Proof (1) Let F and G be series that are [polynomial–exponential, polynomial, exponential] on simple linear subsets. It is clear that for every monomial u and every scalar $\lambda \in K$ , also $\lambda uF$ is of the same form. Thus, it suffices to show that the same is true for $F+G$ . If $\mathcal {S}$ and $\mathcal {T} \subseteq \mathbb {N}^d$ are simple linear subsets, then the intersection $\mathcal {S} \cap \mathcal {T}$ is semilinear. By Proposition 2.3, the intersection $\mathcal {S} \cap \mathcal {T}$ can be represented as a finite disjoint union of simple linear sets. Thus, in representations of the coefficient sequences of F and G as in Definition 3.1, we may assume that the simple linear sets coincide for the two series, by passing to a common refinement. Then the claim about $F+G$ is immediate.

(2) This can again be computed on each simple linear set. Alternatively, it will follow from Theorem 3.10 and Corollary 3.13.

Proof (a) $\,\Leftrightarrow \,$ (b) Clear.

(a) $\,\Leftrightarrow \,$ (c) Consider the family $({{\boldsymbol x}}^{\boldsymbol e_1 c_1 + \cdots + \boldsymbol e_n c_n})_{c_1,\ldots ,c_n \in \mathbb {N}}$ of monomials. The monomials in this family are pairwise distinct, and hence linearly independent over K, if and only if $\boldsymbol e_1$ , $\ldots \,$ , $\boldsymbol e_n$ are linearly independent. However, the linear independence of $({{\boldsymbol x}}^{\boldsymbol e_1 c_1 + \cdots + \boldsymbol e_n c_n})_{c_1,\ldots ,c_n \in \mathbb {N}}$ is equivalent to the algebraic independence of ${\boldsymbol x}^{\boldsymbol e_1}, \ldots , {\boldsymbol x}^{\boldsymbol e_n}$ .

(c) $\,\Leftrightarrow \,$ (d) If $0 \ne P \in K[y_1,\ldots ,y_n]$ , then P vanishes on $(1 - c_1{\boldsymbol x}^{\boldsymbol e_1}, \ldots , 1-c_n{\boldsymbol x}^{\boldsymbol e_n})$ if and only if $P(1 - c_1 y_1, \ldots , 1 - c_n y_n)$ vanishes on $({\boldsymbol x}^{\boldsymbol e_1}, \ldots , {\boldsymbol x}^{\boldsymbol e_n})$ . Hence, ${\boldsymbol x}^{\boldsymbol e_1}$ , $\ldots \,$ , ${\boldsymbol x}^{\boldsymbol e_n}$ are algebraically independent if and only if the polynomials $1 - c_1 {\boldsymbol x}^{\boldsymbol e_1}$ , $\ldots \,$ , $1 - c_n {\boldsymbol x}^{\boldsymbol e_n}$ are algebraically independent.

Now, for any choice of polynomials $f_1$ , $\ldots \,$ , $f_n$ , the field $K(f_1,\ldots ,f_n)$ is an algebraic extension of $K(f_1^{m_1},\ldots ,f_n^{m_n})$ , and therefore the two fields have the same transcendence degree over K. Thus, $1 - c_1 {\boldsymbol x}^{\boldsymbol e_1}$ , $\ldots \,$ , $1 - c_n {\boldsymbol x}^{\boldsymbol e_n}$ is algebraically independent if and only if $(1 - c_1 {\boldsymbol x}^{\boldsymbol e_1})^{m_1}$ , $\ldots \,$ , $(1 - c_n {\boldsymbol x}^{\boldsymbol e_n})^{m_n}$ is algebraically independent.

Proof (a) $\,\Rightarrow \,$ (b). By straightforward substitution.

(b) $\,\Rightarrow \,$ (a). Let $\boldsymbol n=(n_1,\ldots ,n_d) \in \mathcal {S}$ . Since $\boldsymbol b_1$ , $\ldots \,$ , $\boldsymbol b_s$ are linearly independent, there exist unique $m_1$ , $\ldots \,$ , $m_s \in \mathbb {N}^d$ with $\boldsymbol n = \boldsymbol a + \boldsymbol b_1 m_1 + \cdots + \boldsymbol b_s m_s$ . Solving this linear system over $\mathbb {Q}$ , there exists $N \in \mathbb {N}_{\ge 1}$ and, for every $i \in [1,s]$ and $j \in [1,d]$ integers $p_i$ , $q_{i,j}$ such that

$$\begin{align*}m_i = p_i/N + \sum_{j=1}^d n_j q_{i,j}/N \end{align*}$$

for all $\boldsymbol n=(n_1,\ldots ,n_d) \in \mathcal {S}$ and $\boldsymbol m=(m_1,\ldots ,m_s) \in \mathbb {N}^s$ with $\boldsymbol n = \boldsymbol a + \boldsymbol b_1 m_1 + \cdots + \boldsymbol b_s m_s$ .

