Proof. As $(\log (a_r))_{r\ge 1}$ is subadditive, $({\log (a_r)}/{r})_{r\ge 1}$ converges to $\inf _{r \geq 1}(({\log (a_r)}/{r}) \geq 0$ by Fekete’s subadditivity lemma (see [Reference FeketeFek23, Satz II] or [Reference Michael SteeleSte97, Lemma 1.2.1]).

Proof. This follows from Lemma 2.2.1, as

$$ \begin{align*}\deg_{x_{m+1},\ldots,x_n}(f^{r+s})\le \deg_{x_{m+1},\ldots,x_n}(f^r)\cdot \deg_{x_{m+1},\ldots,x_n}(f^s),\end{align*} $$

for all $r,s\ge 1$ .

Proof. For each $r\ge 1$ , we write

$$ \begin{align*} a_r&= \max\{\deg((f^r)_1),\ldots, \deg((f^r)_m)\},\\ b_r&= \max\{\deg((f^r)_{m+1}),\ldots, \deg((f^r)_{n})\},\\ c_r&= \max\{\deg_{x_{m+1},\ldots,x_n}((f^r)_{m+1}),\ldots, \deg_{x_{m+1},\ldots,x_n}((f^r)_{n})\} \\ &= \deg_{x_{m+1}, \ldots, x_n}(f^r). \end{align*} $$

As $b_r\ge c_r$ , we obtain for each $r \geq 1$ ,

$$ \begin{align*}\deg(f^r)=\max\{a_r,b_r\}\ge \max\{a_r,c_r\}.\end{align*} $$

It follows from Corollary 2.2.2 that the limits

exist (and all belong to ${\mathbb R}_{\ge 1}$ ). We obtain

Thus we may assume that $\unicode{x3bb} (f)>\unicode{x3bb} _1$ , which implies that $\lim _{r\to \infty } b_r^{1/r}$ exists and is equal to $\unicode{x3bb} (f)$ . It remains to see that, in this case, $\unicode{x3bb} (f)\le \max \{\unicode{x3bb} _1,\unicode{x3bb} _2\}$ .

For all $r,s \ge 1$ and each $i\in \{m+1,\ldots ,n\}$ , the polynomial $(f^{r+s})_i$ is obtained by replacing $x_1,\ldots ,x_n$ with $(f^{r})_1,\ldots ,(f^r)_n$ in $(f^s)_i$ , so the degree of $(f^{r+s})_i$ is at most

$$ \begin{align*} &\deg_{x_1,\ldots,x_m}((f^s)_i)\cdot \deg((f^r)_1,\ldots,(f^r)_m) \\ &\quad+ \deg_{x_{m+1},\ldots,x_n}((f^s)_i)\cdot \deg((f^r)_{m+1},\ldots,(f^r)_n). \end{align*} $$

This gives $b_{r+s}\le b_s \cdot a_{r}+c_s\cdot b_{r}$ . Then when we choose $s=r$ , we obtain

$$ \begin{align*} b_{2r}\le b_r \cdot (a_{r}+c_r). \end{align*} $$

As $\unicode{x3bb} (f)=\lim \nolimits _{r\to \infty } b_{2r}^{1/2r}$ , we have $\unicode{x3bb} (f)^2=\lim \nolimits _{r\to \infty } b_{2r}^{1/r}$ . The above inequality gives

so $\unicode{x3bb} (f)\le \max \{\unicode{x3bb} _1,\unicode{x3bb} _2\}$ .

Proof. By Lemma 2.4.1, one has $\unicode{x3bb} (f)=\max \{\unicode{x3bb} (g),\unicode{x3bb} _2\}$ , where

It remains to see that $\unicode{x3bb} _2$ is an integer. As $\mathbf {k}[x_1,\ldots ,x_{n-2},f_{n-1},f_{n}]=\mathbf {k}[x_1,\ldots ,x_{n}]$ , one has $K[f_{n-1},f_{n}]=K[x_{n-1},x_n]$ , where $K=\mathbf {k}(x_1,\ldots ,x_{n-2})$ . Hence, one can see the automorphism $(x_1,\ldots ,x_n)\mapsto (x_1,\ldots ,x_{n-1},f_{n-1},f_{n})$ of ${\mathbb A}^n$ as an automorphism $F\in \operatorname {\mathrm {Aut}}_K(\mathbb {A}^2)$ of $\mathbb {A}^2$ defined over K. For each $i \geq 0$ , the automorphism $g^{-i} \circ (x_1,\ldots ,x_{n-1},f_{n-1},f_{n}) \circ g^i$ of ${\mathbb A}^n$ can be seen as an element of $\operatorname {\mathrm {Aut}}_K({\mathbb A}^2)$ that we denote by $F^{g^i}$ , where we identify g with the automorphism $(f_1, \ldots , f_{n-2}, x_{n-1}, x_n) \in \operatorname {\mathrm {Aut}}({\mathbb A}^n)$ . This gives

$$ \begin{align*} \max\{\deg_{x_{n-1},x_n}((f^r)_{n-1}), \deg_{x_{n-1},x_n}((f^r)_n)\}=\deg (G_r), \end{align*} $$

where $G_r=F^{g^{r-1}} \circ \cdots \circ F^{g^2} \circ F^{g} \circ F\in \operatorname {\mathrm {Aut}}_K(\mathbb {A}^2)$ , because $G_r = g^{-r} \circ f^r$ when we consider $G_r$ , g and f as automorphisms of ${\mathbb A}^n$ .

According to the Jung–van der Kulk theorem [Reference JungJun42, Reference van der KulkvdK53], one can write $F=F_1\circ \cdots \circ F_s$ , where each $F_i\in \operatorname {\mathrm {Aut}}_K(\mathbb {A}^2)$ is either triangular or affine. Moreover, one can assume that two consecutive $F_i$ are not both affine or both triangular (as otherwise, one may reduce the description), and get then $\deg (F)=\prod _{i=1}^s \deg (F_i)$ (follows by looking at what happens at infinity or by [Reference van den EssenvdE00, Lemma 5.1.2]). We prove that $\unicode{x3bb} _2$ is an integer by induction on s. If $s=1$ , then F is either affine or triangular; this implies that the set $\{\deg (G_r)\mid {r\ge 1}\}$ is bounded, so $\unicode{x3bb} _2=1$ . If $s>1$ and $F_1$ , $F_s$ are both affine or both triangular, so we replace F with $(F_1)^g \circ F\circ F_1^{-1}$ . This replaces $G_r=F^{g^{r-1}} \circ \cdots \circ F^{g} \circ F$ with $\tilde {G}_r=(F_1)^{g^r}\circ G_r\circ F_1^{-1}$ . As $\deg ((F_1)^{g^r})=\deg (F_1)$ for each $r\ge 1$ , one has

$$ \begin{align*} \frac{1}{\deg(F_1)^2}\deg(G_r)\le \deg(\tilde{G}_r)\le \deg(G_r)\cdot \deg(F_1)^2, \end{align*} $$

so this replacement does not change the value of $\unicode{x3bb} _2$ . As this decreases the value of s, we may assume that $F_1$ and $F_s$ are not both triangular or affine. Hence, for each $r\ge 1$ , $G_r$ is a product of $rs$ elements that are affine or triangular, with no two consecutive in the same group. This gives $\deg (G_r)=\prod _{i=0}^{r-1} \prod _{j=1}^s \deg (F_j^{g^i})=\prod _{i=0}^{r-1} \prod _{j=1}^s \deg (F_j)=\deg (F)^r$ . Hence, $\unicode{x3bb} _2=\deg (F)$ is an integer.

