Proof Let M, P, F, $\chi $ , $\theta $ , $\sigma $ , and G satisfy the hypotheses of the lemma. As discussed at the beginning of this section, without loss of generality, we assume that $\delta =0$ . Let $H=G(\sigma [\theta ])$ . So H is $P(\sigma [\theta ])$ -generic over $M[G_0]$ . We observe that, since F is morass-definable, it is the interpretation in $M[G]$ of a set of terms with strict level, T. For any $\alpha <\omega _2$ , the interpretation of $T\cap M^{P(\alpha )}$ is just $F\cap M[G(\alpha )]$ , and is a subfield of F. Assume $\chi \subseteq (F\cap M[G_0)])$ is linearly dependent over $F\cap M[H]$ and that $\{ y_1,\ldots y_n\} \in M[H]$ (all non-zero) are such that $\sum _{i=1}^n x_i\cdot y_i=0$ in $M[G_0][H]$ . For $1\leq i\leq n$ , let $\mathbf {x}_i \in M^{P(\theta )}$ and $\mathbf {y}_i \in M^{\kern-1pt P(\sigma [\theta ])}$ be terms for $x_i$ and $y_i$ , respectively. Then there is $(p,q)\in G_0 \times H$ such that, in M, $(p,q)\Vdash \sum _{i=1}^n \mathbf {x}_i\cdot \mathbf {y}_i = 0$ . We force beneath $(p,q)$ .

Let $\beta $ be countable, $Q_0$ be a poset for adding countably many generic reals indexed by $\beta \setminus \theta $ and $H_0$ be $Q_0$ -generic over $M[G_0]$ such that $(p,q_0)\in G_0\times H_0$ , and the orthogonal complement of $\mathbf {x}= (x_1,\ldots ,x_n)$ in $F^n \cap M[H_0]$ , $\mathbf {x}_0^{\perp }$ , has maximum possible dimension, m, in $F^n\cap M[H_0]$ , where $0<m<n$ . Let $q_0 \in H_0$ force that the $\mathbf {x}^{\perp }$ has dimension m, where $q_0<q$ . Let $\sigma $ be a splitting map on $Q_0$ with splitting point $\delta =\theta $ . Let $Q_1=\sigma (Q_0)$ , $q_1=\sigma (q_0)<\sigma (q)$ and $\mathbf {x}_1^{\perp }= \sigma (\mathbf {x}_0^{\perp })$ . So $Q_1$ and $Q_0$ are disjoint. If $H_1$ is $Q_1$ -generic over $M[G_0][H_0]$ and $q_1\in H_1$ , then in $M[H_1]$ , $\mathbf {x}_1^{\perp }$ has dimension m. Furthermore, in $M[H_0][H_2]$ , $\mathbf {x}^{\perp }$ has dimension m. Then $\mathbf {x}_0^{\perp }=\mathbf {x}_1^{\perp }$ are elements of mutually generic extensions of M, and are thereby members of M. Therefore $\mathbf {x}^{\perp }$ , computed in M has dimension $m>0$ , and the components of x are linearly dependent over $F\cap M$ .

Proof Let $M[G][H]\models \sum _{i=1}^n x_i\cdot y_i=0$ . Let $A_0$ be the row-reduced echelon form of the orthogonal complement of $\bar {x}, \bar {x}^{\perp }$ , computed in $M[G]$ . We force over ground model, M. Let $(p,q)\in G\times H$ be such that $(p,q)\Vdash \sum _{i=1}^n x_i\cdot y_i=0$ . We force below $(p,q)$ . Let $\theta _1 \geq \sigma (\theta )$ , $Q_1=P(\sigma (\theta _1) \setminus \theta )$ and $H_1$ be $Q_1$ -generic over $M[G]$ be such that the dimension of $\bar {x}^{\perp }$ computed in $M[H_1]$ , $\bar {x}_1^{\perp }$ , is maximal for all possible choices of $\theta _1$ and $H_1$ (where $q\in H_1$ ). Let $A_1\in M[H_1]$ have row space equal to $\bar {x}_1^{\perp }$ and be in row-reduced echelon form. Since we are forcing below $(p,q)$ , $A_1$ has a non-zero row.