Let $\nu \in [1,l]$ . Suppose ${\boldsymbol \beta }_\nu =(\beta _{\nu ,1},\ldots ,\beta _{\nu ,s})$ and pick $\gamma _{\nu ,i} \in K^*$ with $\gamma _{\nu ,i}^N = \beta _{\nu ,i}$ for $i \in [1,s]$ . Set ${\boldsymbol \alpha }_\nu =(\alpha _{\nu ,1}, \ldots \, \alpha _{\nu ,d})$ with

and

If $\boldsymbol n = \boldsymbol a + \boldsymbol b_1 m_1 + \cdots + \boldsymbol b_s m_s$ , then

$$\begin{align*}\begin{aligned} A_\nu(\boldsymbol n) {\boldsymbol\alpha}_\nu^{\boldsymbol n} &= B_\nu(\boldsymbol m) \prod_{i=1}^s \gamma_{\nu,i}^{ p_i} \prod_{j=1}^d \alpha_{\nu,j}^{n_j} = B_\nu(\boldsymbol m) \prod_{i=1}^s \gamma_{\nu,i}^{p_i} \cdot \prod_{j=1}^d \prod_{i=1}^s \gamma_{\nu,i}^{q_{i,j}n_j} \\ & = B_\nu(\boldsymbol m) \prod_{i=1}^s \Big( \gamma_{\nu,i}^{ p_i} \prod_{j=1}^d \gamma_{\nu,i}^{q_{i,j}n_j} \Big) = B_\nu(\boldsymbol m) \prod_{i=1}^s \gamma_{\nu,i}^{m_i N} \\ & = B_\nu(\boldsymbol m) \prod_{i=1}^s \beta_{\nu,i}^{m_i} = B_\nu(\boldsymbol m) \boldsymbol\beta_\nu^{\boldsymbol m}. \end{aligned}\\[-36pt] \end{align*}$$

Proof It suffices to consider $\mathcal {S}=\mathbb {N}^d$ . Let $\Lambda \subseteq K^*$ be a finite set and $e \in \mathbb {N}$ . Let be the K-vector space of rational series spanned by rational functions of the form

$$\begin{align*}\frac{1}{(1-\lambda_1x_1)^{k_1}\cdots (1-\lambda_dx_d)^{k_d}} \end{align*}$$

with $\lambda _1$ , $\ldots \,$ , $\lambda _d \in \Lambda $ and $k_1$ , $\ldots \,$ , $k_d \in [1,e]$ . Then $\dim A_e = \left \lvert {\Lambda } \right \rvert ^d e^d$ .

Now, in the group algebra $K[\boldsymbol x][ \langle \Lambda \rangle ^d]$ , let $V_e$ be the K-vector space consisting of elements of the form

$$\begin{align*}\sum_{j=1}^l C_j(\boldsymbol x)\boldsymbol\gamma_j, \end{align*}$$

with $\boldsymbol \gamma _j \in \Lambda ^d$ and polynomials $C_j \in K[\boldsymbol x]$ with $\deg _{x_i}(C_j) \le e-1$ for all $i \in [1,d]$ and $j \in [1,l]$ . Then $\dim V_{\boldsymbol e} \le \left \lvert {\Lambda ^d} \right \rvert e^d$ .

For $i \in [1,d]$ let $\lambda _i \in \Lambda $ and $k_i \in [1,e]$ . Since

$$\begin{align*}\prod_{i=1}^d (1 - \lambda_i x_i)^{-k_i} = \sum_{n_1,\ldots,n_d=0}^\infty {{k_1+n_1-1}\choose{n_1} } \cdots {{k_d+n_d-1}\choose{n_d} }\lambda_1^{n_1} \cdots \lambda_d^{n_d} x_1^{n_1}\cdots x_d^{n_d}, \end{align*}$$

the vector space $V_e$ maps surjectively onto $A_e$ . Since $\dim V_e \le \dim A_e$ , this is a bijection. This implies the claim.

Proof (a) $\Rightarrow $ (b) Lemma 2.7 allows us to assume that the polynomials $1-c_{i}u_{i}$ for $i \in [1,l]$ are all irreducible. Gathering associated polynomials, and resetting the notation, we may write

$$\begin{align*}F = \frac{P}{(1-c_1u_1)^{e_1} \cdots (1-c_lu_l)^{e_l}}, \end{align*}$$

with irreducible polynomials $1 - c_{i} u_{i}$ that are pairwise coprime and $e_1$ , $\ldots \,$ , $e_l \ge 1$ .

By Proposition 3.9, we may now express

$$\begin{align*}F = \sum_{I \subseteq [1,l]} \frac{P_I}{Q_I}, \end{align*}$$

with $P_I \in K[{\boldsymbol x}]$ and $Q_I = \prod _{i \in I} (1-c_iu_i)^{b_{I,i}}$ where $b_{I,i} \ge 1$ for all $I \subseteq [1,l]$ and $i \in I$ , and the different irreducible factors of $Q_I$ are algebraically independent. By Lemma 3.4, the monomials $\{\, u_i : i \in I \,\}$ are algebraically independent for $I \subseteq [1,l]$ . We can therefore apply Corollary 3.8 followed by (b) $\,\Rightarrow \,$ (a) of Lemma 3.5, to deduce that every $1/Q_I$ is piecewise polynomial–exponential on simple linear subsets of $\mathbb {N}^d$ . By (1) of Lemma 3.3, the same is true for F.

(b) $\Rightarrow $ (a) By assumption, there exists a partition of $\mathbb {N}^d$ into simple linear sets such that on each simple linear set $\mathcal {S}$ , the coefficients of F can be represented as in (a) of Lemma 3.5. Applying (a) $\,\Rightarrow \,$ (b) of Lemma 3.5, followed by Corollary 3.8, we see that F is a K-linear combination of rational functions of the form

$$\begin{align*}\frac{u_0}{(1-c_1u_1)\cdots (1-c_su_s)} \end{align*}$$

with $c_1$ , $\ldots \,$ , $c_s \in K^*$ and monomials $u_0$ , $u_1$ , $\ldots \,$ , $u_s \in K[{\boldsymbol x}]$ .