Proof. The fact that the dynamical degree of any element of $\operatorname {\mathrm {Aut}}(\mathbb {A}^2)$ is an integer follows from Corollary 2.4.2 applied to $n=2$ . If $f\in \operatorname {\mathrm {Aut}}(\mathbb {A}^3)$ is an automorphism that preserves the set of fibres of a linear projection $\mathbb {A}^3\to \mathbb {A}^1$ or $\mathbb {A}^3\to \mathbb {A}^2$ , then one may conjugate by an element of $\operatorname {\mathrm {GL}}_3$ and obtain $f=(f_1,f_2,f_3)$ with either $f_1\in \mathbf {k}[x_1]$ or $f_1,f_2\in \mathbf {k}[x_1,x_2]$ . The fact that $\unicode{x3bb} (f)$ is an integer follows then from Corollary 2.4.2 and Lemma 2.4.1, respectively (in the second case, one uses the fact that the dynamical degree of $(f_1, f_2) \in \operatorname {\mathrm {Aut}}({\mathbb A}^2)$ is an integer). Similarly, in the case of an automorphism of $\mathbb {A}^4$ preserving a linear projection $\mathbb {A}^4\to \mathbb {A}^2$ , one restricts to the case $f=(f_1,\ldots ,f_4)\in \operatorname {\mathrm {Aut}}(\mathbb {A}^4)$ with $f_1,f_2\in \mathbf {k}[x_1,x_2]$ , and applies Corollary 2.4.2.

Proof. The implication condition (2) $\Rightarrow $ (1) is given by choosing $p=x_i$ for $i = 1, \ldots , n$ , so we may assume condition (1) and prove condition (2). It suffices to prove condition (2) for $r=1$ , as the general result follows by induction.

If $p=0$ , then $h(p)=0$ is $\mu $ -homogeneous of any degree. It then suffices to do the case where p is a monomial: we write $p=\zeta x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$ with $\zeta \in \mathbf {k}^*$ , $a_1,\ldots ,a_n\ge 0$ , which is $\mu $ -homogeneous of degree $\deg _\mu (p)=\sum _{i=1}^n a_i\mu _i$ . As $h_i$ is $\mu $ -homogeneous of degree $\theta \mu _i$ , the polynomial $p\circ h=\zeta h_1^{a_1}h_2^{a_2}\cdots h_n^{a_n}$ is $\mu $ -homogeneous of degree $\sum _{i=1}^n a_i \theta \mu _i=\theta \deg _\mu (p)$ .

Now, we assume additionally that $h_i \neq 0$ for each $i \in \{1, \ldots , n\}$ . The equivalence between conditions (1) and (3) follows immediately from the definition of the $\mu $ -degree.

Proof. Condition (1) $\Rightarrow $ (2): For each $i\in \{1,\ldots ,n\}$ , the polynomial $f_i$ is the sum of the ith components of the endomorphisms $g_{\xi }$ . As each of these polynomials has degree $\xi \mu _i\le \theta \mu _i$ , the polynomial $f_i$ is of $\mu $ -degree $\deg _\mu (f_i)\le \theta \mu _i$ .

Condition (2) $\Rightarrow $ (1): As in Remark 2.3.1(1), we write each $f_i$ , $i\in \{1,\ldots ,n\}$ as $f_i=\sum _{0\le \kappa \le \theta \mu _i} p_{i,\kappa }$ , where each $p_{i,\kappa }$ is $\mu $ -homogeneous of degree $\kappa $ .

We define $g_0=(p_{1,0},\ldots ,p_{n,0})\in \operatorname {\mathrm {End}}(\mathbb {A}^n)$ , which is $\mu $ -homogeneous of degree $0$ .

For each $\xi \in {\mathbb R}$ with $0\le \xi \le \theta $ , we define the ith component $(g_\xi )_i$ of $g_\xi $ as follows: if $\mu _i=0$ and $\xi> 0$ , then $(g_\xi )_i=0$ and otherwise, we choose $(g_\xi )_i=p_{i,\xi \mu _i}$ . By construction, $g_\xi $ is $\mu $ -homogeneous of degree $\xi $ .

Moreover, $f_i=\sum _{0\le \kappa \le \theta \mu _i} p_{i,\kappa }=\sum _{0\le \xi \le \theta } (g_{\xi })_i$ for each $i\in \{1,\ldots ,n\}$ with ${\mu _i>0}$ . If $\mu _i=0$ , then $f_i=\sum _{0\le \kappa \le \theta \mu _i} p_{i,\kappa }=p_{i,0}=\sum _{0\le \xi \le \theta } (g_{\xi })_i$ . This yields $f=\sum _{0\le \xi \le \theta } g_\xi $ .here

Proof. Condition (1) $\Rightarrow $ (2): Suppose that $\theta =\deg _\mu (f)<\infty $ . For each $i\in \{1,\ldots ,n\}$ , we get $\deg _\mu (f_i)\le \theta \mu _i$ (Definition 1.4.3). If $\mu _i=0$ , then $\deg _\mu (f_i)=0$ , which means that $f_i$ is a polynomial in the variables $\{x_j\mid j\in \{1,\ldots ,n\},\mu _j=0\}$ .

Condition (2) $\Rightarrow $ (1): It follows from condition (2) that $\deg _\mu (f_i) \leq 0$ for each $i\in \{1,\ldots ,n\}$ such that $\mu _i=0$ . This gives $\deg _\mu (f)=\max \{ \deg _\mu (f_i) / \mu _i \, | \, \mu _i>0 \}<\infty $ .

Proof. As $\mu =(\mu _1,\ldots ,\mu _n)\in (\mathbb {R}_{\ge 0})^n$ is a maximal eigenvector of f, we have $\deg _\mu (f_i)= \theta \mu _i$ for each $i\in \{1,\ldots ,n\}$ , where $\theta $ is the maximal eigenvalue of f. This gives $\deg _\mu (f)=\theta < \infty $ and therefore $\deg _\mu (g_i)=\theta \mu _i=\deg _\mu (f_i)$ for each $i\in \{1,\ldots ,n\}$ . Hence, we get property (1). In case $\mu _i> 0$ , we have $\deg _\mu (g_i) = \theta \mu _i> 0$ and thus $g_i$ is non-constant. In case $\mu _i = 0$ , we have $\deg _\mu (f_i) = \theta \mu _i = 0$ and thus $g_i = f_i$ . As f is dominant, the latter polynomial is non-constant. This shows property (2).