Let the rank of $A_1$ be m. In any generic extension of $M[G]$ by a poset for adding countably many ordinals, the rank of $\bar {x}^{\perp }$ computed in that extension is no greater than m. Let $q_1\in Q_1$ and $q_1\Vdash (A_1$ is in row-reduced echelon form with row space equal to $\bar {x}^{\perp }$ and $Rank(A_1)=m)$ .

Let $\sigma $ be a splitting function on $\theta _1$ with splitting point $\theta $ . Let $Q_2=P(\sigma (\theta ) \setminus \theta )$ and $H_2$ be $Q_2$ generic over $M[G][H_1]$ with $q_2=\sigma (q_1)\in H_2$ . Let $A_2=\sigma (A_1)$ . Then $q_2\Vdash A_2$ is in row-reduced echelon form with row space equal to $\bar {x}^{\perp }$ , and has rank m.

Since m is the maximum possible rank for $\bar {x}^{\perp }$ , and row-reduced echelon form is canonical,

$$\begin{align*}M[G][H_1][H_2] \models A_1=A_2. \end{align*}$$

$A_1$ and $A_2$ are in mutually generic extensions of $M[G]$ , so

$$\begin{align*}A_1=A_2\in M[G]. \end{align*}$$

But $A_1\in M[H_1]$ and $A_2\in M[H_2]$ , so $A_1\in M$ . Then $\bar {y}$ is in the row space of $A_1$ . Hence $\bar {y}$ is in the span of vectors in $F^n\cap M$ .

Proof Let $\chi ^*$ be the multiplicative semi-group generated by the elements of $\chi $ . Then $\chi ^*$ is LI over $M\cap F$ . By Lemma 6.1, $\chi ^*$ is LI over $M[H]\cap F$ . Therefore $\chi $ is AI over $M[H]\cap F$ .

Proof F is morass-closed, so in all $P(\sigma [\theta ])$ -generic extensions of M, $\sigma [\mathbf {\chi }]$ is AI over $M\cap F$ . By Corollary 6.3, $\sigma [\chi ]$ is AI over $M\cap F$ in $M[G]$ . Suppose there is a nontrivial linear combination (over $M\cap F$ ) of distinct elements of the semigroup generated by $\chi \cup \sigma [\chi ]$ that equals 0. By Corollary 6.3, $\chi $ is AI over $M[H]\cap F$ . This implies that there is a nontrivial linear combination (over $M\cap F$ ) of elements of the semigroup generated by $\sigma [\chi ]$ that equals 0. However, $\sigma [\chi ]$ is AI over $M\cap F$ , so $val_G(\mathbf {\chi }\cup \sigma [\mathbf {\chi }])$ is AI over $M\cap F$ .

Proof Without loss of generality we assume that the splitting point of $\sigma $ , ${\delta =0}$ . An element of $X^*$ may be expressed as $\sum _{i=1}^n x_i\cdot \sigma (y_i)$ , for some $n\in \mathbb N$ , and $x_1,\ldots ,x_n,y_1,\ldots ,y_n \in X$ . Let S be the set of expressions of this form. We define a function $\psi :S\to \mathcal {E}$ where

$$\begin{align*}\psi\left(\sum_{i=1}^n x_i\cdot \sigma(y_i)\right) = \sum_{i=1}^n \phi(x_i)\cdot \sigma(\phi(y_i)). \end{align*}$$

Let $\iota :S \to X^*$ be the natural quotient map from the expressions of S to $X^*$ . The kernel of $\iota $ is the set of expressions of S that sum to $0$ in $X^*$ . $\psi $ defines a ring homomorphism on $X^*$ if and only if, for any s in the kernel of $\iota $ ,

$$\begin{align*}\psi(s)=0. \end{align*}$$

For $i\leq n$ let $z_i=\sigma (y_i)$ and $s=\sum _{i=1}^n x_i\cdot z_i$ be in the kernel of $\iota $ . Then in $X^*$ ,

$$\begin{align*}\sum_{i=1}^n x_i\cdot z_i=0. \end{align*}$$

Let

$$\begin{align*}\bar{x}=(x_1,\ldots,x_n)\in X^n \cap M[G], \end{align*}$$

$$\begin{align*}\bar{y}=(y_1,\ldots,y_n) \in X^n\cap M[G], \end{align*}$$

and

$$\begin{align*}\bar{z}=(z_1,\ldots,z_n) \in \sigma([X])^n \cap M[H]. \end{align*}$$