Proof Fix a well-order on $\mathbb {N}^d$ for which $(\mathbb {N}^d,+)$ is an ordered semigroup (for instance, a lexicographical order). We proceed by contradiction and assume that $\boldsymbol m =(m_1,\ldots ,m_d) \in \mathbb {N}_{\ge l_1} \times \cdots \times \mathbb {N}_{\ge l_d}$ is minimal with $f(\boldsymbol m)\ne 0$ . Then $m_i \ge l_i+k$ for all $i \in [1,d]$ . Let $\boldsymbol b \in [0,k]^d$ be the minimum, with respect to the well-order, with $Q_{j,\boldsymbol b} \ne 0$ . Taking $\boldsymbol n = \boldsymbol m + \boldsymbol b$ , we see that equation (4.1) allows us to express $f(\boldsymbol m)$ as a linear combination of certain $f(\boldsymbol m')$ with $\boldsymbol m' < \boldsymbol m$ and $\boldsymbol m' \in \mathbb {N}_{\ge l_1} \times \cdots \times \mathbb {N}_{\ge l_d}$ , showing $f(\boldsymbol m)=0$ , a contradiction.

Proof By assumption, G is a finitely generated group. Therefore, there exists a finitely generated $\mathbb {Z}$ -subalgebra $A \subseteq K$ containing G. By a theorem of Nagata, the integral closure $\overline {A}$ of A is a finitely generated A-module (see [Reference MatsumuraMat80, Theorem 31.H in Chapter 12]). Therefore, $\overline {A}$ is a finitely generated $\mathbb {Z}$ -algebra as well. A theorem of Roquette (see [Reference LangLan83, Corollary 7.5 of Chapter 2]) implies that the group of units $(\overline A)^\times $ is finitely generated. Since $\sqrt {G} \subseteq (\overline A)^{\times }$ , the claim follows.

Proof Let $G'$ be a finitely generated group containing G and all coefficients of the polynomials $Q_j(y+b)$ where $b \in B$ . We may moreover assume $\sqrt {G'}=G'$ . Enlarging G if necessary, we even assume $G'=G=\sqrt {G}$ for notational simplicity.

If $l \in K^*\setminus G$ , then in particular l is not a root of unity, and thus for every $b \in K$ , the set $\{\, l^e + b : e \in \mathbb {N} \,\}$ is infinite. Moreover, if l and b are fixed, then the exponent e is uniquely determined by $l^e +b$ .

We proceed by induction on $ \left \lvert {B} \right \rvert $ . For $B = \emptyset $ , there is nothing to show because then $C=\emptyset $ . So fix $b \in B$ such that $B=B' \uplus \{b\}$ with $ \left \lvert {B'} \right \rvert < \left \lvert {B} \right \rvert $ . Applying the induction hypothesis, there exists $e_0'\in \mathbb {N}$ such that (4.2) holds for all $\omega \in \Omega $ with $\pi (\omega ) \in C'$ , where

Let

For $\omega \in \Omega _{b}$ with $\pi (\omega ) = l^{e} + b$ , define

. For $j \in [1,n]$ , let

$$\begin{align*}Q_j(y+b) = \sum_{s=0}^m q_{j,s} y^s \qquad\text{with}\qquad q_{j,s} \in K.\\[-15pt] \end{align*}$$

Then

$$\begin{align*}0 = \sum_{j=1}^n Q_j(\pi(\omega)) f_j(\omega) = \sum_{j=1}^n \sum_{s=0}^m q_{j,s} l^{\varepsilon(\omega)s} f_j(\omega)\\[-15pt] \end{align*}$$

may be considered as a solution to the unit equation $X_1 + \cdots + X_{n(m+1)}=0$ over the group $\langle G,l\rangle $ . Let $I = [1,n] \times [0,m]$ . For every partition $\mathcal {P}=\{I_1,\ldots , I_p\}$ of I, let $\Omega _{\mathcal {P}} \subseteq \Omega _{b}$ be the set satisfying: for all $\nu \in [1,p]$ ,

• • $\sum _{(j,s) \in I_{\nu }} q_{j,s} l^{\varepsilon (\omega ) s} f_j(\omega ) = 0$ , and

• • $\sum _{(j,s) \in I} q_{j,s} l^{\varepsilon (\omega ) s} f_j(\omega ) \ne 0$ for all $\emptyset \ne I \subsetneq I_\nu $ .

Since the sets $\Omega _{\mathcal {P}}$ , with $\mathcal {P}$ ranging over all partitions, cover $\Omega _{b}$ , it is sufficient to establish the claim of the lemma for each $\Omega _{\mathcal {P}}$ separately. So fix $\mathcal {P}$ . If $\varepsilon (\Omega _{\mathcal {P}})$ is finite, the claim is trivially true by choosing $e_0 \ge e_0'$ sufficiently large. We may therefore assume that $\varepsilon (\Omega _{\mathcal {P}})$ is infinite.

The crucial step lies in showing that in the subsum indexed by $I_\nu $ , all the monomials that occur must be equal. Thus, explicitly, we show: if $(j_1,s_1)$ , $(j_2,s_2) \in I_\nu $ for some $\nu \in [1,p]$ , then $s_{1}=s_{2}$ . Clearly, we only have to consider the case where $ \left \lvert {I_\nu } \right \rvert \ge 2$ . Then $q_{j_1,s_1} l^{\varepsilon (\omega )s_1} f_{j_1}(\omega ) \ne 0$ and $q_{j_2,s_2} l^{\varepsilon (\omega ) s_2} f_{j_2}(\omega ) \ne 0$ for all $\omega \in \Omega _{\mathcal {P}}$ . By construction,

$$\begin{align*}\sum_{(j,s) \in I_{\nu}} q_{j,s} l^{\varepsilon(\omega) s} f_j(\omega) = 0\\[-15pt] \end{align*}$$

is a nondegenerate solution to the unit equation $X_1 + \cdots + X_{ \left \lvert {I_\nu } \right \rvert } = 0$ over $\langle G,l \rangle $ . Thus, there is a finite set Y with

$$\begin{align*}l^{\varepsilon(\omega) (s_1 - s_2)} \underbrace{q_{j_1,s_1} q_{j_2,s_2}^{-1} f_{j_1}(\omega) f_{j_2}(\omega)^{-1}}_{\in G} \in Y.\\[-15pt] \end{align*}$$

for all $\omega \in \Omega _{\mathcal {P}}$ . Then $l^{\varepsilon (\omega ) (s_1 - s_2)} \in \bigcup _{y \in Y} yG$ for all $\omega \in \Omega _{\mathcal {P}}$ .