Proof. As $\deg _\mu (f) = \theta $ , we have $\deg _\mu (f_i)\le \theta \mu _i$ for each $i\in \{1,\ldots ,n\}$ . Moreover, as f is dominant and $\mu \not =0$ , there are $i,j\in \{1,\ldots ,n\}$ such that $\mu _i>0$ and $\deg _{x_i}(f_j)\ge 1$ . This implies that $\deg _\mu (f_j)\ge \mu _i>0$ and thus

$$ \begin{align*} 0<\deg_\mu(f) = \theta. \end{align*} $$

We now observe that $\deg _\mu (f-g)<\theta $ . Indeed, for each $j\in \{1,\ldots ,n\}$ , the jth component $g_j$ of g is the $\mu $ -homogeneous part of $f_j$ of degree $\theta \mu _j\ge \deg _\mu (f_j)$ . If $\mu _j=0$ , then $f_j=g_j$ , and if $\mu _j>0$ , then $\deg _\mu (f_j-g_j)<\theta \mu _j$ .

By Lemma 2.5.3, we can write $f-g$ as a finite sum $f-g=\sum _{0\le \xi <\theta } g_\xi $ , where each $g_\xi \in \operatorname {\mathrm {End}}(\mathbb {A}^n)$ is $\mu $ -homogeneous of degree $\xi $ . This gives assertion (1).

We now prove assertions (2)–(3) by induction on $r\ge 1$ . For $r=1$ , assertion (2) follows from the definition of g. Moreover, assertion (3) is given by hypothesis.

We now assume assertions (2)–(3) for some integer $r\ge 1$ and prove them for $r+1$ . For each $i\in \{1,\ldots ,r\}$ , we write $(f^{r})_i=(g^{r})_i+s_i$ , where $(g^{r})_i$ is $\mu $ -homogeneous of degree $\theta ^r\mu _i$ and $\deg _\mu (s_i)<\theta ^r\mu _i$ . This gives

$$ \begin{align*} (f^{r+1})_i&=((g^{r})_i+s_i) \circ f \\ &\stackrel{\scriptsize({1})}{=} (g^{r+1})_i +s_i \circ g +\displaystyle\sum\limits_{0\le \xi< \theta} ((g^{r})_i+s_i) \circ g_\xi . \end{align*} $$

As g is $\mu $ -homogeneous of degree $\theta $ , the polynomial $(g^{r+1})_i$ is $\mu $ -homogeneous of degree $\theta ^{r+1}\mu _i$ (Lemma 2.5.1). As $s_i$ is a sum of $\mu $ -homogeneous polynomials of degree $<\theta ^r\mu _i$ and $g_\xi $ is $\mu $ -homogeneous of degree $\xi <\theta $ , we have

$$ \begin{align*} \deg_\mu\bigg( s_i \circ g +\sum\limits_{0\le \xi< \theta} ((g^{r})_i+s_i) \circ g_\xi \bigg)< \theta^{r+1}\mu_i \end{align*} $$

(by using Lemma 2.5.1 again). This yields assertions (2)–(3) for $r+1$ .

We now prove assertion (4). We choose $i\in \{1,\ldots ,n\}$ such that $\mu _i=\max \{\mu _1,\ldots , \mu _n\}$ , and observe that for each $r\ge 1$ , there is $j\in \{1,\ldots ,n\}$ such that $\deg _{x_i}((f^r)_j)>0$ (as f is dominant), so $\deg _\mu ((f^r)_j) \ge \mu _i = \max \{\mu _1,\ldots ,\mu _n\}>0$ . This implies that

$$ \begin{align*} 1\le \lim\limits_{r\to\infty} \max\limits_{i\in \{1,\ldots,n\}} \deg_\mu ((f^r)_i)^{1/r} \end{align*} $$

(the limit exists by Remark 2.3.1(2)). Let us write $I_0=\{i\in \{1,\ldots ,n\}\mid \mu _i=0\}$ . For each $i\in I_0$ , we have $\deg _\mu (f_i)\le \theta \mu _i=0$ , so $f_i$ is a polynomial in the variables $\{x_j\mid j\in I_0\}$ . This implies that the same holds for $(f^r)_i$ , for each integer $r\ge 1$ . Hence, $\deg _\mu ((f^r)_i)=0$ for each $i\in I_0$ . Writing $I_{>0}=\{i\in \{1,\ldots ,n\}\mid \mu _i>0\}$ , we get for each $r\ge 1$ ,

$$ \begin{align*} \deg_\mu(f^r)= \max\bigg\{\frac{\deg_\mu((f^r)_i)}{\mu_i} \bigg| i\in I_{>0}\bigg\}. \end{align*} $$

As $\deg _\mu (f^r)\le \theta ^r$ (assertion (3)), we obtain

$$ \begin{align*} \displaystyle\lim_{r\to\infty} \big(\max_{i\in \{1,\ldots,n\}} \deg_\mu (f^r)_i\big)^{1/r}=\lim\limits_{r\to\infty}(\deg_\mu(f^r))^{1/r}\le \theta. \end{align*} $$

It remains to prove assertion (5); for this, we assume that $\theta> 1$ . For each $r\ge 1$ , assertion (3) gives $\deg _\mu (f^r)\le \theta ^r$ , or equivalently $\deg _\mu ((f^r)_i)\le \theta ^r \mu _i$ for each $i\in \{1,\ldots ,n\}$ . The equality $\deg _\mu (f^r) = \theta ^r$ holds if and only if there exists $i\in \{1,\ldots ,n\}$ such that $\mu _i>0$ and $\deg _\mu ((f^r)_i)= \theta ^r \mu _i$ . Because $(g^r)_i$ is the $\mu $ -homogeneous part of $(f^r)_i$ of degree $\theta ^r \mu _i$ (follows from assertion (2)), this gives the equivalence between assertions (ii) and (iii). It remains then to prove assertion (5) $\Leftrightarrow $ (5).

‘Assertion (5) $\Rightarrow $ (5)’: Suppose that for each $r\ge 1$ , there is $i\in \{1,\ldots ,n\}$ such that $\mu _i>0$ and $(g^r)_i\not =0$ . There is then $j\in \{1,\ldots ,n\}$ and an infinite set $I\subset \mathbb {N}$ such that $\mu _j>0$ and $(g^r)_j\not =0$ for each $r\in I$ . Assertion (2) implies that $\deg _\mu ((f^r)_j)\ge \theta ^r\mu _j$ , for each $r\in I$ , which implies that

$$ \begin{align*} \displaystyle\lim\limits_{r\to\infty}\big(\max\limits_{i\in \{1,\ldots,n\}} \deg_\mu (f^r)_i\big )^{1/r}\ge\theta. \end{align*} $$

This, together with assertion (4), gives $\lim _{r\to \infty }(\deg _\mu (f^r))^{1/r} = \theta $ .