By Lemma 6.2, $\bar {z}$ is in the span of elements of $X^n\cap M$ that are orthogonal to $\bar {x}$ . Let $\{ b_1,\ldots ,b_m\}$ be an LI set of vectors of $X^n\cap M$ orthogonal to $\bar {x}$ that contains $\bar {z}$ in its span. Let $\langle \cdot ,\cdot \rangle $ be the dot product and $(\alpha _1,\ldots ,\alpha _m)\in \sigma ([X])^n\cap M[H]$ be such that

$$\begin{align*}\sum_{i=1}^m \alpha_i\cdot b_i=\bar{z}. \end{align*}$$

Let $\bar {\phi }:X^n\to \mathcal {E}^n$ be defined by

$$\begin{align*}\bar{\phi}(s_1,\ldots,s_n)=(\phi(s_1),\ldots,\phi(s_n)). \end{align*}$$

Recall that for a splitting map $\sigma $ , $\sigma (\phi ):\sigma [X]\to \mathcal {E}$ is defined so that $\sigma $ and $\phi $ commute. Then

$$\begin{align*}\psi(\langle \bar{x},\bar{z}\rangle) = \left\langle \bar{\phi}(\bar{x}),\sigma(\bar{\phi})\left(\sum_{i=1}^m \alpha_i \cdot b_i\right) \right\rangle = \sum_{i=1}^m \sigma(\alpha_i)\cdot \langle \bar{\phi}(\bar{x}), \sigma(\bar{\phi})(b_i) \rangle. \end{align*}$$

However, $b_i\in X^n\cap M$ for all $i\leq n$ , so

$$\begin{align*}\sigma(\bar{\phi})(b_i)=\bar{\phi}(b_i). \end{align*}$$

Hence

$$\begin{align*}\sum_{i=1}^m \sigma(\alpha_i)\cdot \langle \bar{\phi}(\bar{x}), \sigma(\bar{\phi}) (b_i) \rangle = \sum_{i=1}^m \sigma(\alpha_i)\cdot \langle \bar{\phi}(\bar{x}),\bar{\phi}(b_i) \rangle = \sum_{i=1}^m \sigma(\alpha_i)\cdot \phi(\langle \bar{x},b_i\rangle). \end{align*}$$

For all $i\leq m$ , $b_i\perp \bar {x}$ . So for all $i\leq m$ ,

$$\begin{align*}\phi(\langle \bar{x},b_i\rangle ) = 0 \end{align*}$$

and $\psi (\langle \bar {x},\bar {z} \rangle )=0$ . Therefore $\psi $ defines a ring homomorphism on $X^*$ , $\phi ^*$ , that extends $\phi \cup \sigma (\phi )$ to $X^*$ .

Assume that $\bar {x}, \bar {y}\in M[G]\cap X^n$ , $\bar {z}=\sigma (\bar {y})$ and

$$\begin{align*}\langle \bar{x},\bar{z} \rangle \neq 0. \end{align*}$$

We consider X as a vector space over $M\cap X$ . Let B be a basis for X over ${M\cap X}$ . There is a finite subset of basis vectors, $X'=\{x_1',\ldots ,x_m'\}\subseteq B$ so that every component of $\bar {x}$ is in the span of $X'$ . Hence $\langle \bar {x},\bar {y} \rangle $ may be expressed as a linear combination of $X'$ over $\sigma [X]$ . For $1\leq i\leq m$ , let $z^{\prime }_i$ be the sum of coefficients of $x_i'$ in the linear combination over $X'$ . Then $z_i'\in \sigma [X]$ and

$$\begin{align*}\sum_{i=1}^m x_i'\cdot z_i'=\sum_{i=1}^n x_i\cdot z_i \neq 0. \end{align*}$$

Then $\{ \phi (x_1'),\ldots ,\phi (x_m')\}$ is LI over $\phi [X] \cap M$ . By Lemma 6.1, $\{ \phi (x_1'),\ldots ,\phi (x_m')\}$ is LI over $\mathcal {E} \cap M[H]$ . Therefore

$$\begin{align*}\sum_{i=1}^n \phi^*(x_i\cdot z_i)=\sum_{i=1}^m \phi^*(x_i'\cdot z_i')\neq 0. \end{align*}$$

Thus $\phi ^*$ is a monomorphism.

