Fact 3.9. The following properties of field extensions are invariant under base extension: separable extensions, relatively perfect extensions, primary extensions and regular extensions.

Proof. The results are well known, so we only sketch the proofs. Let ${K}/{k}$ be a field extension and ${M}/{m}$ a base extension (i.e, we have $M=K\otimes _{k}m$ ).

Separable extensions. We want to prove that ${K}/{k}$ is separable if and only if ${M}/{m}$ is so. If M is separable over m, then by definition, the fields M and $m^{p^{-\infty }}$ are linearly disjoint over m. By transitivity of linear disjointness, the fields K and $m^{p^{-\infty }}$ are linearly disjoint over k. In view of the inclusion $k^{p^{-\infty }}\subseteq m{}^{p^{-\infty }}$ , it follows in particular that K and $k^{p^{-\infty }}$ are linearly disjoint over k, which is to say that K is separable over k.

Now assume conversely that K is separable over k; we want to prove that M is separable over m. By finite character of separable extensions and the discussion of 3.0.5, we may assume that K is finitely generated over k. So K is separably generated over k (i.e, there is a transcendence basis x of K over k such that K is separably algebraic over $k\left (x\right )$ ). It then follows immediately from the definition of linear disjointness that x is also a separating transcendence basis of M over m.

Primary extensions. First, assume that M is primary over m. Then M and $m^{{\mathrm {sep}}}$ are linearly disjoint over m. By transitivity of linear disjointness, the fields K and $m^{{\mathrm {sep}}}$ are linearly disjoint over k; hence, in particular, K and $k^{{\mathrm {sep}}}$ are linearly disjoint over k.

Now assume that K is primary over k; we show that M is primary over m. We will give a model theoretic argument. By finite character, we may assume that $K=k\left (\alpha \right )$ for a finite sequence $\alpha $ . Recall that the theory of algebraically closed fields eliminates imaginaries in the field language and that the definable closure of a subfield is the perfect hull. It is moreover a stable theory. So K is primary over k if and only if the type ${\mathrm {tp}}_{{\mathrm {ACF}}}\left ({\alpha }/{k}\right )$ admits a unique global nonforking extension. Now in the theory ${\mathrm {ACF}}$ , forking is characterized by a drop in the transcendence degree. Since K and m are linearly disjoint over k, a transcendence basis of K over k lifts to a transcendence basis of M over m. So as ${\mathrm {tp}}\left ({\alpha }/{m}\right )$ does not fork over k, we find that it admits a unique global nonforking extension too. So ${\mathrm {tp}}\left ({\alpha }/{m}\right )$ is stationary, whence M is primary over m.

Regular extensions. This follows from the fact that an extension is regular if and only if it is both primary and separable.

Relatively perfect extensions. We want to prove that ${K}/{k}$ is relatively perfect if and only if ${M}/{m}$ is so. Note that relatively perfect extensions are in particular separable; we have already shown invariance of separable extensions, so assume that one (and hence both) of these extensions is separable.

Let us assume first that k and m are both perfect. By definition, a relatively perfect extension of a perfect field is just a perfect extension. Thus, we must show that if one of K or M is perfect, then so is the other. If K is perfect, then M is perfect, since it is the tensor product of the perfect fields K and m. Conversely, assume that M is perfect. The field M is separable over K as it is obtained from the separable extension ${m}/{k}$ via base extension. Since no separable extension of an imperfect field can possibly be perfect, we find that K is perfect.

Next, consider the case where $m=k^{p^{-\infty }}$ is the perfect hull of k. If K is relatively perfect over k, then by definition, M is the perfect hull of K, and hence perfect; so it is a relatively perfect extension of m. Conversely, assume M is perfect. The field M is purely inseparably algebraic over K, since it is obtained from the purely inseparably algebraic extension ${m}/{k}$ via base extension. Since it is perfect, it is also the maximal purely inseparably algebraic extension of K (i.e, we have $M=K^{p^{-\infty }}$ , which means that ${K}/{k}$ is relatively perfect).