‘Assertion (5) $\Rightarrow $ (5)’: Conversely, suppose that there exists $s\ge 1$ such that $(g^s)_i=0$ for each $i\in \{1,\ldots ,n\}$ with $\mu _i>0$ . For all such i, we obtain $\deg _\mu ((f^s)_i)< \theta ^s \mu _i$ (by assertions (2) and (3)). As $\theta> 1$ , there exists then $\theta '\in {\mathbb R}$ with $1<\theta '< \theta $ such that

$$ \begin{align*} \deg_\mu((f^s)_i)\le{\theta^{\prime}}^s\mu_i \end{align*} $$

for each $i\in \{1,\ldots ,n\}$ . Applying the inequality of assertion (4) for $f^s$ , we obtain

$$ \begin{align*} \displaystyle\lim\limits_{r\to\infty}\big(\max\limits_{i\in \{1,\ldots,n\}} (\deg_\mu (f^{sr})_i)^{1/r}\big)\le {\theta^{\prime}}^s, \end{align*} $$

which gives, by taking the sth root,

$$ \begin{align*} \displaystyle\lim\limits_{r\to\infty}\Bigg(\max\limits_{i\in \{1,\ldots,n\}} (\deg_\mu (f^r)_i)^{1/r}\Bigg)\le \theta'<\theta.\\[-42pt] \end{align*} $$

Proof of Proposition A

As $\mu \in (\mathbb {R}_{> 0})^n$ , we have $\theta : = \deg _\mu (f) < \infty $ (Lemma 2.5.6).

Using Remark 2.3.1(2), we get

$$ \begin{align*} \unicode{x3bb}(f) = \lim_{r \to \infty} \max\limits_{i\in \{1,\ldots,n\}} (\deg_\mu (f^r)_i)^{1/r}. \end{align*} $$

By definition, g is the $\mu $ -leading part of f. Now, Lemma 2.6.1(4) implies that $1 \leq \unicode{x3bb} (f) \leq \theta $ . If $\theta> 1$ , we moreover obtain

$$ \begin{align*} \unicode{x3bb}(f)=\theta\Leftrightarrow\deg_\mu(f^r)=\theta^r \text{ for each }r\ge 1 \Leftrightarrow g^r\not=0 \text{ for each }r\ge 1 \end{align*} $$

(by Lemma 2.6.1(4) and Lemma 2.6.1(5)).

Proof. Assertion (1) follows from the fact that $\deg _\mu (f)<\infty $ and the choice of m (Lemma 2.5.6(2)).

Lemma 2.4.1 then gives $\unicode{x3bb} (f)=\max \{\unicode{x3bb} (\hat {f}), \lim _{r\to \infty }\deg _{x_{m+1},\ldots ,x_n}(f^r)^{1/r}\}$ . By using the equality $\lim _{r\to \infty }\deg _{x_{m+1},\ldots ,x_n}(f^r)^{1/r}=\lim _{r\to \infty }\deg _{\mu }(f^r)^{1/r}$ (see Remark 2.3.1(2) and Lemma 2.6.1(4)), we obtain

$$ \begin{align*}\unicode{x3bb}(f)=\max\{\unicode{x3bb}(\hat{f}), \lim\limits_{r\to \infty}\deg_{\mu}(f^r)^{1/r}\}.\end{align*} $$

Moreover, Lemma 2.6.1(4) implies that $\lim _{r\to \infty }\deg _{\mu }(f^r)^{1/r}\le \deg _{\mu }(f)=\theta.$ This provides assertion (2). To show assertion (3), we assume that $\unicode{x3bb} (\hat {f})<\theta $ and obtain $\unicode{x3bb} (f)=\theta \Leftrightarrow \lim _{r\to \infty }\deg _{\mu }(f^r)^{1/r} =\theta $ . This is equivalent to ask that $ f$ is $\mu $ -algebraically stable, by Lemma 2.6.1(5) (note that $1 \leq \unicode{x3bb} (\hat {f})$ , since f, and thus $\hat {f}$ , is dominant).

Proof. Denote by g the $\mu $ -leading part of f. As $\mu \in ({\mathbb R}_{>0})^n$ , no component of g contains any constant. Hence, g is also the $\mu $ -leading part of $\tau \circ f$ . By Lemma 2.6.1(5), f (respectively $\tau \circ f$ ) is $\mu $ -algebraically stable if and only if for each $r\ge 1$ , there is $i\in \{1,\ldots ,n\}$ such that $(g^r)_i\not =0$ .

Proof. The equivalence between conditions (1) and (3) follows from [Reference LindLin84, Theorem 3, p. 291], and the equivalence between conditions (2) and (3) follows from Lemma 3.2.2. The equivalence between conditions (1) and (4) can be found, for instance, in [Reference SchinzelSch97, Lemma 4] or [Reference BrunotteBru13, Theorem 2].

Proof. Note that the endomorphism $f_M \in \operatorname {\mathrm {End}}({\mathbb A}^n)$ restricts to an endomorphism $h_M \in \operatorname {\mathrm {End}}(({\mathbb A}^1 \setminus \{ 0 \})^n)$ .

If $\det (M)=0$ , any non-zero element of the kernel of the transpose of M gives rise to a non-constant element p in the Laurent polynomial ring $\mathbf {k}[x_1^{\pm }, \ldots , x_n^{\pm }]$ such that $p \circ h_M$ is constant, so $h_M$ , and thus $f_M$ , is not dominant. We then assume that $\det (M)\not =0$ . This implies that $h_M \in \operatorname {\mathrm {End}}(({\mathbb A}^1 \setminus \{0\})^n)$ is surjective on $\overline {\mathbf {k}}$ -points and thus $f_M$ is dominant. In particular, $\unicode{x3bb} (f_M) \geq 1$ . Thus we only have to show that $\unicode{x3bb} (f_M)=\rho (M)$ . By the Perron–Frobenius theorem (Theorem 3.2.3), there exists an eigenvector $\mu \in (\mu _1,\ldots ,\mu _n)\in ({\mathbb R}_{\ge 0})^n$ of M to the eigenvalue $\rho (M)$ . Because the spectral radius of M and the dynamical degree of $f_M$ do not change if we conjugate M with a permutation matrix, we may assume that there is $m < n$ such that $\mu _1 = \cdots = \mu _m = 0$ and $\mu _i> 0$ for each $i \geq m+1$ . Because $(f_M)^r = f_{M^r}$ , we get for each $r \geq 1$ and each $i \in \{1, \ldots , n\}$ that $\deg _\mu (((f_M)^r)_i) = (M^r \mu )_i = \rho (M)^r \mu _i$ . This implies that $\deg _\mu ((f_M)^r)=\rho (M)^r$ for each $r \geq 1$ . Thus $f_M$ is $\mu $ -algebraically stable and $\deg _\mu (f_M) = \rho (M) < \infty $ . By Corollary 2.6.2(1), we may write

$$ \begin{align*} M = \left( \begin{array}{c|c} \hat{M} & 0 \\ \hline \ast & \ast \end{array} \right), \end{align*} $$

where $\hat {M} \in \mathrm {Mat}_m({\mathbb N})$ with $\det (\hat {M}) \neq 0$ . By induction, the endomorphism $f_{\hat {M}} \in \operatorname {\mathrm {End}}({\mathbb A}^m)$ satisfies $\unicode{x3bb} (f_{\hat {M}}) = \rho (\hat {M}) \leq \rho (M)$ . By Corollary 2.6.2(2), (3), we get then $\unicode{x3bb} (f_M) = \deg _\mu (f_M) = \rho (M)$ .