We show that $\phi ^*$ is $\mathbb R$ -linear. Let $r\in \mathbb R\cap X^*$ . Since X is a subring of finite elements of ${\mathbb R}^{\omega }/U$ , every element of X may be expressed as the sum of a real number and an infinitesimal.

Then there are $x_1\dots ,x_n,y_1, \dots y_n\in X$ such that

$$\begin{align*}r=\sum_ {i=1}^n x_i\cdot \sigma(y_i). \end{align*}$$

For $1\leq i\leq n$ , each $x_i$ and $y_i$ may be written as $r_i+\delta _i$ and $s_i+\epsilon _i$ , resp., where for $r_i,s_i\in \mathbb R$ and $\delta _i$ , $\epsilon _i$ are infinitesimal. So

$$\begin{align*}\sum_{i=1}^n r_i\cdot \sigma(s_i)=r \end{align*}$$

and

$$\begin{align*}\sum_{i=1}^n r_i\cdot \sigma(\epsilon_i)+\sigma(s_i)\cdot \delta_i+\delta_i\cdot \sigma(\epsilon_i)=0. \end{align*}$$

Since $\phi $ is extendible, for $1\leq i\leq n$ , $\phi (r_i)=r_i$ and $\phi (\delta _i)$ is infinitesimal in  $\mathcal {E}$ . Therefore $r_i\in X$ . Similarly, $s_i\in X$ . Hence r is in the ring generated by $(\mathbb R\cap X) \cup (\sigma [\mathbb R\cap X])$ . So $\phi ^*(r)=r$ , and $\phi ^*$ is $\mathbb R$ -linear.

We show that $\phi ^*$ is order-preserving. For $n\in \mathbb N$ and $x_1,\ldots ,x_n,y_1,\ldots ,y_n \in X$ , we show that $\sum _{i=1}^n x_i\cdot \sigma (y_i)>0$ iff $\sum _{i=1}^n \phi ^*(x_i)\cdot \phi ^*(\sigma (y_i))>0$ .

If $n=1$ , then the sign of $x_1\cdot \sigma (y_1)$ is the sign of the product of the leading coefficients of $x_1$ and $\sigma (y_1)$ , which are preserved by $\phi ^*$ . So

$$\begin{align*}x_1\cdot \sigma(y_1)>0 \iff \phi(x_1)\cdot \sigma(\phi(y_1))>0. \end{align*}$$

Assume that $n \geq 2$ , and

$$\begin{align*}\sum_{i=1}^n x_i\cdot \sigma(y_i)>0. \end{align*}$$

Let

$$\begin{align*}y=\phi^*\left(\sum_{i=1}^n x_i\cdot \sigma(y_i)\right)=\sum_{\lambda<\gamma} \alpha_{\lambda}x^{a_{\lambda}}. \end{align*}$$

By assumption, $\phi [X]$ is a ring of finite elements of $\mathcal {E}$ that is closed under partial sums and contains all coefficients that appear in members of the range of $\phi $ . Since $\phi $ is extendible, there are $b_0\in M[G]\cap G_{\omega _1}$ and $c_0\in M[H]\cap G_{\omega _1}$ such that $x^{b_0} \in \phi [X]$ , $x^{c_0}\in \sigma [\phi [X]]$ , $b_0, c_0\geq e$ (in $G_{\omega _1}$ ) and