Since $\varepsilon (\omega )$ takes infinitely many values, there exist $y \in Y$ and $\omega $ , $\omega ' \in \Omega _{\mathcal {P}}$ with $\varepsilon (\omega ) \ne \varepsilon (\omega ')$ , and

$$\begin{align*}l^{\varepsilon(\omega)(s_1 - s_2)},\ l^{\varepsilon(\omega')(s_1 - s_2)} \in yG.\\[-15pt] \end{align*}$$

But then, $l^{(\varepsilon (\omega ) - \varepsilon (\omega '))(s_1 - s_2)} \in G$ . By choice of l, this implies $s_{1}=s_{2}$ .

Taking the union over all $I_\nu $ containing a fixed $s \in [0,m]$ in the second coordinate, we conclude

$$\begin{align*}l^{\varepsilon(\omega) s} \sum_{j=1}^n q_{j,s} f_j(\omega) = 0,\\[-15pt] \end{align*}$$

and therefore $\sum _{j=1}^n q_{j,s} f_j(\omega )=0$ for all $s \in [0,m]$ . But then,

$$\begin{align*}0 = \sum_{j=1}^n \sum_{s=0}^m q_{j,s} {y}^{s} f_j(\omega) = \sum_{j=1}^n Q_j(y+b) f_j(\omega)\\[-15pt] \end{align*}$$

for all $\omega \in \Omega _{\mathcal {P}}$ . Substituting $y=-b$ into this polynomial, we obtain

$$\begin{align*}\sum_{j=1}^n Q_j(0) f_j(\omega) = 0 \qquad\text{for all}\qquad{\omega \in \Omega_{\mathcal{P}}}.\\[-40pt] \end{align*}$$

Proof Working in a subfield, we may assume that K is a finitely generated field. Let $P = \sum _{i=0}^t c_i y^i$ with $t \in \mathbb {N}$ and $c_0$ , $\ldots \,$ , $c_t \in K$ . Fix $n_0 \in \mathbb {N}$ such that $P(n) \in rG_0$ for $n \ge n_0$ , and let $g_1$ , $\ldots \,$ , $g_r \colon \mathbb {N} \to G_0$ be such that

$$\begin{align*}P(n) - g_1(n) - \cdots - g_r(n) = 0 \qquad\text{for}\ n \ge n_0.\\[-15pt] \end{align*}$$

We will apply a simple special case of Lemma 4.5 to this equation. To do so, we choose $\Omega =\mathbb {N}$ with $\pi \colon \mathbb {N} \to K$ given by $\pi (n)=n$ , $Q_1=P$ , $f_1=1$ , and $f_j=g_{j-1}$ , $Q_j=-1$ for $j \in [2,r+1]$ . For the set B, we choose $B=\{0\}$ . By Lemma 4.5, there exists $l \in \mathbb {N}_{\ge 1} \setminus \sqrt {G}$ and $e_0 \in \mathbb {N}$ such that, for all $e\ge e_0$ ,

$$\begin{align*}P(0) = g_1(l^e) + \cdots + g_r(l^e).\\[-15pt] \end{align*}$$

This implies $P(0)=P(l^e)$ for $e \ge e_0$ . Since a nonconstant polynomial can take a fixed value at most $\deg (P)-1$ times, it follows that P is constant.

Proof By Corollary 3.8, we have $F=P/Q$ with $Q=(1-\alpha _1 y_1) \cdots (1- \alpha _s y_s)$ where $\alpha _1$ , $\ldots \,$ , $\alpha _d \in K^*$ and $y_1$ , $\ldots \,$ , $y_s \in \{x_1,\ldots ,x_d\}$ (repetition is allowed). We proceed by contradiction. Suppose that one of the polynomials $Q_j$ is not constant. Then a factor in Q occurs with multiplicity $e \ge 2$ , again by Corollary 3.8. Multiplying F by some polynomial, the Bézivin property is preserved, and hence we may assume $F=P/(1-\alpha x_1)^2$ where P is not divisible by $1-\alpha x_1$ (reindexing the variables if necessary).

Through successive polynomial division, write $P = B_2(1-\alpha x_1)^2 + B_1(1-\alpha x_1) + B_0$ with $B_2 \in K[{\boldsymbol x}]$ and $B_1$ , $B_0 \in K[x_2,\ldots ,x_d]$ . Since $(1-\alpha x_1)$ does not divide P, we have $B_0 \ne 0$ . Then

$$\begin{align*}\frac{P}{Q} = B_2 + \frac{B_1}{(1-\alpha x_1)} + \frac{B_0}{(1-\alpha x_1)^2} = B_2 + \sum_{n=0}^\infty B_1 \alpha^n x_1^n + (n+1)B_0 \alpha^n x_1^n. \end{align*}$$

Let $\boldsymbol u \in K[x_2,\ldots ,x_d]$ be a monomial in the support of $B_0$ . For all sufficiently large n,

$$\begin{align*}\alpha^{-n} [x_1^{n}\boldsymbol u]P = (n+1)[\boldsymbol u]B_0 + [\boldsymbol u]B_1. \end{align*}$$

Let . By choice of $\boldsymbol u$ , the polynomial A has degree $1$ . However, enlarging G to ensure $\alpha \in G$ , there exists $r \ge 0$ such that $A(n) \in rG_0$ for all sufficiently large n. This contradicts Lemma 4.7.