Proof. By the Perron–Frobenius theorem (Theorem 3.2.3), there exists an eigenvector $\mu =(\mu _1,\ldots ,\mu _n)\in ({\mathbb R}_{\ge 0})^n$ of M to the eigenvalue $\rho (M)$ . Hence, $\sum _{j=1}^n m_{i,j}\mu _j=\rho (M)\mu _i$ for each $i\in \{1,\ldots ,n\}$ . By choosing an integer $r\in \{1,\ldots ,n\}$ such that $\mu _r=\max \{\mu _1,\ldots ,\mu _n\}$ , we obtain

$$ \begin{align*}\rho(M)\mu_r=\sum_{j=1}^n m_{r,j}\mu_j\le \mu_r\sum_{j=1}^n m_{r,j}.\end{align*} $$

The coefficient of the monomial $\prod _{j=1}^n x_j^{m_{r,j}}$ in $f_r$ is non-zero (as M is contained in f, see Definition 1.4.8). This monomial has degree $\sum _{j=1}^n m_{r,j}$ , so $\deg (f)\ge \sum _{j=1}^n m_{r,j}$ . As $\mu _r>0$ , this gives $\rho (M)\le \deg (f)$ .

Proof. Let $\unicode{x3bb} \in {\mathbb R}_{> 0}$ be a Handelman number. There exists $(a_0, \ldots , a_{n-1}) \in {\mathbb Z}^n \setminus \{ 0 \}$ such that $\unicode{x3bb} $ is a root of $P(x) = x^n - \sum _{i=0}^{n-1} a_i x^i \in {\mathbb Z}[x]$ . By Lemma 3.2.7, all roots of P, except $\unicode{x3bb} $ , are either non-real or real and non-positive. Because P is the characteristic polynomial of the matrix

$$ \begin{align*} A = \begin{pmatrix} a_{n-1} & \cdots & a_1 & a_0 \\ 1 & \cdots & 0 & 0 \\ \vdots & \ddots & \vdots & \vdots \\ 0 & \cdots & 1 & 0 \end{pmatrix} \in \mathrm{Mat}_n({\mathbb R}_{\geq 0}), \end{align*} $$

it follows by the Perron-Frobenius theorem (Theorem 3.2.3) that the spectral radius of A is equal to $\unicode{x3bb} $ . This implies that $\unicode{x3bb} $ is a weak Perron number (Theorem 3.2.4).

Notation 3.3.1. Let $n,N\ge 1$ . We denote by $\widehat {\mathcal {M}}_{n,N}\subset \mathrm {Mat}_n({\mathbb R})^N$ the ${\mathbb R}$ -vector subspace of N-tuples $(M_1,\ldots ,M_N)$ that have the following property:

For each $i,j\in \{1,\ldots ,N\}$ and each $l \in \{1,\ldots ,n\}$ , the replacement of the lth row of $M_i$ with the lth row of $M_j$ gives a matrix which lies in $\{M_1,\ldots ,M_N\}$ .

We then denote by $\mathcal {M}_{n,N}\subset \widehat {\mathcal {M}}_{n,N}$ the subset that consists of the N-tuples $(M_1,\ldots ,M_N)$ , where $M_1,\ldots ,M_N$ are N distinct matrices with non-negative coefficients.

Proof. The result being trivially true for $N=1$ , we will assume $N\ge 2$ . For each $i\in \{1,\ldots ,n\}$ , we denote by $\Gamma _i \subset {\mathbb R}^n$ the finite set of ith rows of the matrices $M_1,\ldots ,M_N$ :

$$ \begin{align*} \Gamma_i=\{r\in {\mathbb R}^n \mid r\text{ is the}\ i\text{th row of one of the matrices }M_1,\ldots,M_N\}.\end{align*} $$

We then write $\Gamma _i=\{r_{i,1},\ldots ,r_{i,s_i}\}$ , where $s_i\ge 1$ is the cardinality of $\Gamma _i$ .

As all matrices $M_1,\ldots ,M_N$ are pairwise distinct and as one can ‘exchange rows’ (see Notation 3.3.1), we have $N=s_1\cdot \cdots s_n$ , and obtain a unique ${\mathbb R}$ -linear map

$$ \begin{align*} \varphi\colon \prod_{i=1}^n ({\mathbb R}^n)^{s_i}\to \widehat{\mathcal{M}}_{n,N} \end{align*} $$

with the following properties.

1. (1) For each $k \in \{1, \ldots , N \}$ , the composition of $\varphi $ with the projection map $\pi _k \colon \mathrm {Mat}_n({\mathbb R})^N\to \mathrm {Mat}_n({\mathbb R})$ onto the kth factor is of the form

   $$ \begin{align*}\begin{array}{rccc} \pi_k\circ \varphi\colon& \prod_{i=1}^n ({\mathbb R}^n)^{s_i}&\to& \mathrm{Mat}_n({\mathbb R})\\ &(v_{i,j})_{1\le i\le n, 1\le j\le s_i} & \mapsto & \begin{pmatrix} v_{1, j_1} \\ \vdots \\ v_{n, j_n} \end{pmatrix}, \end{array}\end{align*} $$
   where $j_i\in \{1,\ldots ,s_i\}$ for each $i\in \{1,\ldots ,n\}$ .

2. (2) $(M_1,\ldots ,M_N)=\varphi ((r_{i,j})_{1\le i\le n, 1\le j\le s_i})$ .

Indeed, the possibilities for maps $\pi _k \circ \varphi $ as in property $(1)$ are parametrized by the N possible choices of $j_i\in \{1,\ldots ,s_i\}$ for each $i\in \{1,\ldots ,n\}$ , and by property $(2)$ , the image of $(r_{i,j})_{1\le i\le n, 1\le j\le s_i}$ by the maps $\pi _1 \circ \varphi , \ldots , \pi _N \circ \varphi $ give the matrices $M_1,\ldots ,M_N$ ; this gives the existence and the unicity of $\varphi $ .

We now identify $\prod _{i=1}^n ({\mathbb R}^n)^{s_i}$ with the real locus $X({\mathbb R})$ of the affine space $X=\mathbb {A}^{n\sum s_i}$ .

For any two matrices $A,B\in \mathrm {Mat}_n({\mathbb R})$ , the resultant of the characteristic polynomials $\chi _A$ and $\chi _B$ is denoted by $r(A,B)$ . Recall that $r(A, B) = 0$ if and only if A and B have a common eigenvalue. Hence, for any distinct $a,b \in \{1,\ldots ,N\}$ , the set

$$ \begin{align*} Z_{a,b}=\bigg\{x\in \prod_{i=1}^n ({\mathbb R}^n)^{s_i} \bigg| \begin{array}{l}\text{the matrices }\pi_{a}(\varphi(x))\text{ and }\pi_{b}(\varphi(x))\\ \text{have a common eigenvalue}\end{array}\!\!\bigg\} \end{align*} $$

corresponds to the elements of $X({\mathbb R})$ that satisfy one polynomial equation $P_{a,b}\in {\mathbb R}[X]$ .