The general case follows from transitivity, combining the two special cases considered above together.

Fact 3.10. Let $K\subseteq M\subseteq L$ be a tower of fields and assume M is relatively perfect over K. Then L is separable over K if and only if it is separable over M.

Proof. Let us assume first that L is separable over M. We must prove that L is separable over K. Since ${M}/{K}$ is relatively perfect, it is in particular separable. The field extension ${L}/{K}$ is therefore separable as the composition of such.

Let us assume conversely that L is separable over K and that M is relatively perfect over K; we must prove that L is separable over M. Let $\overline {K}=K^{p^{-\infty }}$ be the perfect hull of K. Let us set $\overline {M}=M\otimes _{K}\overline {K}$ and likewise $\overline {L}$ . Since separable and relatively perfect extensions are both invariant under base extension (Fact 3.9), replacing everything by the overlined data, we may assume that K is perfect. But then it follows immediately from the definition of relatively perfect extensions that M is perfect. In this situation, the field L is automatically separable over M, as wanted.

Proof. This follows from formal properties of linear disjointness. Let us prove, for example, that the class of transformally separable extensions descends along base extension; we use the notation of (4).

Let us assume first that ${L_{0}}/{K_{0}}$ is transformally separable. We will show that ${L}/{K}$ is transformally separable. The assumption that ${L_{0}}/{K_{0}}$ is transformally separable means that $L_{0}$ and $K_{0}^{\sigma ^{-\infty }}$ are linearly disjoint over $K_{0}$ . By transitivity of linear disjointness, the fields L and $K_{0}^{\sigma ^{-\infty }}$ are linearly disjoint over K. In view of the embedding $K^{\sigma ^{-\infty }}\subseteq K_{0}^{\sigma ^{-\infty }}$ , it follows in particular that L and $K^{\sigma ^{-\infty }}$ are linearly disjoint over K, which is to say that ${L}/{K}$ is transformally separable, as wanted.

Let us assume conversely that ${L}/{K}$ is transformally separable; we must show that $L_0$ is transformally separable over $K_0$ . By transitivity of linear disjointness and Remark 4.2, it is enough to prove that L is linearly disjoint from $(K_0)^{\sigma ^{-1}}$ over K.

So let $x_1,\dots ,x_n\in L$ be linearly dependent over $(K_0)^{{\sigma }^{-1}}$ , that is, there are $m_i^{\sigma ^{-1}}\in (K_0)^{\sigma ^{-1}}$ , not all zero, such that the equality $\sum m_i^{\sigma ^{-1}}x_i=0$ holds. We must show that the $x_i$ are linearly dependent over K already. Applying $\sigma $ , we get that $\sum m_ix_i^{\sigma }=0$ . By linear disjointness of L from $K_0$ over K, there are $k_i\in K$ with $\sum k_ix_i^{\sigma }=0$ , and hence, $\sum k_i^{\sigma ^{-1}}x_i=0$ . So the $x_i$ are linearly dependent over $K^{\sigma ^{-1}}$ ; since L is transformally separable over K, they are linearly dependent over K already, and hence in particular over $K_0$ .

Proof. (1) Let $\widetilde {K}$ be the splitting field of the polynomials of the form $X^{p^{n}}-x^{\sigma }$ over K, for $x\in K$ and $n\in \mathbf {N}$ varying. Then $\widetilde {K}$ is a normal purely inseparably algebraic extension of K. By construction, if the polynomial $f\in K\left [x\right ]$ splits in $\widetilde {K}$ , then the same is true of the polynomial obtained by applying $\sigma $ to the coefficients. By uniqueness of splitting fields, the endomorphism $\sigma $ of K lifts to $\widetilde {K}$ . Since $\widetilde {K}$ is purely inseparably algebraic over K, the lifting is unique. We claim that $\widetilde {K}$ is closed under twists, and that it embeds uniquely over K in every other extension with this property.