Proof of Theorem 4.1

We may assume that K is finitely generated. Let $F=\sum _{\boldsymbol n \in \mathbb {N}^d} f(\boldsymbol n) {\boldsymbol x}^{\boldsymbol n}$ be a D-finite power series all of whose coefficients are in $rG_0$ for some $r \ge 0$ and a finitely generated subgroup $G \le K^*$ . Since the sequence $f\colon \mathbb {N}^d \to K$ is P-recursive, there exist $Q_{j,\boldsymbol a}$ as in equation (4.1). For every $j \in [1,d]$ , there exists $\boldsymbol a \in [0,k]^d$ with $Q_{j,\boldsymbol a}(0) \ne 0$ . Let

Then $H=PF$ with a nonzero polynomial P, and it suffices to show that H is rational. Since H is D-finite as well, the sequence of coefficients of H, denoted by h, is P-recursive of some size $k'$ ; without restriction, $k' \ge k$ .

For each $j \in [1,d]$ , we will now apply Lemma 4.5 in the following way. We choose $\Omega = \mathbb {N}^d$ and $\pi \colon \mathbb {N}^d \to \mathbb {N} \subseteq K$ to be the projection on the jth coordinate. We set $B = [0,k'd]$ and consider the equation

$$\begin{align*}0 = \sum_{\boldsymbol a \in [0,k]^d} Q_{j,\boldsymbol a}(n_j) f(\boldsymbol n - \boldsymbol a) = \sum_{\boldsymbol a \in [0,k]^d} Q_{j,\boldsymbol a}(\pi(\boldsymbol n)) f_{\boldsymbol a}(\boldsymbol n), \end{align*}$$

where $f_{\boldsymbol a}\colon \mathbb {N}^d \to rG_0$ is defined by $f_{\boldsymbol a}(\boldsymbol n) = f(\boldsymbol n - \boldsymbol a)$ . By assumption on f, this equation holds for $\boldsymbol n \in \mathbb {N}_{\ge k'}^d$ . Enlarging $G=G'$ if necessary, Lemma 4.5 (after expanding $f_{\boldsymbol{a}}(\vec n)=g_{\boldsymbol{a},1}(\vec n)+\cdots +g_{\boldsymbol{a},r}(\vec n)$ , analogously to the proof of Lemma 4.7) implies that, for any choice of $l \in \mathbb {N}_{\ge 1} \setminus G$ , there exists $e_0 \in \mathbb {N}$ such that

$$\begin{align*}\sum_{\boldsymbol a \in [0,k]^d} Q_{j,\boldsymbol a}(0) f(\boldsymbol n - \boldsymbol a) &= \sum_{\boldsymbol a \in [0,k]^d} Q_{j,\boldsymbol a}(0) f_{\boldsymbol a}(\boldsymbol n) = 0 \quad\text{for}\quad\\&\quad \boldsymbol n \in \mathbb{N}_{\ge k'}^{j-1} \times (l^{e_0+\mathbb{N}}+B) \times \mathbb{N}_{\ge k'}^{d-j}. \end{align*}$$

Further enlarging $G=G'$ if necessary, we may assume that the same finitely generated group G works for all $j \in [1,d]$ . Then we can also take the same $l \in \mathbb {N}_{\ge 1} \setminus G$ for all $j \in [1,d]$ , and finally, taking $e_0$ large enough, we can also assume the same $e_0$ works for all $j \in [1,d]$ . As a further convenience, we may enlarge $e_0$ to ensure $l^{e_0} \ge k'd$ .

For $j \in [1,d]$ , define

. Then, for all $j \in [1,d]$ and $\boldsymbol n \in \mathbb {N}_{\ge k'd}^{j-1} \times C_j \times \mathbb {N}_{\ge k'd}^{d-j}$ ,

Because h is P-recursive of size $k'$ , Lemma 4.3 implies that h vanishes on $\mathbb {N}_{\ge m}^d$ with $m = l^{e_0} + k'(d-1)$ .

For a subset $\emptyset \ne I \subseteq [1,d]$ and a tuple $(j_i)_{i\in I} \in [0,m]^{I}$ , let

We may write

$$\begin{align*}H = \sum_{\emptyset \ne I \subseteq [1,d]} \sum_{\substack{(j_i)_{i \in I}\\ j_i \in [0,m]}} u_{I,(j_i)_{i \in I}} H_{I,(j_i)_{i \in I}} \big( (x_\mu)_{\mu \in [1,d]\setminus I}\big), \end{align*}$$

with suitable power series $H_{I,(j_i)_{i \in I}}$ in $d- \left \lvert {I} \right \rvert $ variables. Note that in each summand of the form

$$\begin{align*}u_{I,(j_i)_{i \in I}} H_{I,(j_i)_{i\in I}} \big((x_\mu)_{\mu \in [1,d]\setminus I}\big), \end{align*}$$

the exponents of $x_i$ , for $i \in I$ , are fixed at some $j_i \in [0,m]$ , whereas the exponents of $x_{i}$ are greater than m for $i \in [1,d]\setminus I$ . It follows that these summands have pairwise disjoint support, and hence the $H_{I,(j_i)_{i \in I}}$ have their coefficients in $rG_0$ . These coefficient sequences being shifted sections of h, moreover the $H_{I,(j_i)_{i \in I}}$ , are D-finite. Because $I \ne \emptyset $ , and by induction on d, we may assume that the $H_{I,(j_i)_{i \in I}}$ are rational, and therefore so is H.