$$\begin{align*}a_0=b_0+c_0. \end{align*}$$

Let $u,v\in X$ be such that

$$\begin{align*}\phi(u)=x^{b_0} \end{align*}$$

and

$$\begin{align*}\sigma(\phi(v))=x^{c_0}. \end{align*}$$

Let $\hat {X}$ be the ring generated by X, $1/u$ and $1/v$ in ${\mathbb R}^{\omega }/U$ . Let $\psi :\hat {X} \to \mathcal {E}$ be the unique ring homomorphism extending $\phi $ satisfying $\psi (1/u)=1/\phi (u)=x^{-b_o}\in \mathcal {E}$ and $\psi (\sigma (1/v))=x^{-c_0}\in \mathcal {E}$ . Then $\psi $ is extendible. Hence the extension of $\psi $ to the ring homomorphism, $\psi ^*$ , on the ring generated by $\hat {X}\cup \sigma [\hat {X}]$ is an $\mathbb R$ -linear monomorphism. Furthermore, since $\phi $ is an $\mathbb R$ -linear order-monomorphism, $\phi $ extends uniquely to an $\mathbb R$ -linear order-preserving field monomorphism on the field generated by $\hat {X}$ . Hence $\psi $ and $\sigma [\psi ]$ are $\mathbb R$ -monomorphisms on $\hat {X}$ and $\sigma [\hat {X}]$ , resp. Then

$$\begin{align*}\psi^*\left((1/u)\cdot (\sigma(1/v))\cdot \left(\sum_{i=1}^n x_i\cdot \sigma(y_i)\right)\right)= \end{align*}$$

$$\begin{align*}\alpha_0+x^{-a_0}\cdot \phi^*\left(\sum_{i=1}^n x_i\cdot \sigma(y_i)\right) = \end{align*}$$

$$\begin{align*}\alpha_0+x^{-a_0}\cdot \left(\sum_{\lambda=1}^{\gamma} \alpha_{\lambda} x^{a_{\lambda}}\right) = \end{align*}$$

$$\begin{align*}\alpha_0 + \sum_{\lambda=1}^{\gamma} \alpha_{\lambda} x^{a_{\lambda} - a_0}. \end{align*}$$

For $1\leq i\leq n$ , $(1/u)\cdot (\sigma (1/v)) \cdot x_i\cdot \sigma (y_i)$ is finite. Therefore, for $1\leq i\leq n$ ,

$$\begin{align*}(1/u)\cdot (\sigma(1/v)) \cdot x_i\cdot \sigma(y_i) = r_i \cdot \sigma(s_i) +\delta_i\cdot \sigma(s_i) +r_i\cdot \sigma(\epsilon_i)+\delta_i\cdot \sigma(\epsilon_i), \end{align*}$$

where $r_i, s_i \in \mathbb R\cap X$ and $(1/u)\cdot \delta _i, \sigma (1/v)\cdot \sigma (\epsilon _i) \in \hat {X}$ are infinitesimal. If the ith term is infinitesimal, then $r_i=0=s_i$ . Hence,

$$\begin{align*}\alpha_0=\sum_{i=1}^n r_i\cdot s_i.\end{align*}$$

Since $\psi ^*$ is $\mathbb R$ -linear, the real part of $(1/u)\cdot (\sigma (1/v))\cdot \sum _{i=1}^n x_i\cdot \sigma (y_i)$ is $\alpha _0$ , and therefore $\alpha _0>0$ . Furthermore, $\psi ^*((1/u)\cdot \sigma (1/v))>0$ . Hence

$$\begin{align*}\psi^*\left(\sum_{i=1}^n x_i\cdot \sigma(y_i)\right)=\phi^*\left(\sum_{i=1}^n x_i\cdot \sigma(y_i)\right)>0. \end{align*}$$

Therefore $\phi ^*$ is order-preserving and $\phi ^*$ is extendible.

Proof If $\nu =\bar {\nu }+1$ , then the result follows from Lemma 6.5.

If there is no limit ordinal $\lambda $ , $\bar {\nu }<\lambda \leq \nu $ . Then there is $n\in \omega $ such that

$$\begin{align*}\nu=\bar{\nu}+n. \end{align*}$$

By Lemma 6.5, for any extendible $\phi :D\to \mathcal {E}$ and splitting function $\sigma $ , the ring monomorphism on the ring generated by $D\cup \sigma [D]$ extending $\phi \cup \sigma [\phi ]$ is extendible. By n iterated applications of Lemma 6.5, there is a unique extendible $\mathbb R$ -monomorphic extension of $\phi $ , $\phi ^*\supset \bigcup _{\sigma \in \mathcal {F}_{\bar {\nu } \nu }} \sigma [\phi ]$ , to the ring generated by $\bigcup _{\sigma \in \mathcal {F}_{\bar {\nu } \nu }} \sigma [D]$ .