First, we prove that $\widetilde {K}$ is closed under twists. Fix an element $a\in \widetilde {K}$ ; we must show that $a^{\sigma }$ has all p-th power roots in $\widetilde {K}$ . It is enough to prove this after replacing a by $a^{p^{n}}$ for $n\gg 0$ . Since $\widetilde {K}$ is purely inseparably algebraic over K, we may therefore assume without loss of generality that the element a lies in K. But then it follows from the construction that the polynomials $X^{p^{n}}-a^{\sigma }$ all split in $\widetilde {K}$ .

Now let L be an extension of K closed under twists. By Remark 4.7, for every $x\in K$ and every $n\in \mathbf {N}$ , the polynomial $X^{p^{n}}-x^{\sigma }$ splits in L. By uniqueness of splitting fields, there is an abstract field embedding of $\widetilde {K}$ in L over K; since $\widetilde {K}$ is purely inseparably algebraic over K, this embedding is unique and compatible with the action of $\sigma $ .

(2) The lift of $\sigma $ is unique if it exists, as L is purely inseparably algebraic over K. Moreover, by assumption, we have the series of inclusions $K\subseteq L\subseteq K^{p^{-\infty }}\subseteq K^{\sigma ^{-1}}$ ; if $a\in L$ , then $a^{\sigma }$ lies in K, and hence in L. So L is a difference field. Moreover, we have the inclusions $L\subseteq L^{p^{-\infty }}=K^{p^{-\infty }}\subseteq K^{\sigma ^{-1}}\subseteq L^{\sigma ^{-1}}$ , so L is closed under twists.

(3) Stability under directed unions is clear. Next, assume that L is relatively perfect over K and that K is closed under twists; we want to prove that L is closed under twists as well. By assumption, we have $L^{\sigma \cdot p^{-\infty }}=L^{\sigma }\otimes _{K^{\sigma }}K^{\sigma \cdot p^{-\infty }}$ ; so the inclusion $K^{\sigma \cdot p^{-\infty }}\subseteq K$ gives an inclusion $L^{\sigma \cdot p^{-\infty }}\subseteq L$ . Finally, assume that L is algebraic over K. There is a canonical factorization $K\subseteq M\subseteq L$ where ${M}/{K}$ is separably algebraic and ${L}/{M}$ purely inseparable; in view of (2), we may assume that L is separably algebraic over K. In this situation, the field L is relatively perfect over K; this case has already been dealt with, so we are done.

Proof. We may assume that K is closed under twists, by Proposition 4.10.

Let $K^{\sigma ^{-\infty }}$ denote the inversive hull of K, and let E be the separable algebraic closure of K inside. Then E is a difference field: if $a\in K^{\sigma ^{-\infty }}$ is separably algebraic over K, then $a^{\sigma }$ is separably algebraic over $K^{\sigma }$ , and hence over K. We claim that E is advertised; for this, we must prove E is a model of ${\mathrm {FE}}$ which embeds uniquely over K in any other.

First, we check that E is a model of the theory ${\mathrm {FE}}$ . By construction, the field E is separably algebraically closed in $K^{\sigma ^{-\infty }}$ ; however, we have $E^{\sigma ^{-\infty }}=K^{\sigma ^{-\infty }}$ , so E is separably algebraically closed in its inversive hull. Since E is an algebraic extension of K, which is closed under twists, it is again closed under twists, using Proposition 4.10 again. It follows that E is a model of the theory ${\mathrm {FE}}$ .

To finish, we must prove that E embeds uniquely over K in every model L of ${\mathrm {FE}}$ over K. So fix such a model L. The embedding of K in L lifts uniquely to an embedding of $K^{\sigma ^{-\infty }}$ in $L^{\sigma ^{-\infty }}$ . Since the field L is a model of ${\mathrm {FE}}$ , it is separably algebraically closed in its inversive hull; so the restriction of this embedding to E factors through L, by definition of the relative separable algebraic closure. So E embeds uniquely over K in L, and the proof is finished.