We now prove that $P_{a,b}\not =0$ , or equivalently that $Z_{a,b}\not = X({\mathbb R})=\prod _{i=1}^n ({\mathbb R}^n)^{s_i}$ , by showing that $\pi _{a}(\varphi (x))$ and $\pi _{b}(\varphi (x))$ have no common eigenvalue for at least one $x\in X({\mathbb R})$ . We consider $j_1,\ldots ,j_n$ and $j_1',\ldots ,j_n'$ so that $\pi _a\circ \varphi $ and $\pi _b\circ \varphi $ are respectively given by

$$ \begin{align*}\begin{array}{ccc} \prod_{i=1}^n ({\mathbb R}^n)^{s_i}&\to& \mathrm{Mat}_n({\mathbb R})\\ (v_{i,j})_{\scriptsize\begin{array}{l} 1\le i\le n\\ 1\le j\le s_i\end{array}} & \mapsto & \begin{pmatrix} v_{1, j_1} \\ \vdots \\ v_{n, j_n} \end{pmatrix} \end{array} \qquad \textrm{and} \qquad \begin{array}{ccc} \prod_{i=1}^n ({\mathbb R}^n)^{s_i}&\to& \mathrm{Mat}_n({\mathbb R})\\ (v_{i,j})_{\scriptsize\begin{array}{l} 1\le i\le n\\ 1\le j\le s_i\end{array}} & \mapsto & \begin{pmatrix} v_{1, j_1'} \\ \vdots \\ v_{n, j_n'} \end{pmatrix}. \end{array} \end{align*} $$

Because the matrices $M_a$ and $M_b$ are distinct, the linear maps $\pi _a \circ \varphi $ and $\pi _b\circ \varphi $ are also distinct. Thus, there is $l\in \{1,\ldots ,n\}$ such that $j_l\not =j_l'$ . Suppose first that $l=1$ , that is, $j_1\not =j_1'$ . We may choose $x\in X({\mathbb R})$ such that

$$ \begin{align*} \pi_{a}(\varphi(x))=\left(\begin{array}{c|c} 0 & 1 \\ \hline I_{n-1} & 0 \end{array}\right) \quad \textrm{and} \quad \pi_{b}(\varphi(x))=\left(\begin{array}{c|c} 0 & 0 \\ \hline I_{n-1} & 0 \end{array}\right)\!. \end{align*} $$

These matrices have characteristic polynomials $t^n-1$ and $t^n$ , respectively. If $l>1$ , we simply consider conjugation of the above matrices by permutations. In all cases, we find an $x\in X({\mathbb R})$ such that $\pi _{{a}}(\varphi (x))$ and $\pi _{{b}}(\varphi (x))$ are matrices without common eigenvalue in $\mathbb {C}$ . This shows that $Z_{a,b} \neq X({\mathbb R})$ , that is, $P_{a,b} \neq 0$ .

The product of all polynomials $P_{a,b}$ with distinct $a,b \in \{1,\ldots ,n\}$ gives a non-zero polynomial $P\in {\mathbb R}[X]$ . Thus we can take a real affine linear map $\ell \colon \mathbb {A}^1 \to X =\mathbb {A}^{n\sum s_i}$ such that $\ell (0)=(r_{i,j})_{1\le i\le n, 1\le j\le s_i}$ , such that the coordinates of $\ell ({\mathbb R}_{\ge 0})$ are non-negative and such that the restriction of P to $\ell ({\mathbb R})$ is non-zero. We obtain that $P(\ell ({1}/{n})) \neq 0$ for any sufficiently large positive integer n. It suffices then to fix a sufficiently large $c\ge 1$ and to define $D_t=\varphi (\ell ({1}/(t+c)))$ for each integer $t\ge 0$ .

Proof. Assume for contradiction that there is $i\in \{2,\ldots ,N\}$ such that $S_i v \not \leq \unicode{x3bb} v$ . Denote by $v_j$ the jth component of v for each $j \in \{1, \ldots , n\}$ . Because we may replace each row $R_j$ in $S_i$ , such that $R_j v < \unicode{x3bb} v_j$ , with the jth row from $S_1$ and still get an element in $\{S_1, \ldots , S_N\}$ , we may assume that $S_i v \geq \unicode{x3bb} v \geq 0$ . As the coefficients of v and $S_i$ are non-negative, we obtain by induction that $(S_i)^r v \geq \unicode{x3bb} ^r v \geq 0$ for each $r\ge 1$ . In particular,

$$ \begin{align*} \lVert (S_i)^r\rVert \geq \frac{\lVert (S_i)^r v \rVert}{\lVert v \rVert} \geq \unicode{x3bb}^r, \end{align*} $$

and we obtain $\rho (S_i)=\lim _{r\to \infty } \lVert (S_i)^r \rVert ^{1/r} \geq \unicode{x3bb} $ . This contradicts the assumption that $\unicode{x3bb}>\rho (S_i)$ .

Proof. Let $S = (S_1, \ldots , S_N) \in \mathcal {M}_{n,N}$ . By Lemma 3.3.3, there exists a sequence ${(D_t)}_{t\in \mathbb {N}}$ of elements $D_t=(D_{t,1},\ldots ,D_{t,N})\in \mathcal {M}_{n,N}$ that converges towards S and such that for each $t\in \mathbb {N}$ , there is no complex number which is an eigenvalue of two elements of $D_{t,1},\ldots ,D_{t,N}$ . In particular, $\rho (D_{t, i}) \neq \rho (D_{t, j})$ for distinct $i, j$ by the Perron–Frobenius theorem (Theorem 3.2.3).

By possibly replacing this sequence with a subsequence, we may assume that there is a $j\in \{1,\ldots ,N\}$ such that $\rho (D_{t, j})> \rho (D_{t, i})$ for all $i \in \{1,\ldots ,N\}\setminus \{j\}$ and each $t\in \mathbb {N}$ . After exchanging the ordering of $S_1,\ldots ,S_N$ , we may assume that $j = 1$ . For each $i\in \{1,\ldots ,N\}$ , the sequence $(D_{t,i})_{t\in \mathbb {N}}$ converges towards $S_i$ , so $(\rho (D_{t,i}))_{t\in \mathbb {N}}$ converges towards $\rho (S_i)$ [Reference OstrowskiOst73, Theorem in Appendix A]. In particular, $\rho (S_1)=\unicode{x3bb} =\max \{\rho (S_1),\ldots ,\rho (S_n)\}$ . By the Perron–Frobenius theorem (Theorem 3.2.3), there is for each $t\in \mathbb {N}$ an eigenvector $v_t \geq 0$ of $D_{t,1}$ to the eigenvalue $\rho (D_{t,1})$ . Lemma 3.3.4 then gives for each $i\in \{1,\ldots ,N\}$ and each $t\in \mathbb {N}$ ,

$$ \begin{align*} D_{t,i} v_t \leq \rho(D_{t,1}) v_t. \end{align*} $$

Now, we may assume that $\lVert v_t \rVert = 1$ for all t (after normalizing $v_t$ ). Let

$$ \begin{align*} \mathbb{S}^{n-1} = \{{w \in {\mathbb R}^n} \ | \ {\lVert w \rVert =1}\}. \end{align*} $$