So assume there is a limit ordinal $\lambda $ , $\nu _{\bar {\alpha }}<\lambda \leq \nu _{\alpha }$ . Let $\lambda $ be the least limit ordinal greater than $\nu _{\bar {\alpha }}$ . Let

$$\begin{align*}D_{\lambda}=\bigcup_{\sigma \in \mathcal{F}_{\bar{\nu}, \lambda}} \sigma[D]. \end{align*}$$

Let $D^*_{\lambda }$ be the ring generated by $D_{\lambda }$ , We show that there is an extendible $\mathbb R$ -monomorphism of $D^*_{\lambda }$ which, for any $\sigma \in \mathcal {F}_{\bar {\nu } \lambda }$ , extends $\sigma [\phi ]$ .

Let F be a finitely generated subring of $D^*_{\lambda }$ . Let $\{ d_1,\ldots ,d_n\}$ generate F, $\sigma _1, \ldots ,\sigma _n \in \mathcal {F}_{\bar {\nu },\lambda }$ and for all $i\leq n$ , $c_i\in D$ be such that

$$\begin{align*}d_i=\sigma_i(c_i). \end{align*}$$

By condition P4 in the definition of the simplified morass, there is $N\in \omega $ , $g\in \mathcal {F}_{\nu _{\bar {\alpha }}+N, \lambda }$ and $f_1,\ldots ,f_n \in \mathcal {F}_{\nu _{\bar {\alpha }}, \nu _{\bar {\alpha }}+N}$ such that, for $i\leq n$ ,

$$\begin{align*}\sigma_i=g\circ f_i. \end{align*}$$

For each $m<n$ , let $h_m$ be the splitting function of $\mathcal {F}_{\bar {\nu }+m, \bar {\nu }+m+1}$ . By Lemma 6.5, $\phi \cup h_1[\phi ]$ may be extended to an extendible $\mathbb R$ -monomorphism. Furthermore this ring monomorphism may be extended by the splitting functions $h_2$ through $h_m$ . Let $\psi $ be the function on $\bigcup _{\sigma \in \mathcal {F}_{\bar {\nu } \lambda }} \sigma [D]$ resulting after the n splits. Then $\psi $ is extendible and $f_i(c_i)$ is in the domain of $\psi $ for all $i\leq n$ . Therefore $g\circ \psi $ is an $\mathbb R$ -linear order monomorphism and is the restriction of $\phi ^*$ to a ring containing F. Thus

$$\begin{align*}\phi^*_{\lambda}=\bigcup_{\sigma \in \mathcal{F}_{\bar{\nu} \lambda}} \sigma[\phi] \end{align*}$$

is a well-defined extendible $\mathbb R$ -monomorphism of $D^*_{\lambda }$ .

By an inductive argument on $\nu $ , invoking condition P2, and the results above at limits and Lemma 6.5 at successor ordinals, it is straightforward to show that $\bigcup _{\sigma \in \mathcal {F}_{\lambda , \nu }} \sigma [\phi ^*_{\lambda }]$ has a unique extension to an extendible $\mathbb R$ -monomorphism of the ring generated by $\bigcup _{\sigma \in \mathcal {F}_{\lambda , \nu }} \sigma [D^*_{\lambda }]$ .

Proof Since $\phi $ is extendible, X is full and for all $r_0\in \mathbb R \cap X$ , $\phi (r_0)=r_0\cdot x^e$ (where e is the group identity of $\mathcal {E}$ ). Then the real closure of X is full and the set of real numbers of the real-closure of X is the real closure of $\mathbb R\cap X$ . Let $X^*$ be the real closure of X and $\phi ^*:X^* \to \mathcal {E}$ be the unique $\mathbb R$ -monomorphism extending $\phi $ to $X^*$ . Then for all $r_0\in \mathbb R \cap X^*$ , $\phi (r_0)=r_0\cdot x^e$ . If r is transcendental over X, then r is transcendental over $X^*$ . The real closure of Y and $X^*$ contain precisely the same real numbers and are full. By Lemma 3.4, there is an $\mathbb R$ -monomorphic extension of $\phi ^*$ , $\psi ^*$ , to the real closure of the field generated by $X[r]$ . Let $\psi =\psi ^*\upharpoonright _{X[r]}$ . Then $\psi $ is extendible.