Proof. The following elegant proof is due to the anonymous referee. By a limit argument, we may assume that M is finitely generated over K; since M is transformally algebraic over K, the transcendence degree ${\mathrm {tdeg}}_{K}{M}$ is finite. Since K is inversive, we have the inclusions $K \subseteq M^{\sigma } \subseteq M$ ; the claim follows by comparing transcendence degrees in the displayed tower.

Proof. Let ${L}/{K}$ be transformally separable with K closed under twists. Then $K^{\sigma ^{-1}}$ and L are linearly disjoint over K. Since K is closed under twists, we have the inclusions $K\subseteq K^{p^{-1}}\subseteq K^{\sigma ^{-1}}$ ; by formal properties of linear disjointness, we find that $K^{p^{-1}}$ and L are linearly disjoint over K, whence ${L}/{K}$ is algebraically separable.

Proof. Everything here is invariant under base extension (see Fact 3.9 and Proposition 4.5), so assume that K is inversive (Remark 4.8). By Lemma 4.29, the field L is algebraic over $L^{\sigma }$ , and hence in particular algebraic over $L^{\sigma \cdot p^{-\infty }}$ . But algebraic extensions of perfect fields are perfect, so L is perfect, as wanted.

Proof. If K is a model of ${\mathrm {FE}}$ , then it is closed under twists; a transformally separable extension is then algebraically separable by Lemma 4.30. Moreover, by Remark 4.25, an extension which is transformally separable is almost transformally separable, no matter the assumption on the ground field. Thus, an extension of a model of ${\mathrm {FE}}$ which is transformally separable is algebraically separable and almost transformally separable; it remains to show the converse.

We first reduce this to the case where K is perfect, using the discussion of 3.0.4. By elementary properties of transformally separable and almost transformally separable extensions (Proposition 4.28 and Proposition 4.5), the hypothesis and the conclusion are both invariant under base extension. Furthermore, by Lemma 4.16, a purely inseparably algebraic extension of a model of ${\mathrm {FE}}$ is a model of ${\mathrm {FE}}$ . We may therefore assume without loss of generality that K is perfect.

It remains to be shown that if K is a perfect model of ${\mathrm {FE}}$ , then an extension is transformally separable if and only if it is almost transformally separable. Since K is a model of ${\mathrm {FE}}$ , the field $K^{\sigma ^{-1}}$ is a primary extension of K; our assumption that K is perfect implies that $K^{\sigma ^{-1}}$ is in fact regular over K. By Fact 4.23, linear disjointness form $K^{\sigma ^{-1}}$ over K is equivalent to algebraic disjointness, which is precisely what we needed.

Proof. It is enough to prove that ${L}/{M}$ is almost transformally separable; the other directions follow formally from Proposition 4.28.

Let us assume first that K is inversive. Since M is transformally algebraic over K, it follows from Lemma 4.29 that the field M is an algebraic extension of $M^{\sigma }$ . So L is automatically almost transformally separable over M, as wanted.

It remains to reduce to the case where K is inversive. Let $L^{{\mathrm {alg}}}$ be an algebraic closure of L in the category of difference fields; replacing L by $L^{{\mathrm {alg}}}$ and $M,K$ by the algebraic closures inside $L^{{\mathrm {alg}}}$ does not affect the statement and the conclusion descends, using Remark 4.27; thus, we may assume that K is algebraically closed, and hence that ${L}/{K}$ is transformally separable. Let $K_{0}$ be the inversive hull of K. Since ${L}/{K}$ is transformally separable, the difference field L is linearly disjoint from $K_{0}$ over K.

Let us set $M_{0}=M{\mathrm {\otimes }}_{K}K_{0}$ and $L_{0}=L{\mathrm {\otimes }}_{K}K_{0}$ . By Proposition 4.28, the class of almost transformally separable extensions is invariant under base extensions; one checks, moreover, that the same is true of the class of transformally algebraic extensions. It is therefore harmless to replace $K,M$ and L by $K_{0},M_{0}$ and $L_{0}$ , respectively, which brings us back to the situation already dealt with.

Proof. (1) This is a special case of Lemma 4.31, since models of ${\mathrm {FE}}$ are in particular closed under twists.