Because $\mathbb {S}^{n-1}$ is compact (with respect to the Euclidean topology), we may take a subsequence and assume that $(v_t)_{t\in \mathbb {N}}$ converges to a $v \geq 0$ in $\mathbb {S}^{n-1}$ . Thus we get

$$ \begin{align*} \unicode{x3bb} v=\rho(S_1) v = \lim_{t\to\infty} \rho(D_{t, 1}) v_t = \lim_{t \to \infty} D_{t, 1} v_t = S_1 v \end{align*} $$

and for each $i\in \{1,\ldots ,N\}$ ,

$$ \begin{align*} S_i v = \lim_{t\to\infty} D_{t, i} v_t \leq \lim_{t\to\infty} \rho(D_{t, 1}) v_t = \rho(S_1) v =\unicode{x3bb} v. \end{align*} $$

This finishes the proof of the proposition.

Proof of Proposition B

By Remark 3.3.2, there exists $(S_1, \ldots , S_N) \in \mathcal {M}_{n, N}$ such that $\{ S_1, \ldots , S_N \}$ is the set of matrices contained in f. By Proposition 3.4.1, there exists $j \in \{1, \ldots , N\}$ and an eigenvector $\mu =(\mu _1,\ldots ,\mu _n) \in ({\mathbb R}_{\geq 0})^n \setminus \{ 0 \}$ of $S_{j}$ to the eigenvalue $\theta =\max \{\rho (S_1),\ldots ,\rho (S_N)\}$ such that $S_i \mu \leq S_{j} \mu = \theta \mu $ for each $i \in \{1,\ldots ,N\}$ . We now prove that this implies that $\deg _\mu (f_l) = \theta \mu _l$ for each $l \in \{1, \ldots , n\}$ , which shows that $\mu =(\mu _1,\ldots ,\mu _n)$ is a maximal eigenvector of f, and thus proves assertion (1). For each monomial $m=\chi x_1^{r_1}\cdots x_n^{r_n}$ of $f_l$ with $\chi \in \mathbf {k}^*$ , there is a matrix $S_i$ with its lth line equal to $(r_1\ r_2\ \cdots \ r_n)$ . The lth component of $S_i \mu $ is equal to $r_1\mu _1+\cdots +r_n\mu _n=\deg _\mu (m)$ . The inequality $S_i\mu \le \theta \mu $ then yields $\deg _\mu (m)\le \theta \mu _l$ . As this holds for each monomial of $f_l$ , we obtain $\deg _\mu (f_l) \leq \theta \mu _l$ . The equality follows from $S_j\mu =\theta \mu $ , because the monomial m that corresponds to the lth row of $S_j$ has $\mu $ -degree equal to $\theta \mu _l$ .

We now prove assertion (2). The dominance of f implies that $1\le \deg (f^r)$ for each r and this in turn gives $1 \leq \unicode{x3bb} (f)$ . The inequality $\theta \le \deg (f)$ follows from Corollary 3.2.6, so we only need to prove $\unicode{x3bb} (f)\le \theta $ . This is done by induction on n. If $n = 1$ , then $\mu \in ({\mathbb R}_{>0})^1$ and the statement follows from Proposition A(2). Now, let $n> 1$ . We may assume (after a permutation of the coordinates) that $\mu _1 \leq \mu _2 \leq \cdots \leq \mu _n$ . Now, let $m \in \{0, \ldots , n-1\}$ with $\mu _i = 0$ for $i \leq m$ and $\mu _i> 0$ for $i> m$ . From Remark 2.3.1(2), we get

From Lemma 2.5.6, we get that for each $i \in \{1, \ldots , m \}$ , the element $f_i$ is a polynomial in the variables $\{x_1, \ldots , x_m \}$ . Thus we get from Lemma 2.4.1 that $\unicode{x3bb} (f) = \max \{\unicode{x3bb} _1, \unicode{x3bb} _2 \}$ , where

Because $m \leq n-1$ , by induction hypothesis we have

Note that each eigenvalue of a matrix that is contained in $\hat {f}$ is an eigenvalue of a matrix that is contained in f. Thus we get $\theta _1 \leq \theta $ . From Lemma 2.6.1(4), it follows that $\unicode{x3bb} _2 \leq \theta $ . In summary, we proved that $\unicode{x3bb} (f) = \max \{\unicode{x3bb} _1, \unicode{x3bb} _2\} \leq \theta $ , that is, assertion (2) holds for n.

We now prove assertion (3). We take a maximal eigenvector $\mu $ of f. As $\deg _\mu (f_i)= \theta \mu _i$ for each $i\in \{1,\ldots ,n\}$ , we have $\deg _\mu (f)=\theta $ . If $\theta =1$ , (i) follows from assertion (2) and (ii) is trivially true, so we may assume that $\theta>1$ . If f is $\mu $ -algebraically stable, then Lemma 2.6.1(5) gives $\unicode{x3bb} _2 = \theta $ and thus $\unicode{x3bb} (f) = \theta $ , so (i) is proven. Conversely, if $\mu \in ({\mathbb R}_{>0})^n$ and $\unicode{x3bb} (f)=\theta>1$ , then f is $\mu $ -algebraically stable by Proposition A(3). This achieves the proof of (3). As $\theta = \deg _\mu (f) \in {\mathbb R}_{\geq 0}$ (that is, is not equal to $+\infty $ ), (3) is a direct consequence of Lemma 2.6.1(5).

Proof. We take $\alpha \in \operatorname {\mathrm {Aff}}(\mathbb {A}^n)$ and $\tau \in \operatorname {\mathrm {TEnd}}(\mathbb {A}^n)$ and show that we can conjugate $f=\alpha \circ \tau $ to a permutation-triangular endomorphism by an element of $\operatorname {\mathrm {Aff}}(\mathbb {A}^n)$ .

Let $p=(p_1,\ldots ,p_n)\in \mathbb {A}^n$ be the point such that $\alpha (p)=0$ and consider the translation $\tau _p=(x_1+p_1,\ldots ,x_n+p_n) \in \operatorname {\mathrm {Aff}}(\mathbb {A}^n)\cap \operatorname {\mathrm {TAut}}(\mathbb {A}^n)$ . Then $\alpha '=\alpha \circ \tau _p \in \operatorname {\mathrm {Aff}}(\mathbb {A}^n)$ fixes the origin $(0,\ldots ,0)\in \mathbb {A}^n$ . We then replace $\alpha $ with $\alpha '$ and $\tau $ with $\tau _p^{-1} \circ \tau $ , and may assume that $\alpha $ belongs to the subgroup $\operatorname {\mathrm {GL}}_n=\operatorname {\mathrm {GL}}_n(\mathbf {k})\subset \operatorname {\mathrm {Aff}}(\mathbb {A}^n)$ of elements that fix the origin.