Proof Let $m = \min \{ \mathsf v(\alpha _1), \ldots , \mathsf v(\alpha _s) \}$ and $m'= \min \{\mathsf v(\beta _1),\ldots ,\mathsf v(\beta _s)\}$ . We may without restriction replace $F_n$ by $x^{-m'-nm} F_n$ , and may thus assume $m=m'=0$ . After reindexing,

$$\begin{align*}0 = \mathsf v(\alpha_1)=\cdots =\mathsf v(\alpha_k) < \mathsf v(\alpha_{k+1}) \le \cdots \le \mathsf v(\alpha_s) \qquad\text{for some}\ k \in [1,s]. \end{align*}$$

For $i \in [1,k]$ , we have $\alpha _i = c_i + \theta _i$ with $c_i \in K^*$ and $\theta _i \in L$ where $\mathsf v(\theta _i)> 0$ . We proceed with a proof by contradiction, and may therefore assume $\theta _i \ne 0$ for all $i \in [1,k]$ .

Reindexing the $\alpha _1$ , $\ldots \,$ , $\alpha _k$ again if necessary, we may partition

$$\begin{align*}[1,k]= [k_1,k_2-1] \uplus [k_2,k_3-1] \uplus \cdots \uplus [k_{p},k_{p+1}-1], \end{align*}$$

with $1=k_1 < k_2 < \cdots < k_{p} < k_{p+1}=k+1$ , such that $c_i=c_j$ if and only if i, $j \in [k_{\nu },k_{\nu +1}-1]$ for some $\nu $ .

Let

Then $\mathsf v(T_n)> 0$ for all n, because the constant terms cancel.

Step 1. There exists $m \in H$ with $m> 0$ such that $\mathsf v(T_n) \le m$ for infinitely many $n \ge 0$ .

Let

$$\begin{align*}A_n = \begin{pmatrix} \alpha_1^n & \cdots & \alpha_k^n & c_{k_{1}}^n & \cdots & c_{k_p}^{n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \alpha_1^{n+k-1+p} & \cdots & \alpha_k^{n+k-1+p} & c_{k_{1}}^{n+k-1+p} & \cdots & c_{k_p}^{n+k-1+p} \\ \end{pmatrix}. \end{align*}$$

Extracting scalars, these are Vandermonde matrices with

$$\begin{align*}\det(A_n) = \alpha_1^n \cdots \alpha_k^n c_{k_{1}}^n \cdots c_{k_p}^n \prod_{1 \le i < j \le k} (\alpha_j - \alpha_i) \prod_{1 \le i <j \le p} (c_{k_j} - c_{k_i}) \prod_{\substack{1 \le i \le k \\ 1 \le j \le p}} (c_{k_j} - \alpha_i). \end{align*}$$

By construction, $\det (A_n) \ne 0$ . Since $\mathsf v(\alpha _i)=0$ for $i \in [1,k]$ , moreover $\mathsf v(\det (A_n))$ is in fact independent of n. Let and $\boldsymbol w = (w_1,\,\ldots ,w_{k+p})^T$ with

$$\begin{align*}w_i = \begin{cases} \beta_i, & \text{if}\ 1 \le i \le k, \\ -\sum_{j=k_\nu}^{k_{\nu+1}-1} \beta_j, & \text{if}\ i=k+\nu\ \text{with}~ 1 \le \nu \le p. \end{cases} \end{align*}$$

Since $\beta _1 \ne 0$ , we have $w_1 \ne 0$ . Let . The ith entry of $A_n\boldsymbol w$ is $T_{n+i-1}$ . Hence, using the subscript $1$ to denote the first coordinate of a vector,

$$\begin{align*}\det(A_n)w_{1} = \big( \operatorname{adj}(A_n)A_n \boldsymbol w )_{1} = \big(\operatorname{adj}(A_n) (T_n, \ldots, T_{n+k+p-1})^T \big)_{1} = \sum_{j=0}^{k+p-1} \gamma_j T_{n+j} \end{align*}$$

for some $\gamma _j$ with $\mathsf v(\gamma _j) \ge 0$ . Since $\mathsf v(\det (A_n)w_{1}) = m_1 + m_2$ , we must have $\mathsf v(T_{n+\delta (n)}) \le m_1 + m_2$ for some $\delta (n) \in [0,k+p-1]$ .

Step 2. There exist $N_0 \ge 0$ such that for all $n \ge N_0$ ,

$$\begin{align*}\mathsf v\bigg( T_n - \sum_{i=1}^k \sum_{l=1}^{N_0} {n \choose l} \beta_i c_i^{n-l} \theta_i^l \bigg)> m. \end{align*}$$

Substituting $\alpha _i=c_i + \theta _i$ into the definition of $T_n$ and expanding, for $0 \le N_0 \le n$ ,

$$\begin{align*}T_n = \sum_{i=1}^k \sum_{l=1}^{N_0} {n \choose l} \beta_i c_i^{n-l} \theta_i^l + \sum_{i=1}^k \sum_{l=N_0+1}^n {n \choose l} \beta_i c_i^{n-l} \theta_i^l. \end{align*}$$

There exists $N_0$ such that $\mathsf v(\beta _i\theta _i^l) =l\mathsf v(\theta _i)+\mathsf v(\beta _i)> m$ for all $l \ge N_0$ , and choosing such $N_0$ establishes the claim of Step 2.