(2) Assume that K is inversive; we show that L is inversive, or equivalently that the inclusion $L^\sigma \subseteq L$ is an equality. For this, it is enough to show that L is transformally separable over $L^\sigma $ . By the criterion of Proposition 4.32, in order to prove that L is transformally separable over $L^{\sigma }$ , it will be sufficient to prove that L is algebraically separable and almost transformally separable over $L^{\sigma }$ .

First, we show that L is algebraically separable over $L^{\sigma }$ . By Remark 4.8, inversive models of ${\mathrm {FE}}$ are perfect; so using (1), we find that $L^{\sigma }$ is perfect, as the isomorphic image of the perfect field L. So L is automatically algebraically separable over $L^{\sigma }$ . Since K is inversive, the extension L of K is almost transformally separably algebraic; this class is transitive in towers, using Proposition 4.33, so L is almost transformally separable over $L^{\sigma }$ . This concludes the proof.

Proof. By formal properties of transformally separable extensions as in Proposition 4.5, it will be sufficient to prove the following: if ${L}/{K}$ is a transformally separable extension of models of ${\mathrm {FE}}$ , and $K\subseteq M\subseteq L$ is a model of ${\mathrm {FE}}$ transformally separably algebraic over K, then ${L}/{M}$ is transformally separable. Using the criterion of Proposition 4.32, it is enough to prove that ${L}/{M}$ is almost transformally separable and algebraically separable.

The fact that ${L}/{M}$ is almost transformally separable follows from Proposition 4.33; so we must show that ${L}/{M}$ is algebraically separable. By Proposition 4.34, the perfect hull of M splits over K, so the claim follows from Fact 3.10.

Proof. $(1)\implies (2)$ We may assume here that a is a single element. By assumption, the element a is transformally algebraic over K, so for some $n\in \mathbf {N}$ , the sequence $a,\ldots ,a^{\sigma ^{n}}$ is algebraically dependent over K; let n be chosen minimal with respect to this property. Then by minimality of n the sequence, $a,\ldots ,a^{\sigma ^{n-1}}$ is algebraically independent over K. Since L is almost transformally separable over K, the sequence $a^{\sigma },\ldots ,a^{\sigma ^{n}}$ must then likewise be algebraically independent over K; by Steinitz exchange, the element a is algebraic over $K\left (a^{\sigma },\ldots ,a^{\sigma ^{n}}\right )$ , and hence in particular algebraic over $K\left (a^{\sigma \cdot \mathbf {N}\left [\sigma \right ]}\right )$ .

$(2)\implies (1)$ Since the tuple a is finite, there is some large natural number N with $a\in K\left (a^{\sigma },\ldots ,a^{\sigma ^{N}}\right )^{{\mathrm {alg}}}$ . Since $\sigma $ is an endomorphism of fields, it follows inductively that for all $n\in \mathbf {N}$ , we have $a^{\sigma ^{n}}\in K^{\sigma ^{n}}\left (a^{\sigma ^{n+1}},\ldots ,a^{\sigma ^{n+N}}\right )^{{\mathrm {alg}}}$ , and so in particular that

$$\begin{align*}a^{\sigma^{n}}\in K\left(a^{\sigma^{n+1}},\ldots,a^{\sigma^{n+N}}\right)^{\operatorname{\mathrm{alg}}}.\end{align*}$$

Enlarging a by the tuple $a\ldots a^{\sigma ^{N-1}}$ does not affect the statement; using the analysis of the previous paragraph, we then have that $a\in K\left (a^{\sigma }\right )^{{\mathrm {alg}}}$ and inductively that $a,\ldots , a^{\sigma ^{n-1}}\in K\left (a^{\sigma ^{n}}\right )^{{\mathrm {alg}}}$ for all $n\in \mathbf {N}$ . So the transcendence degree of $K\left (a^{\mathbf {N}\left [\sigma \right ]}\right )$ over K is finite; namely, it is bounded by the length of the tuple a. This means that a is transformally algebraic over K.

