ on $K$ (not necessarily henselian!) such that $\mathcal {O}_w \subseteq \mathcal {O}_{v_K}$ . Let $\psi (X,\overline {c})$ be the formula defining $w$ . Denote by $w^*$ the valuation corresponding to $\psi (K^{\mathcal U},\overline {c})$ . Note that since $(K,v_K) \preceq (K^{\mathcal U}, v_K^{\mathcal U})$ , then we still have that $w^*$ is a proper refinement of $v_K^{\mathcal U}$ .

As in the proof of [Reference Engler and Prestel4, Theorem 2.3.4], given two valuation rings $\mathcal {O}_1$ and $\mathcal {O}_2$ , we let

Note that $\mathcal {O}_1 \cdot \mathcal {O}_2$ contains both $\mathcal {O}_1$ and $\mathcal {O}_2$ and is always a valuation ring. Indeed, it is the finest valuation ring containing both: if $\mathcal {O}_1, \mathcal {O}_2 \subseteq \mathcal {O}_3$ , then for any $x \in \mathcal {O}_1$ and $y \in \mathcal {O}_1 \setminus \mathfrak {m}_2$ , one has that $y \notin \mathfrak {m}_3$ , so $\frac {1}{y} \in \mathcal {O}_3$ and thus $\frac {x}{y} \in \mathcal {O}_3$ . We now consider the $\mathcal {L}_{\mathrm {ring}}(\overline {c})$ -definable set

$$\begin{align*}X = \left\{\overline{b} \in (K^{\mathcal U})^\ell \mid \varphi(K^{\mathcal U},\overline{b}) \; \text{is a valuation ring and} \; \mathcal{O}_{w^*}\cdot\varphi(K^{\mathcal U},\overline{b}) \neq K^{\mathcal{U}}\right\}. \end{align*}$$

For each $\overline {b} \in X$ , we consider the $\mathcal {L}_{\mathrm {ring}}(\overline {c},\overline {b})$ -definable valuation ring

with corresponding valuation $v_{\overline {b}}$ . As all coarsenings of a given valuation are comparable, any $v_{\overline {b}}$ for $\overline {b} \in X$ is comparable with $\mathcal {O}_{v_K^{\mathcal {U}}}$ .

Since $v_K^{\mathcal U}$ is a tame valuation with divisible value group, Lemma 3.4 implies that $v_{\overline {b}}$ cannot be a proper coarsening. In particular, then, $\mathcal {O}_{v_{\overline {b}}} \subseteq \mathcal {O}_{v_K^{\mathcal U}}$ . It follows that the union

is a non-trivial $\mathcal {L}_{\mathrm {ring}}(\overline {c})$ -definable (note the parameters!) valuation ring on $K^{\mathcal U}$ .

To show that $\mathcal {O}$ is henselian, we will show that $\mathcal {O}$ contains $\varphi (K^{\mathcal {U}},\overline {\beta })$ , i.e., that $\overline {\beta } \in X$ . It is enough to show that ${\mathcal {O}_{\overline {\beta }} \neq K^{\mathcal U}}$ . Since $\varphi (K^{\mathcal U},\overline {\beta })$ is comparable with $v_K^{\mathcal U}$ , we have two possible cases. Either $\varphi (K^{\mathcal U},\overline {\beta })$ is coarser than $v_K^{\mathcal U}$ , in which case $\mathcal {O}_{\overline {\beta }} = \varphi (K^{\mathcal {U}},\overline {\beta })$ , or $\varphi (K^{\mathcal U},\overline {\beta })$ is finer than $v_K^{\mathcal U}$ , in which case $\mathcal {O}_{\overline {\beta }}$ is also a refinement of $v_K^{\mathcal {U}}$ . Either way, $\mathcal {O}_{\overline {\beta }}$ is non-trivial.

We have just shown that $\mathcal {O}$ is a coarsening of $\varphi (K^{\mathcal {U}},\overline {\beta })$ , thus it is henselian. In particular, the same formula defines a non-trivial henselian valuation on $K$ , as needed.

