Proof. We set $G=vK$ . Let $\varepsilon \in K$ and let $n\in \mathbb {N}$ such that $v(\varepsilon )$ is not n-divisible 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 left-sided limit point of G, i.e., for any positive $h\in {G}^{\mathrm {div}}$ the half-open 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

$$ \begin{align*}\exists y\ (y^n-y^{n-1}=\varepsilon x^n)\end{align*} $$

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 $y^n-y^{n-1}=\varepsilon x^n$ , 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 $y^{n-1}(y-1)=\varepsilon x^n$ . By applying v, we obtain

$$ \begin{align*}(n-1)v(y)+v(y-1) = v(\varepsilon)+nv(x)\end{align*} $$

and distinguish three cases.

Case 1: $v(y)>0$ . Then $v(y-1)=0$ and thus $0<(n-1)v(y)=v(\varepsilon )+nv(x)=v(\varepsilon x^n)$ , as required.

Case 2: $v(y)=0$ . Then $v(y-1)=v(\varepsilon )+nv(x)$ . If $v(y-1)>0$ , then we are done immediately. Otherwise, $v(y-1)=0$ and we obtain $\tfrac {v(\varepsilon )}{n} = {{-}\hspace{1pt}v}(x)$ , a contradiction to the fact that $v(\varepsilon )$ is not n-divisible 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 n-divisible.

Now set

$$ \begin{align*}\Psi_\varepsilon = \{\varepsilon x^n\mid x\in \Phi_\varepsilon\}=\{\varepsilon x^n\mid x\in K, v(\varepsilon x^n)>0\}.\end{align*} $$

Again, $\Psi _\varepsilon $ is $\mathcal {L}_{\mathrm {r}}$ -definable only using the parameter $\varepsilon $ . Finally, set

$$ \begin{align*}\Omega_\varepsilon=\left\{ x^n-x^{n-1}\mid x\in K, K\models\exists y\ \exists (z\in \Psi_\varepsilon)\ z(y^n-y^{n-1})=x^n-x^{n-1} \right\}\!.\end{align*} $$

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 $x^n-x^{n-1} = a$ . Then by definition of $\Omega _\varepsilon $ there are $y\in K$ and $z\in \Psi _\varepsilon $ such that $z(y^n-y^{n-1})=a$ . 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

$$ \begin{align*} 0 &> v(x^{n-1}(x-1))=(n-1)v(x)+v(x-1)=nv(x)\\ \text{and} \quad 0 &> v(z)+(n-1)v(y)+v(y-1)>(n-1)v(y)+v(y-1)=nv(y). \end{align*} $$

Since

$$ \begin{align*}\varepsilon=zr^{-n}=(x^n-x^{n-1})[(y^n-y^{n-1})r^n]^{-1},\end{align*} $$

we obtain

$$ \begin{align*}v(\varepsilon)=v(x^n-x^{n-1})-v(y^n-y^{n-1})-nv(r)=n(v(x)-v(y)-v(r)),\end{align*} $$

contradicting the choice of $\varepsilon $ . If $v(a)=0$ , then $0=(n-1)v(x)+v(x-1)$ and thus $v(x)=v(x-1)=0$ . Moreover, $0=v(z)+(n-1)v(y)+v(y-1)$ and thus $0>{{-}\hspace{1pt}v}(z)=(n-1)v(y)+v(y-1)=nv(y)$ . This implies

$$ \begin{align*}v(\varepsilon)=v(x^n-x^{n-1})-v(y^n-y^{n-1})-nv(r)=n(v(y)-v(r)),\end{align*} $$

which again contradicts the fact that $v(\varepsilon )$ is not n-divisible. 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 $x^n-x^{n-1}=a$ . Since $\tfrac {v(\varepsilon )}{n}$ is a left-sided limit point of G, we obtain some $b \in K^\times $ with

$$ \begin{align*}\frac{v(\varepsilon)}{n}-\frac{v(a)}{n}<v(b)< \frac{v(\varepsilon)}{n}.\end{align*} $$

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 $y^n-y^{n-1}=\frac {a}{z}$ . This yields $a\in \Omega _\varepsilon $ and therefore $\Omega _\varepsilon = \mathcal {M}_v$ , as required. ⊣