It remains to be shown that ${L}/{K}$ is almost transformally separable: in other words, that the fields K and $L^{\sigma }$ are algebraically disjoint over $K^{\sigma }$ . By assumption, the field L is algebraic over $KL^{\sigma }$ . Moreover, we have shown that L is transformally algebraic over K, so the transcendence degree ${\mathrm {tdeg}}_{K}L$ is finite. We then have

$$\begin{align*}\operatorname{\mathrm{tdeg}}_{K}L=\operatorname{\mathrm{tdeg}}_{K}KL^{\sigma}\leq\operatorname{\mathrm{tdeg}}_{K^{\sigma}}L^{\sigma}. \end{align*}$$

But $\sigma $ is a field endomorphism, so ${\mathrm {tdeg}}_{K^{\sigma }}L^{\sigma }={\mathrm {tdeg}}_{K}L$ ; the displayed inequalities must therefore be equalities, so $L^{\sigma }$ and K are algebraically disjoint over $K^{\sigma }$ .

Proof. $(1)\implies (2)$ Fix an element $a\in L$ ; we must show that it is a root of a simple twisted difference polynomial over K. Since L is transformally algebraic over K, there is a finite nonempty set $I\subseteq \mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]$ and nonzero elements $c_{\nu }\in K$ such that $\sum c_{\nu }a^{\nu }=0$ ; indeed, this is precisely what it means for L to be transformally algebraic over K. If $fx=\sum c_{\nu }x^{\nu }$ , then $f'$ is nonzero precisely in the event that some $\nu \in I$ has constant term prime to p; let us show how to arrange this.

First, assume that every element $\nu \in I$ has constant term zero. Then $\sum c_{\nu }^{\frac {1}{\sigma }}a^{\frac {\nu }{\sigma }}=0$ , so the elements $a^{\frac {\nu }{\sigma }}$ are linearly dependent over $K^{\sigma ^{-1}}$ ; since L is transformally separable over K, they must have been linearly dependent over K already. Thus, inductively, we can assume that some $\nu \in I$ has a nonzero constant term. Now assume that the constant term of every $\nu \in I$ is divisible by p; then the elements $a^{\frac {\nu }{p}}$ are linearly dependent over $K^{p^{-1}}$ . The latter is contained in $K^{\sigma ^{-1}}$ as K is a model of ${\mathrm {FE}}$ , and by assumption, L is transformally separable over K; thus we may assume that at least one $\nu \in I$ has constant term prime to p.

We have found a nonzero twisted difference polynomial f such that $f'\neq 0$ and $fa=0$ . If $f'a\neq 0$ , then we are done. Otherwise, replace f by $f'$ and argue inductively; this cannot go on forever.

$(2) \implies (3)$ This is clear.

$(3) \implies (4)$ Let us assume that L is generated as a twisted difference field by simple roots of twisted difference polynomials over K. Then L is the directed union of difference fields $K\left (a^{\mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]}\right )$ , where $a=\left (a_{1},\ldots ,a_{n}\right )$ is a finite tuple and each $a_{i}$ is a simple root of a twisted difference polynomials over K. Then each $a_{i}$ is separably algebraic over $K\left (a_{i}^{\sigma \cdot \mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]}\right )$ by Remark 4.39; it follows in particular that a is separably algebraic over $K\left (a^{\sigma \cdot \mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]}\right )$ , hence the claim.

$(4)\implies (1)$ Since transformally separably algebraic extensions are of finite character, we may assume that $L=K\left (a^{\mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]}\right )$ , where a is a finite tuple separably algebraic over $K\left (a^{\sigma \cdot \mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]}\right )$ . We have already seen in Lemma 4.40 that L is then almost transformally separable over K. By Proposition 4.32, it is enough to prove that ${L}/{K}$ is algebraically separable. We will show a stronger statement – namely, that L is relatively perfect over K.