Fact 6.6 [Reference Anscombe and Jahnke1, Proposition 4.13].

Let $p$ be a prime or zero. There exists a non-henselian t-henselian field of characteristic $p$ which is:

1. (1) not separably closed, and

2. (2) of divisible-tame type.

Proof. As $(K,v_t)$ is the increasing union of henselian fields, $v_t$ is henselian. However, it is not algebraically maximal. Indeed, the equation $X^p-X-\frac {1}{t} = 0$ has no solution in $K$ , however it admits a solution in $K_0(\!(\mathbb Q)\!)$ , namely $a = \sum _{n \geq 0} t^{-\frac {1}{p^n}}$ . Then, $K \subsetneq K(a) \subseteq K_0(\!(\mathbb {Q})\!)$ is a tower of immediate extensions, thus $K \subsetneq K(a)$ is a proper algebraic immediate extension. Note that if $K_0$ is perfect, then $K_0(\!(t^{\frac {1}{n}})\!)^{\frac {1}{p}} = K_0(\!(t^{\frac {1}{np}})\!)$ for every $n$ , and thus the union $K$ is perfect.

Proof. We follow the proof of [Reference Anscombe and Jahnke1, Lemma 4.8], but we substitute the generalized power series with the Puiseux series. Indeed, inside the Puiseux series , endowed with the restriction $v_t$ of the $t$ -adic valuation on $K(\!(\mathbb Q)\!)$ , consider the subfield . Consider . Note that since $L$ is henselian, so is $\widetilde {F}$ . Then, arguing as in [Reference Anscombe and Jahnke1, proof of Lemma 4.8], there is a subgroup $G \leq {\operatorname {Gal}({F^{\mathrm {sep}}} | {F})}$ with $G \cong \mathbb Z_q$ .

Let $E = \operatorname {Fix}(G) \subseteq F^{\mathrm {sep}}$ , and consider . Then, $(K',v_t)$ has residue field $K$ and value group $\mathbb Q$ , and it is perfect. Moreover, the roots of $X^p-X-\frac {1}{t}$ give rise to immediate extensions of degree $p$ over $(K',v_t)$ . Now, arguing as in [Reference Anscombe and Jahnke1, Claim 4.8.1], $(K',v_t)$ is not $q$ -henselian. Similarly, arguing as in [Reference Anscombe and Jahnke1, Claim 4.8.2], $(K',v_t)$ is $n_{\leq }$ -henselian (note that they argue that $(K',v_t)$ is $(n!^2!)_{\leq }$ -henselian, using that in their case $q> n!^2!$ ).

As for the last point, note that $L$ admits an immediate extension of degree $p$ , namely given by any root of the Artin–Schreier polynomial $X^p-X-\frac {1}{t}$ . In particular, then, $\widetilde {F}$ also admits an immediate extension of degree $p$ , and thus so does $K'$ .

Proof. For each $n \geq 0$ , set $k_n = n+p+1$ and choose a prime $q_n> k_n$ . Let $K_0 = \mathbb F_p^{\mathrm {alg}}$ . Then, using Lemma 6.13, build a sequence $\{(K_{n+1},\overline {v_n}) \mid n \geq 0\}$ of valued fields such that each $(K_{n+1},\overline {v_{n}})$ has value group $\mathbb Q$ , residue field $K_{n}$ , is not $q_n$ -henselian, but it is ${(k_n)}_{\leq }$ -henselian, and it admits a proper degree $p$ immediate extension.