Proof If $x^*\in \mathbb R$ , then apply Lemma 6.7. Assume $x^*\notin \mathbb R$ . If $x^*$ is not infinitesimal, then $x^*=r+\delta $ , where $r\in \mathbb R$ and $\delta $ is infinitesimal. If $r\notin X$ , then we may extend $\phi $ to an extendible function, $\phi ^*:X[r]\to Y[r]$ . We consider the case in which $r\in X$ . Since X is extendible, $x^*-r=\delta \in X$ . Therefore we may assume that $x^*$ is infinitesimal. Let $(l,u)$ be the gap formed by $x^*$ in X, $L=\phi [l]$ and $U=\phi [u]$ . The Esterle algebra is an $\eta _1$ -ordering, so there is $y\in \mathcal {E}$ that witnesses the gap $(L,U)$ . By application of Johnson’s Lemma 3.5, there is $y\in \mathcal {E}$ that witnesses the gap $(L,U)$ and such that the real closure of the field extending $Y[y]$ is full. Although the existence of such an element can be used to advantage, the element y may fail to have some of the properties we require for a morass construction.

Let $\mu < \omega _1$ be the strict level of $x^*$ . We seek an element of $\mathcal {E}$ that witnesses the gap $(L,U)$ and is a candidate for the image of $x^*$ . The candidate must have strict level $\mu $ and be transcendental over Y, among other requirements. The Esterle algebra is level dense and upward level dense, so by Lemma 4.5 of [Reference Dumas3] there is $y\in \mathcal {E}$ , with the strict level $\mu $ , such that for all $z\in X$ ,

$$\begin{align*}x^*<z \iff y<\phi(z). \end{align*}$$

Let Z be the countable set of real coefficients appearing in power series of $Y[y]$ . Then by iterated applications of Lemma 6.7 there is an extendible extension of $\phi $ to $\phi ^*: X[Z] \to Y[Z]$ . Let $X^*=X[Z]$ and $Y^*=Y[Z]$ . If $x^*$ is in the real-closure of $X^*$ , let $\bar {X}=X^*$ and $\psi :\bar {X} \to \mathcal {E}$ be the unique ring monomorphism extending $\phi ^*$ . Then $\psi $ is extendible. So we assume that $x^*$ is transcendental over $X^*$ .

It is possible that $Y^*[y]$ is not closed under partial sums. We show that there is $y^*\in \mathcal {E}$ that witnesses $(L,U)$ and such that $Y^*[y^*]$ is full and closed under partial sums.

Case 1: There is a largest partial sum of y that is a member of $Y^*$ .

We include in this case that the first term of y is a monomial not in $Y^*$ . Let t be the largest partial sum of y that is also a member of $Y^*$ and $s=(\phi ^*)^{-1}(t)\in X^*$ . Then $y-t \notin Y^*$ and $x^*-s\notin X^*$ . We shift our attention to the gap formed by $x^*-s$ in $X^*$ . Then for all $z\in X^*$

$$\begin{align*}x^*-s<z \iff y-t<\phi(z). \end{align*}$$

If $x^*-s$ were algebraic over $X^*$ , then it would be in the real-closure of $X^*$ . Since $X^*$ is a ring, and $s\in X^*$ , this would imply that $x^*$ is in the real-closure of $X^*$ , contrary to assumption that $x^*$ is transcendental over $X^*$ . Therefore $x^*-s$ is transcendental over $X^*$ . We claim that $x^*-s$ must have an Archimedean valuation distinct from the Archimedean valuations of members of $X^*$ . Let $y-t \in \mathcal {E}$ have leading term $\alpha x^a$ . Then $\alpha \in Z \subseteq Y^*$ . If $x^a$ were in the range of $\phi ^*$ , then $t+ \alpha x^a \in Y^*$ would be a partial sum of y, contrary to assumption. So a must be a valuation distinct from the valuations of $Y^*$ . Therefore $x^*-s$ must have a valuation distinct from the valuations of members of $X^*$ .