By Proposition 4.32 and Lemma 4.40, the extension $L^{\sigma \cdot p^{-\infty }}$ of $K^{\sigma \cdot p^{-\infty }}$ is transformally separable. Let

$$\begin{align*}E=K\operatorname{\mathrm{\otimes}}_{K^{\sigma\cdot p^{-\infty}}}L^{\sigma\cdot p^{-\infty}}\end{align*}$$

be the displayed difference field. The field extension $L^{\sigma \cdot p^{-\infty }}$ of $K^{\sigma \cdot p^{-\infty }}$ is relatively perfect; by descent, the field E is relatively perfect over K. Since L is separably algebraic over E, it follows from elementary properties of relatively perfect extensions that L is relatively perfect over K, which is what we wanted.

Proof. The truth of (1) and (2) follows immediately from Proposition 4.41. Namely, let $\mathcal {X}$ be the set of elements of E which are simple roots of twisted difference polynomials over F. The smallest model $\widetilde {F}$ of ${\mathrm {FE}}$ inside E containing $\mathcal {X}$ is transformally separably algebraic over F; however, all elements of $\widetilde {F}$ are simple roots over F, so $\mathcal {X}=\widetilde {F}$ .

The truth of (3) follows from Proposition 4.36. In (4), the equivalence is given by associating to a model E of ${\mathrm {FE}}$ transformally separably algebraic over F its inversive hull $E^{\sigma ^{-\infty }}$ . Indeed, the inversive hull of a model of ${\mathrm {FE}}$ is again one, and by Proposition 4.34, all models of ${\mathrm {FE}}$ transformally algebraic over $F^{\sigma ^{-\infty }}$ are inversive. We can recover E inside $E^{\sigma ^{-\infty }}$ as the relative transformal separable algebraic closure of F inside, since $E^{\sigma ^{-\infty }}$ is by construction purely transformally inseparably algebraic over E. The proof of (5) is given in Proposition 4.46, and finally, (6) is Proposition 4.20.

Fact 5.7.

1. (1) The class of models of ${\mathrm {VFA}}$ admits a model companion $\widetilde {{\mathrm {VFA}}}$ .

2. (2) In $\widetilde {{\mathrm {VFA}}}$ , the residue field and the value group are stably embedded and fully orthogonal; the induced structure is that of a model of ${\mathrm {ACFA}}$ and a pure model of $\widetilde {\omega {\mathrm {OGA}}}$ , respectively.

3. (3) Let K be a model of ${\mathrm {VFA}}$ which is algebraically Henselian and assume that K has no nontrivial finite $\sigma $ -invariant Galois extensions – that is, no difference field extensions which are finite and Galois as abstract field extensions. Then the theory of models of $\widetilde {{\mathrm {VFA}}}$ over K is complete, and in particular, K admits a unique algebraic closure in the category of models of ${\mathrm {VFA}}$ , up to a (generally non unique) isomorphism. The converse holds if the various powers $\sigma ^n$ of $\sigma $ for $0 < n \in \mathbf {N}$ are considered (that is the lack of such extensions is equivalent to the uniqueness of the lifts of the maps $\sigma ^n$ for all n).

4. (4) The theory $\widetilde {{\mathrm {VFA}}}$ is precisely the asymptotic theory of algebraically closed and nontrivially valued Frobenius transformal valued fields – that is, the set of sentences true over all algebraically closed and nontrivially valued Frobenius transformal valued fields outside a finite set of exceptional prime powers.

5. (5) Let K be a model of ${\mathrm {VFA}}$ which is algebraically closed and transformally Henselian (see Definition 5.14). Then the category of models of ${\mathrm {VFA}}$ admits linearly disjoint amalgamation over K; that is, if L and M are models of ${\mathrm {VFA}}$ over K, then there is a model N of ${\mathrm {VFA}}$ over K in which L and M are jointly embedded over K and linearly disjoint over K. Moreover, the residue fields in the amalgamation can be taken linearly disjoint, and the transformal divisible hulls of the value groups rationally independent.