Proof. Note that since G contains a convex subgroup that is not n-divisible, we have that G is a non-divisible ordered abelian group. Assume, for a contradiction, that G is closed in ${G}^{\mathrm {div}}$ . Since C is not n-divisible, 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

$$ \begin{align*}\left(\frac{a}{n}-\delta,\frac{a}{n}+\delta\right) \cap G =\emptyset.\end{align*} $$

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

(1) $$ \begin{align} \left(\frac{a}{n}-\frac{b}{n},\frac{a}{n}+\frac{b}{n}\right) \cap G =\emptyset. \end{align} $$

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 $(a-b,a+b)$ is an infinite convex subset of C. By n-regularity of C, there is some $z\in C$ such that $a-b \leq nz \leq a+b$ . This is equivalent to

$$ \begin{align*}\frac{a-b}{n} \leq z \leq \frac{a+b}{n},\end{align*} $$

a contradiction to (1). Hence, G is not closed in ${G}^{\mathrm {div}}$ .⊣

Proof. Let $h_n$ be the $\mathcal {L}_{\mathrm {vf}}$ -sentence stating that v is a non-trivial 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., $L\equiv K$ .⊣

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.⊣

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

$$ \begin{align*} f(T) = T^m + a_{m-1}T^{m-1}+ \cdots + a_0 \in L[T], \end{align*} $$

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 $f(T)=T^m + a_{m-1}T^{m-1}+ \cdots + a_0 \in L[T]$ . By construction of L, we have $\deg _K(a_i)\in M$ for any $i\in \{0,\ldots ,m-1\}$ . Moreover, since $[K(a_0,\ldots ,a_{m-1}):K]$ divides the product $[K(a_0):K]\ldots [K(a_{m-1}):K]=\deg _K(a_0)\ldots \deg _K(a_{m-1})$ , we obtain $[K(a_0,\ldots ,a_{m-1}):K] \in M$ . Suppose that $\alpha \in {K}^{\mathrm {h}}$ is a zero of f. Since also $[K(a_0,\ldots ,a_{m-1},\alpha ):K(a_0,\ldots ,a_{m-1})]\leq m\leq p$ , we obtain

$$ \begin{align*} &[K(a_0,\ldots,a_{m-1},\alpha):K]\\ &\qquad = [K(a_0,\ldots,a_{m-1},\alpha):K(a_0,\ldots,a_{m-1})]\cdot [K(a_0,\ldots,a_{m-1}):K]\in M.\end{align*} $$

Finally, since $\deg _K(\alpha )=[K(\alpha ):K]$ divides $[K(a_0,\ldots ,a_{m-1},\alpha ):K]$ , 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

$$ \begin{align*}a = t^{(b_1,0)} + t^{(0,b_2)} + 1 \in k[A_1\amalg A_2],\end{align*} $$

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

$$ \begin{align*}g(T) = T^q + a T^{q-1} + a t^{(b_1,0)} T^{q-2} + \dots + a t^{(b_1,0)} T + a t^{(b_1,0)} \in K[T]. \end{align*} $$

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 $gv_{\min } (T)=T^q+T^{q-1}\in k[T]$ . 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.⊣

Construction 4.5 (Main Construction)

Let $\{A_\gamma \mid \gamma \in \omega \}$ be a family of non-trivial 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 non-trivial Henselian valuation. In particular, K is t-Henselian (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 $(K_{m-1},<)$ has already been defined. We denote the valuation given by $s\mapsto \min \mathrm {supp} (s)$ on $K_{m-1}\!\left (\!\left ( A_{2m-1}\amalg A_{2(m-1)} \right )\!\right )$ by $v_{\min }^m$ . By Lemma 4.4, there is an ordered valued field $K_m$ with

$$ \begin{align*} &K_{m-1}\left(A_{2m-1}\amalg A_{2(m-1)}\right) \subseteq K_m, \\ \subsetneq\: &{K_{m-1}\left(A_{2m-1}\amalg A_{2(m-1)}\right)}^{\mathrm{h}}\subseteq K_{m-1}\!\left(\!\left( A_{2m-1}\amalg A_{2(m-1)} \right)\!\right), \end{align*} $$

where the ordering is inherited from $K_{m-1}\!\left (\!\left ( A_{2m-1}\amalg A_{2(m-1)} \right )\!\right )$ and the valuation is $v_{\min }^m$ , such that any zero in ${K_{m-1}\!\left (A_{2m-1}\amalg A_{2(m-1)}\right )}^{\mathrm {h}}$ 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 $K_mv_{\min }^m =K_{m-1}$ . Let $\mathcal {O}_m'$ and $\mathcal {M}_m'$ be the valuation ring and valuation ideal of $v_{\min }^m$ in $K_m$ and let $\pi _m\colon \mathcal {O}_m' \to K_{m-1}$ 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

$$ \begin{align*}\psi_{i,i} = \mathrm{id}_{K_i \cup \{\infty\}} \text{ and }\psi_{i,i-1} (x) = \begin{cases} \pi_i (x),& \text{ if } x\in \mathcal{O}_{i}',\\ \infty,& \text{ if } x\notin\mathcal{O}_{i}'. \end{cases} \end{align*} $$