For any $n> m \geq 0$ , then, write and denote by $\mathcal {O}_{n,m}$ the corresponding valuation ring on $K_n$ . Denote by $\pi _{n,m} \colon \mathcal {O}_{n,0} \to \mathcal {O}_{m,0}$ the restriction of the residue map $\mathcal {O}_{n,m} \to K_m$ . Then, the valuation rings $(\mathcal {O}_{n,0})_{n \geq 0}$ form a projective system together with the maps $\pi _{n,m}$ . The limit $\mathcal {O}$ is again a valuation ring, together with natural projections $\pi _{\infty ,n} \colon \mathcal {O} \to \mathcal {O}_{n,0}$ . For each $n \geq 0$ , consider the localization . Each $\mathcal {O}_{v_n}$ is a valuation ring on $K = \operatorname {Frac}(\mathcal {O}) = \bigcup _{n \geq 0} \mathcal {O}_{v_n}$ with residue field $K_n$ and non-trivial divisible value group. Indeed, for each $n \geq 0$ , $\mathcal {O}_{v_n} \subseteq \mathcal {O}_{v_{n+1}}$ , and $v_{n}$ induces precisely $\overline {v_n}$ on $K_{n+1} = Kv_{n+1}$ .

We now let $(K^*,v^*)$ be an ultraproduct of the family $(K,v_n)_{n \geq 0}$ . Then, $K^*$ is perfect, and $v^*$ is henselian with divisible value group. Moreover, $K$ is not henselian and thus not separably closed.

Now, we diverge from the proof of [Reference Anscombe and Jahnke1, Proposition 4.13] and argue that $(K^*,v^*)$ admits a degree $p$ immediate extension, in particular a defect extension. It is enough to show that each $(K,v_n)$ admits one such. We know that $(K_{n+1},\overline {v_n})$ admits a degree $p$ immediate extension generated by a root $\alpha $ of an Artin–Schreier polynomial $X^p-X-\xi $ , for some $\xi \in K_{n+1}$ . Denote by $\overline {u}$ the prolongation of $\overline {v_n}$ to $K_{n+1}(\alpha )$ that makes $(K_{n+1}(\alpha ),\overline {u})$ immediate over $(K_{n+1},\overline {v_n})$ . For some $z \in K$ with $\operatorname {\mathrm {res}}_{v_{n+1}}(z) = \xi $ , we consider the Artin–Schreier polynomial $X^p-X-z$ . Let $w$ be a prolongation of $v_{n+1}$ to $K^{\mathrm {alg}}$ : then there is $a \in K^{\mathrm {alg}}$ with $a^p-a-z = 0$ and $\operatorname {\mathrm {res}}_{w}(a) = \alpha $ , by henselianity of $w$ . Then, if we denote by $w$ again the restriction of $w$ to $K(a)$ , we have that $K(a)w = Kw(\alpha ) = K_{n+1}(\alpha )$ . We now let $u = \overline {u} \circ w$ . Then, $K(a)u = K_{n+1}(\alpha )\overline {u} = Kv_n$ . Moreover, $v_nK$ is divisible, so $uK(a) = v_nK$ .

Proof. We distinguish two cases. If $v$ is a (possibly non-proper) coarsening of $v_L$ , then we are done, since defect goes up in coarsenings. If $v$ is a proper refinement of $v_L$ , then $Lv_L$ is separably closed, and since $L$ is perfect, it is in particular algebraically closed. Thus, the valuation $\overline {v}$ induced by $v$ on $Lv_L$ is defectless. But then, since $v = \overline {v} \circ v_L$ has defect if and only at least one of $v_L$ and $\overline {v}$ has defect [Reference Anscombe and Jahnke2, Lemma 2.9], $v_L$ must have defect.

Question 6.18. Let $K$ be a non-henselian, t-henselian of divisible-tame type field of characteristic $p$ which is not separably closed. Suppose that $L \equiv K$ admits a non-trivial henselian valuation. Must $(L,v_L)$ be defectless? This is the defectless version of Proposition 6.2, and it would yield that if $K_0$ is non-henselian, t-henselian of divisible-tame type, not separably closed, then $K_0(\!(\mathbb Q)\!)$ is an example for $\neg 1 \land \neg 2 \land \neg 3 \land \neg 4 \land \neg 5 \land \neg 6$ (and thus admits no non-trivial definable henselian valuation).