The group $B = \operatorname {\mathrm {TAut}}({\mathbb A}^n) \cap \operatorname {\mathrm {GL}}_n$ is a Borel subgroup of $\operatorname {\mathrm {GL}}_n$ . It consists of all lower triangular matrices. The so-called Bruhat decomposition of $\operatorname {\mathrm {GL}}_n$ :

$$ \begin{align*} \operatorname{\mathrm{GL}}_n = B \operatorname{\mathrm{Sym}}(\mathbb{A}^n) B \end{align*} $$

yields $\beta , \gamma \in B$ and $\sigma \in \operatorname {\mathrm {Sym}}(\mathbb {A}^n)$ such that $\alpha = \beta \circ \sigma \circ \gamma $ . This gives

$$ \begin{align*} \beta^{-1} \circ f \circ \beta =\beta^{-1} \circ \alpha \circ \tau \circ \beta = \sigma \circ \gamma \circ \tau \circ \beta, \end{align*} $$

where $\gamma \circ \tau \circ \beta \in \operatorname {\mathrm {TEnd}}({\mathbb A}^n)$ . This achieves the proof.

Proof. We write $h=\sigma \circ \tau $ , where $\sigma \in \operatorname {\mathrm {Sym}}(\mathbb {A}^{n+1})$ and $\tau \in \operatorname {\mathrm {EAut}}(\mathbb {A}^{n+1})$ . We may choose $\alpha = (\alpha _1, \ldots , \alpha _{n+1})\in \operatorname {\mathrm {Sym}}(\mathbb {A}^{n+1})$ such that $\alpha _{n+1} = x_{n+1}$ and $\alpha \circ \sigma \circ \alpha ^{-1}$ induces the following cyclic permutation on the last coordinates

$$ \begin{align*} (\alpha \circ \sigma \circ \alpha^{-1})_{m+1} = x_{n+1}, \ (\alpha \circ \sigma \circ \alpha^{-1})_{m+2} = x_{m+1}, \ \ldots, \ (\alpha \circ \sigma \circ \alpha^{-1})_{n+1} = x_{n}, \end{align*} $$

for some integer m with $0\le m\le n$ . This gives

$$ \begin{align*} \alpha \circ \sigma \circ \alpha^{-1} = (f_1, \ldots, f_m, x_{n+1}, x_{m+1}, \ldots, x_n), \end{align*} $$

where $\{ x_1, \ldots , x_m \} = \{f_1, \ldots , f_m \}$ . As $\alpha _{n+1}=x_{n+1}$ , we obtain

$$ \begin{align*} \alpha \circ \tau \circ \alpha^{-1} = (x_1, \ldots, x_{n}, \xi x_{n+1} + p(x_1, \ldots, x_n)) \end{align*} $$

for some $\xi \in \mathbf {k}^\ast $ and $p \in \mathbf {k}[x_1, \ldots , x_n]$ . This implies that $\alpha \circ h \circ \alpha ^{-1}$ is equal to

$$ \begin{align*} &(\alpha \circ \sigma \circ \alpha^{-1}) \circ (\alpha \circ \tau \circ \alpha^{-1}) \\ &\qquad= (f_{1},\ldots,f_{m}, \xi x_{n+1}+p(x_1,\ldots,x_{n}),x_{m+1},\ldots,x_{n}). \\[-40pt] \end{align*} $$

Proof. For each $r\ge 1$ , we write $g^r=((g^r)_1,\ldots ,(g^r)_{n+1})$ . The result is true by assumption when $r=1$ . For each $r\ge 1$ and $1 \leq i \leq m$ , we have $(g^r)_i = (\hat {f}^r)_i \not =0$ .

As $(f_1,\ldots ,f_m)\in \operatorname {\mathrm {Aut}}(\mathbb {A}^m)$ , we also have $(f_1,\ldots ,f_m,x_{m+1},\ldots ,x_{n})\in \operatorname {\mathrm {Aut}}(\mathbb {A}^{n})$ . In particular, g is dominant if $q\not \in \mathbf {k}[x_1,\ldots ,x_n]$ , that is, if $\deg _{x_{n+1}}(q)\ge 1$ . Thus we assume that $q \in \mathbf {k}[x_1, \ldots , x_n] \setminus \{0\}$ .

Suppose first that $m=n$ , in which case $g=(f_1,\ldots ,f_m,q)$ . For each $r\ge 2$ , we get $g^r=((g^r)_1,\ldots ,(g^r)_{m},q((g^{r-1})_1,\ldots ,(g^{r-1})_{m}))$ . As $\hat {f}$ is dominant and q is not the zero polynomial, every component of $g^r$ is not zero.

We then assume that $n>m$ and prove the result by induction on $n-m$ . As $(f_1,\ldots ,f_m,x_{m+1},\ldots ,x_{n})\in \operatorname {\mathrm {Aut}}(\mathbb {A}^{n})$ , there is a polynomial $h\in \mathbf {k}[x_1,\ldots ,x_n]$ such that $h(f_1,\ldots ,f_m,x_{m+1},\ldots ,x_n)=q$ , because $q \in \mathbf {k}[x_1,\ldots ,x_n]$ . We denote by $\phi \colon \mathbb {A}^n\hookrightarrow \mathbb {A}^{n+1}$ the closed embedding that is given by

$$ \begin{align*} (x_1,\ldots,x_n)\mapsto (x_1,\ldots,x_m,h(x_1,\ldots,x_n),x_{m+1},\ldots,x_n) \end{align*} $$

and we write $\tau =(f_1,\ldots ,f_m,h,x_{m+1},\ldots ,x_{n-1})\in \operatorname {\mathrm {End}}(\mathbb {A}^n)$ . We now prove that $g\circ \phi =\phi \circ \tau $ :

$$ \begin{align*} &g\circ \phi(x_1,\ldots,x_n) \\ &=(f_1,\ldots,f_m,q(x_1,\ldots,x_m,h,x_{m+1},\ldots,x_{n-1}),h,x_{m+1},\ldots,x_{n-1})\\ &=(f_1,\ldots,f_m,h(f_1,\ldots,f_m,h,x_{m+1},\ldots,x_{n-1}),h,x_{m+1},\ldots,x_{n-1})\\ &=\phi\circ \tau(x_1,\ldots,x_n). \end{align*} $$

Hence, $g^r\circ \phi =\phi \circ \tau ^r$ for each $r\ge 1$ . By induction, every component of $\tau ^r$ is non-zero, so every component of $g^r$ is non-zero, except maybe the $(m+1)$ th component. However, if the $(m+1)$ th component of $g^r$ were zero, then the $(m+2)$ th of $g^{r+1}$ would be zero, which is impossible as the $(m+2)$ th component of $g^{r+1}\circ \phi =\phi \circ \tau ^{r+1}$ is not equal to zero.