Let $(L^*,U^*)$ be the gap formed by $y-t$ in $Y^*$ . If $\alpha>0$ , then it is sufficient to show that there is $c\in G_{\omega _1}$ such that $x^c$ has strict level $\mu $ and witnesses the gap $(L^*,U^*)$ . The case $\alpha <0$ is altogether similar. Let $y_0=x^a$ . The strict level of $y_0$ equals the strict level of $a\in G_{\omega _1}$ . It is straightforward to see that there is $b\in G_{\omega _1}$ such that:

1. 1. b is positive.

2. 2. Any element of the support of b is greater than any element of the support of any exponent occurring in any power series of $Y^*$ .

3. 3. b has strict level $\mu $ .

The exponent $b\in G_{\omega _1}$ is greater than $0$ but less than any positive exponent in $Y^*$ . Furthermore, b is greater than the constant $0$ function, the additive identity of $G_{\omega _1}$ (and the valuation of standard reals in $\mathcal {E}$ ). However b is greater than any positive valuation occurring in $Y^*$ . Consequently $x^b$ is less than any infinitesimal of $Y^*$ . Let $c=a+b$ . Then,

1. 1. $x^{c}$ witnesses the gap $(L,U)$ .

2. 2. $x^{c}$ is transcendental over X.

3. 3. $x^{c}$ has strict level $\mu $ .

4. 4. The subring of $\mathcal {E}$ generated by $Y^*\cup \{ x^{c}\}$ is full and is closed under partial sums.

Let $\psi :X^*[x^*]\to Y^*[x^{c}]$ be the unique $\mathbb R$ -monomorphism extending $\phi $ such that

$$\begin{align*}\psi(x^*)=t+x^{c}. \end{align*}$$

Then $\psi $ is extendible.

Case 2: There is no largest partial sum of y that is a member of $Y^*$ .

Let D be the well-ordering by ascending valuation of the exponents of y. Then D has a countable order-type. Let $D'$ be the smallest initial segment of D such that $y\upharpoonright _{D'}$ (the partial sum of y with exponents from $D'$ ) is not a member of Y. Let $y'=y\upharpoonright _{D'}$ . Then $y'$ witnesses the gap $(L,U)$ . The strict level of $y'$ is no greater than $\mu $ . If the strict level of $y'$ equals $\mu $ , let $\psi :X^*[x^*]\to Y^*[y']$ be the unique $\mathbb R$ -monomorphism extending $\phi $ such that

$$\begin{align*}\psi(x^{*})=y'. \end{align*}$$

Otherwise, let $a\in G_{\omega _1}$ be an exponent of $\mathcal {E}$ that has strict level $\mu $ and is greater than all exponents occurring in $Y^*$ . Let

$$\begin{align*}y^*=y'+x^{a}. \end{align*}$$

Then $y^*$ witnesses the gap $(L,U)$ and has strict level $\mu $ . Let $\psi ^*:X^*[x^*]\to Y^*[y^*]$ be the unique $\mathbb R$ -monomorphism extending $\phi $ such that

$$\begin{align*}\psi(x^*)=y^*. \end{align*}$$

Then $\psi ^*$ satisfies the conditions for an extendible function, except closure of the range of $\psi ^*$ under partial sums. In particular $y'$ and $x^a$ are not in the range of $\psi ^*$ . Let $(L^*,U^*)$ be the gap formed by $x^a$ in $Y^*[y^*]$ . We observe that $x^a$ is infinitesimal with respect to every member of $Y^*[y^*]$ . Since F is level dense and upward level dense, there is a positive element of F, $\epsilon $ , with strict level $\mu $ that is infinitesimal with respect to all elements of $X^*[x^*]$ . Therefore there is an $\mathbb R$ -monomorphism, $\psi :X^*[x^*,\epsilon ]\to Y^*[y^*, x^a]$ , extending $\psi ^*$ and such that

$$\begin{align*}\psi(\epsilon)=x^a. \end{align*}$$

We note that $y'\in Y^*[y^*,x^a]$ and

$$\begin{align*}Y^*[y^*, x^a]=Y^*[y', x^a]. \end{align*}$$

Then $Y^*[y', x^a]$ is closed under coefficients and partial sums, so $\psi $ is extendible.