6. (6) Let K be an algebraically closed model of ${\mathrm {VFA}}$ and let $L=K\left (x^{\mathbf {N}[\sigma ]}\right )$ . Assume the residue class of x is transformally transcendental over k or else the valuation of x is not in $\Gamma \otimes \mathbf {Q}\left (\sigma \right )$ . Then the inversive, algebraically Henselian hull of L has no nontrivial finite $\sigma $ -invariant Galois extension. Moreover, the quantifier free type of x is determined in the first case by the fact that the residue class is transformally transcendental, and in the second case by the cut of $vx$ over $\Gamma \otimes \mathbf {Q}\left (\sigma \right )$ .

7. (7) Let K be a model of $\widetilde {{\mathrm {VFA}}}$ . Then K is transformally algebraically maximal; that is, if L is a model of ${\mathrm {VFA}}$ transformally algebraic over K with the same value group and residue field, then $K = L$ . Furthermore, there is an elementary immediate extension of K which is spherically complete.

8. (8) The theory $\widetilde {{\mathrm {VFA}}}$ admits the amalgamation property over an algebraically closed base. Thus, if K is a model of ${\mathrm {VFA}}$ , then completions of the theory $\widetilde {{\mathrm {VFA}}}_{K}$ of models of $\widetilde {{\mathrm {VFA}}}$ over K are determined by an isomorphism type of an extension of the transformal valued field structure to an algebraic closure of K and $\widetilde {{\mathrm {VFA}}}$ eliminates quantifiers in the language of transformal valued fields provided that formulas ranging over the field theoretic algebraic closure are permitted.

Proof. It is clear that k is closed under twists; for this, we do not need K to be algebraically Henselian. Furthermore, the embedding $K^{\sigma }\hookrightarrow K$ is a primary extension of Henselian valued fields; the induced extension $k^{\sigma }\hookrightarrow k$ of residue fields must therefore also be primary, by Hensel lifting.

Proof. Let $fx\in \mathcal {O}_{K}\left [x^{\mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]}\right ]$ be a twisted difference polynomial and let $a\in \mathcal {O}$ be such that $vfa>0$ and $vf'a=0$ ; we must find $b\in \mathcal {O}_{K}$ with $fb=0$ and whose residue class coincides with that of a. Since L is transformally Henselian, such b can be found in $\mathcal {O}_{L}$ . By assumption, however, the field K is transformally separably algebraically closed in L; since b is a simple root of a twisted difference polynomial with coefficients in K, the element b lies in K by Theorem 4.48(2). It follows that K is transformally Henselian.

We have seen that K is transformally Henselian; we now show that k is transformally separably algebraically closed in l. Fix an element $\alpha \in l$ which is transformally separably algebraic over k; then $g\alpha =0$ and $g'\alpha \neq 0$ for some twisted difference polynomial $gx\in k\left [x^{\mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]}\right ]$ . Let $fx\in \mathcal {O}_{K}\left [x^{\mathbf {N}\left [\frac {\sigma }{p^{\infty }}\right ]}\right ]$ be a twisted difference polynomial lifting $gx$ . Since L is transformally Henselian, there is an element $a\in \mathcal {O}_{L}$ with residue class equal to $\alpha $ such that $fa=0$ and $f'a\neq 0$ . Since K is transformally separably algebraically closed in L, the element a lies in K; hence, its residue class lies in k, as wanted.

Proof. It is enough to prove this when L is the inversive hull of K. First, assume that K is transformally Henselian; then L is the directed union of copies of K and hence transformally Henselian as the directed union of transformally Henselian models of ${\mathrm {VFE}}$ . The converse follows from Proposition 5.16.

Proof. Let us assume first that F is inversive. In this situation, an algebraically closed, transformally Henselian hull of F is unique up to isomorphism, using [Reference Dor and Hrushovski11, Corollary 5.4]. Thus, F admits an inversive, algebraically closed, transformally Henselian hull E in the category of models of ${\mathrm {VFE}}$ which is unique up to isomorphism. Using Lemma 5.18, the relative transformal separable algebraic closure of F in E is a separably algebraically closed, transformally Henselian hull of F.