For $0\leq j<i-1$ we set $\psi _{i,j} = \psi _{j+1,j} \circ \dots \circ \psi _{i,i-1}$ . 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:

$$ \begin{align*} I = \left\{\left.\! (x_m) \in \prod_{m\in\omega} \left(K_m\cup \{\infty\}\right)\ \right\vert\ \psi_{i,j}(x_i)=x_j \text{ for all } i,j\in \omega \text{ with }j\leq i \right\}\!. \end{align*} $$

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

(2) $$ \begin{align} x_k \in \mathcal{O}^\prime_k \setminus \mathcal{M}^\prime_k, \end{align} $$

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. $

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 $j\leq i$ :

(3) $$ \begin{align} \psi_{i,j} \circ \psi_{i} = \psi_j. \end{align} $$

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 $a\in \mathcal {M}_{n+1}$ :

$$ \begin{align*} \psi_n(a) = \psi_{n+1,n}(\psi_{n+1}(a)) = \psi_{n+1,n}(0) = 0. \end{align*} $$

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 $n\in \mathbb {N}$ .

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

$$ \begin{align*}f(T) = T^{ 2n+1} + T^{ 2n} + a^{( 2n-1)}T^{ 2n-1} + \dots + a^{(0)} \in K[T],\end{align*} $$

where $a^{( 2n-1)},\dots ,a^{(0)}\in \mathcal {M}_n$ . We need to find a zero $(y_m)\in K$ of f. For any $i\in \omega $ , set $a_i^{( 2n-1)} = \psi _i\big (a^{( 2n-1)}\big ),\dots ,a_i^{(0)}=\psi _i\big (a^{(0)}\big ) \in K_i$ and

$$ \begin{align*}f_i(T) = T^{ 2n+1} + T^{ 2n} + a_i^{( 2n-1)}T^{ 2n-1} + \dots + a_i^{(0)}\in K_i[T].\end{align*} $$

First let $i\leq n$ . Then $a_i^{( 2n-2)}= \dots = a_i^{(0)}=0$ 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 $y_{i-1}\in K_{i-1}$ is a simple zero of $f_{i-1}$ with $\psi _{i-1,j}(y_{i-1})=y_j$ for $0\leq j\leq i-1$ . Applying $\pi _i$ to the coefficients of $f_i(T)\in K_i[T]$ we obtain the residue polynomial

$$ \begin{align*}T^{ 2n+1} + T^{ 2n} + a_{i-1}^{( 2n-1)} T^{ 2n-1} + \dots + a_{i-1}^{(0)}=f_{i-1}(T)\in K_{i-1}[T].\end{align*} $$

Thus, $f_i$ has a simple zero $y_i$ in the henselisation of $K_i$ with $\pi _i(y_i)=y_{i-1}$ . 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.

Proof. Proof of Claim 3

Assume for a contradiction that $v_n$ is Henselian. Let $m\in \mathbb {N}$ and let

$$ \begin{align*} g(X)=X^m + X^{m-1} + a_{m-2}X^{m-2} + \dots + a_0 \in K_{n+1}[X] \end{align*} $$

for some $a_{m-2},\dots ,a_0\in \mathcal {M}^{\prime }_{n+1}$ . Let $a^{(m-2)},\dots ,a^{(0)}\in K$ such that $\psi _{n+1}\!\big (a^{(i)}\big )=a_i$ for $0\leq i \leq m-2$ . Then $a^{(m-2)},\dots ,a^{(0)}\in \psi _n^{-1}(\{0\})=\mathcal {M}_n$ . As $v_n$ is Henselian, there exists $x\in \mathcal {O}_n$ such that $x^m + x^{m-1} + a^{(m-2)}x^{m-2} + \dots + a^{(0)} = 0$ . 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 non-Henselian but $( 2n+1)_\leq $ -Henselian valuation $v_n$ on K. It remains to show that K does not admit a non-trivial 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

$$ \begin{align*} \mathbb{R}(A_1\amalg A_0)\subseteq K_1\subseteq \mathbb{R}\!\left(\!\left( A_1\amalg A_0 \right)\!\right)\subseteq \mathbb{R}\!\left(\!\left( G \right)\!\right)\kern-1.5pt, \end{align*} $$