Enlarging $N_0$ if necessary, we may also assume $\mathsf v(\alpha _i^{N_0})> m$ for $i \in [k+1,s]$ by Lemma 2.5. Let

Fix $h \in H$ with $0 < h \le m$ such that $[x^h]P_n \ne 0$ for some $n \ge N_0$ . The—crucial—existence of such an h is guaranteed by Steps 1 and 2. Substituting Hahn series expressions for $\beta _i$ and $\theta _i$ , we can express $[x^h]P_n$ as

$$\begin{align*}[x^h]P_n = \sum_{i \in I} p_{h,i}(n) c_i^n, \end{align*}$$

where $I \subseteq [1,k]$ is such that the $c_i$ are pairwise distinct, and $p_{h,i} \in K[x]$ are polynomials. Since l is always at least $1$ in the expression for $P_n$ , the polynomials $p_{h,i}$ are either zero or nonconstant as a consequence of Lemma 5.1. By choice of h, there must be at least one i with $p_{h,i} \ne 0$ .

For $n \ge N_0$ ,

$$\begin{align*}[x^h]P_n = [x^h]T_n = [x^h]F_n - \sum_{i=1}^k [x^h]\beta_i c_i^n - \underbrace{[x^h] \sum_{i=k+1}^s \beta_i \alpha_i^n}_{=0}. \end{align*}$$

Enlarging G, we may assume $c_i$ , $-[x^h]\beta _i \in G$ , and hence $[x^h]P_n \in (r+k)G_0$ . Now, the univariate case of Lemma 4.8 implies that the $p_{h,i}$ are constant, and by our construction therefore $p_{h,i}=0$ for all $i \in I$ . This is a contradiction to $[x^h]P_n \ne 0$ for some $n \ge N_0$ .

Proof Let $\boldsymbol a_i = (a_{i,1},\ldots ,a_{i,d})$ for $i \in [1,s]$ . Let $N \in \mathbb {N}$ be such that $N a_{i,j} \in \mathbb {Z}$ for all $i \in [1,s]$ and $j \in [1,d]$ . Since

$$\begin{align*}1 - c_i^N{\boldsymbol x}^{N\boldsymbol a_i} = (1-c_i {\boldsymbol x}^{\boldsymbol a_i}) \sum_{j=0}^{N-1} c_i^j {\boldsymbol x}^{j\boldsymbol a_i}, \end{align*}$$

there exists $F \in K (\mkern -3mu( x_1^{\mathbb {Q}},\ldots ,x_d^{\mathbb {Q}} )\mkern -3mu)$ , having finite support, such that $PF=Q$ with $Q=(1- c_1^N {\boldsymbol x}^{N\boldsymbol a_1}) \cdots (1- c_s^{N} {\boldsymbol x}^{N \boldsymbol a_s})$ . Since P and Q are Laurent polynomials, the quotient $F=Q/P$ is a rational function in ${\boldsymbol x}$ . Therefore, $Q/P$ has a Laurent series expansion at the origin, which coincides with the Hahn series F. Thus, F is in fact a Laurent series in ${\boldsymbol x}$ . Since F has finite support, it is even a Laurent polynomial. We conclude that P divides Q in the Laurent polynomial ring $K[{\boldsymbol x}^{\pm 1}]$ .

The Laurent polynomial ring is a factorial domain, arising from $K[{\boldsymbol x}]$ by inverting the prime elements $x_1$ , $\ldots \,$ , $x_d$ . By Lemma 2.7, we know that all factors of Q in $K[{\boldsymbol x}^{\pm 1}]$ are—up to units of $K[{\boldsymbol x}^{\pm 1}]$ —of the form $1-c'{\boldsymbol x}^{\boldsymbol b}$ with $c' \in K^*$ and $\boldsymbol b \in \mathbb {Z}^d \setminus \{\boldsymbol 0\}$ . Therefore,

$$\begin{align*}P = (1-c_1'{\boldsymbol x}^{\boldsymbol b_1})\cdots (1-c_t'{\boldsymbol x}^{\boldsymbol b_t}) \end{align*}$$

with $\boldsymbol b_1$ , $\ldots \,$ , $\boldsymbol b_t \in \mathbb {Z}^d \setminus \{\boldsymbol 0\}$ , and this factorization is unique up to order and associativity in $K[{\boldsymbol x}^{\pm 1}]$ .

Suppose now that P is a polynomial and $P(\boldsymbol 0)=1$ . Then $P=Q_1\cdots Q_r$ with irreducible polynomials $Q_i \in K[{\boldsymbol x}]$ . Since $P(\boldsymbol 0)=1$ , we have $Q_1(\boldsymbol 0)\cdots Q_r(\boldsymbol 0)=1$ . Replacing each $Q_i$ by $Q_i/Q_i(\boldsymbol 0)$ , we can therefore assume $Q_i(\boldsymbol 0)=1$ for each $i \in [1,r]$ . In particular, no $Q_i$ is a monomial. Therefore, $Q_1$ , $\ldots \,$ , $Q_r$ remain prime elements in the localization $K[{\boldsymbol x}^{\pm 1}]$ . Since $K[{\boldsymbol x}^{\pm 1}]$ is factorial, this implies $r=t$ and that, after reindexing the polynomials $Q_i$ if necessary, we may assume that $Q_i$ is associated with $1- c_i'{\boldsymbol x}^{\boldsymbol b_i}$ for each $i \in [1,r]$ . Explicitly, this means $Q_i = q_i{{\boldsymbol x}^{\boldsymbol e_i}}(1-c_i'{\boldsymbol x}^{\boldsymbol b_i})$ with $\boldsymbol e_i \in \mathbb {Z}^d$ and $q_i \in K^*$ . Then $Q_i \in K[\boldsymbol x]$ forces $\boldsymbol e_i \in \mathbb {N}^d$ , and so there are two possibilities: either $\boldsymbol e_i=\boldsymbol 0$ and $q_i=1$ , in which case necessarily $\boldsymbol b_i \in \mathbb {N}^d$ , or $q_i c_i' {\boldsymbol x}^{\boldsymbol e_i+\boldsymbol b_i}=1$ . In either case, $Q_i$ is of the desired form.