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

$$ \begin{align*}\mathbb{R}(A_1\amalg A_0)(A_3\amalg A_2)\subseteq K_2\subseteq \mathbb{R}\!\left(\!\left( A_1\amalg A_0 \right)\!\right)\!\left(\!\left( A_3\amalg A_2 \right)\!\right)\subseteq \mathbb{R}\!\left(\!\left( G \right)\!\right)\kern-2pt.\end{align*} $$

The last inclusion is given via the isomorphismFootnote ⁸

$$ \begin{align*}\mathbb{R}\!\left(\!\left( A_1\amalg A_0 \right)\!\right)\!\left(\!\left( A_3\amalg A_2 \right)\!\right)\cong \mathbb{R}\!\left(\!\left( (A_3\amalg A_2)\amalg(A_1\amalg A_0) \right)\!\right)\end{align*} $$

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

$$ \begin{align*}\mathbb{R}\left(\coprod_{\gamma=-(2m-1)}^0 A_{-\gamma}\right) \subseteq K_m \subseteq \mathbb{R}\!\left(\!\left( \coprod_{\gamma=-(2m-1)}^0 A_{-\gamma} \right)\!\right)\subseteq \mathbb{R}\!\left(\!\left( G \right)\!\right)\kern-1.7pt.\end{align*} $$

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 $v_{\min } K_m=\coprod _{\gamma =-(2m-1)}^0 A_{-\gamma }$ . 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 =-(2m-1)}^0 A_{-\gamma }$ and $v_{\min }^m K_m = \left (A_{2m-1} \amalg A_{2(m-1)}\right )$ . Thus, $v_{\min }^m$ is the coarsening of $v_{\min }$ induced by the convex subgroup $\coprod _{\gamma =-(2m-3)}^0 A_{-\gamma }$ 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 $ .

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

$$ \begin{align*}x_k = \sum_{g \in G} c_g t^g \in \mathbb{R}\!\left(\!\left( G \right)\!\right)\kern-1.8pt.\end{align*} $$

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

$$ \begin{align*}x_{k+1} = c_{g_0} t^{g_0} + \sum_{g>g_0} c_gt^g + \varepsilon,\end{align*} $$

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

$$ \begin{align*}\varepsilon = \sum_{g\in \coprod^0_{\gamma = -(2k+1)} A_{-\gamma}} \varepsilon_g t^g.\end{align*} $$

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

$$ \begin{align*} v((x_m)):=v_{\min}(x_i) \in G \end{align*} $$

for any $x_i \notin \{0, \infty \}$ .

Proof. Proof of Claim 6

Note that by Claim 5, v is well-defined. 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

$$ \begin{align*} v((x_m)(y_m))=v_{\min}(x_iy_i)=v_{\min}(x_i)+v_{\min}(y_i)=v((x_m))+v((y_m)). \end{align*} $$

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 8. For any $n\in \omega $ we have $v_0(\mathcal {O}_n^\times ) = \coprod _{\gamma = -2n+1}^0 A_{-\gamma }$ and therefore

(4) $$ \begin{align} v_nK=\left(\coprod_{\gamma>-\infty}^{0}\!A_{-\gamma}\right) / \left(\coprod_{\gamma = -2n+1}^0 A_{-\gamma}\right) = \coprod_{\gamma>-\infty}^{-2n} A_{-\gamma}. \end{align} $$

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 $u_{2n+1}K=\coprod _{\gamma>-\infty }^{-2n-1} A_{-\gamma }$ , as required.⊣

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 non-trivial $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 non-trivial, $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 non-trivial $0$ - $\mathcal {L}_{\mathrm {or}}$ -definable Henselian valuation in $(L,<)$ corresponds to a non-trivial $0$ - $\mathcal {L}_{\mathrm {or}}$ -definable Henselian valuation in $(K, <)$ . However, since $(K,<)$ does not admit any non-trivial Henselian valuation, this shows the required conclusion.⊣