Proof Let and . Since Q is irreducible in $K[x_1,\ldots ,x_d]$ and is not contained in R, it is also irreducible as a univariate polynomial in $R[x_d]$ . By Gauss’s lemma, then Q is irreducible in $L_0[x_d]$ . Since L is algebraically closed, Q can be expressed as a product of linear factors as above.

Suppose now without restriction that $\lambda _1^{-1}= a {\boldsymbol x}^{\boldsymbol e}$ with $a \in K^*$ and $\boldsymbol e \in \mathbb {Q}^{d-1}$ . Let $N \in \mathbb {Z}_{\ge 1}$ with $N{\boldsymbol e} \in \mathbb {Z}^{d-1}$ . Then $\lambda _1^{-1}$ is a root of $P = x_d^{N} - a^{N}{\boldsymbol x}^{\boldsymbol{e}N} \in L_0[x_d]$ . However, $P= \prod _{j=0}^{N-1} (x_d - a\zeta ^j {\boldsymbol x}^{\boldsymbol e})$ in $L[{\boldsymbol x}]$ , where $\zeta \in L^*$ is a primitive Nth root of unity. Since all roots of P are monomials, and Q divides P by irreducibility, all roots of $Q \in L[x_d]$ are monomials.

Proof We proceed by induction on d. If $d=0$ , then the claim holds trivially with $s=0$ . Suppose $d \ge 1$ and that the claim holds for $d-1$ .

Let $Q = Q_1 \cdots Q_n$ where $Q_1$ , $\ldots \,$ , $Q_n \in K[{\boldsymbol x}]$ are irreducible ( $n \ge 0$ ). For each $j \in [1,n]$ , the series $P/Q_j = FQ_1\cdots Q_{j-1}Q_{j+1}\cdots Q_n$ is also a rational Bézivin series. It therefore suffices to show: if $F=P/Q$ with an irreducible polynomial $Q \in K[{\boldsymbol x}]$ , not dividing P, and $Q(\boldsymbol 0)=1$ and F is a Bézivin series, then Q is of the form $1-cu$ with $c \in K^*$ and a monomial $u \in K[{\boldsymbol x}]$ .

By reindexing, we may assume that Q has degree at least one in the variable $x_d$ . Then, since Q is irreducible and does not divide P, there are polynomials A and B in $K[{\boldsymbol x}]$ such that

. Since $FA$ and B are both rational Bézivin series, the same is true for $C/Q=FA+B$ . Now, since Q has degree at least one in $x_d$ and since it has distinct roots as a polynomial in $x_d$ , we may use the fact that

is algebraically closed to factor Q and write

$$\begin{align*}C/Q = CQ_0^{-1} (1-\lambda_1 x_d)^{-1}\cdots (1-\lambda_s x_d)^{-1}, \end{align*}$$

where $Q_0 \in K[x_1,\ldots ,x_{d-1}]$ and $\lambda _1$ , $\ldots \,$ , $\lambda _s\in L^*$ are pairwise distinct. Using partial fractions, we may write this as

$$\begin{align*}C/Q=\alpha + \sum_{i=1}^s \frac{\beta_i}{1-\lambda_i x_d}, \end{align*}$$

where $\alpha $ , $\beta _i\in L$ and the $\beta _i$ are nonzero. Then, if we expand, we have that for $n\ge 1$ , the coefficient of $x_d^n$ in $C/Q$ is

Applying Proposition 5.2 to this family of Hahn series, one of the $\lambda _i$ is of the form $\lambda _i=c_i u_i$ with $c_i \in K^*$ and $u_i \in L$ a monomial, that is, of the form $u_i = x_1^{e_{i,1}}\cdots x_{d-1}^{e_{i,d-1}}$ with $e_{i,j} \in \mathbb {Q}$ . By Lemma 5.4, then all $\lambda _i$ are of the form $\lambda _i=c_iu_i$ with $c_i \in K^*$ and $u_i \in L$ a monomial.

Since $CQ_0^{-1} = CQ^{-1}(1-c_1u_1x_d)\cdots (1-c_su_sx_d)$ , it follows that the rational series $CQ_0^{-1}$ is a Bézivin series. The induction hypothesis implies $Q_0=(1-c_{s+1}u_{s+1})\cdots (1-c_{t}u_t)$ with $c_j \in K^*$ and monomials $u_{s+1}$ , $\ldots \,$ , $u_t \in K[x_1,\ldots ,x_{d-1}]$ . Replacing $u_i$ by $u_ix_d$ for $i \in [1,s]$ , we have

$$\begin{align*}Q = (1-c_1u_1) \cdots (1-c_t u_t) \in K[{\boldsymbol x}]. \end{align*}$$

By Lemma 5.3, we can rewrite this product in such a way that the Hahn monomials $u_j$ are monomials with nonnegative integer exponents, that is, monomials in the polynomial ring $K[{\boldsymbol x}]$ . Then $t=1$ by irreducibility of Q and the claim is shown.