在(e)中对$1\le i\le m$令$g'_i = -g_i$可得$f(-g_1,\cdots,-g_m) = (-1)^Nf(g_1,\cdots,g_m) + O(2N')$.

设$f\in A[[X_1,\cdots,X_m]]$满足$f = O(N)$, $g_1,\cdots,g_m\in A[[X_1,\cdots,X_n]]$满足对任意$1\le j\le m$有$g_j = O(N')$. 由形式幂级数复合的结合律、线性性和命题\ref{2.1.3}下面的注记 (ii) (分别令$\bm{l} = (1,0,\cdots,0),\cdots,(0,\cdots,0,1)$)有 \begin{align*} &f(ag_1,\cdots,ag_m) - a^Nf(g_1,\cdots,g_m) \\ = &f(aX_1,\cdots,aX_m)(g_1,\cdots,g_m) - (a^Nf)(X_1,\cdots,X_m)(g_1,\cdots,g_m) \\ = &(f(aX_1,\cdots,aX_m) - (a^Nf)(X_1,\cdots,X_m))(g_1,\cdots,g_m), \end{align*} 其中若$A$无单位元, 则$ (aX_j)^{m}_{j=1}$的含义见第\pageref{prod}页. 容易证明$f(aX_1,\cdots,aX_m) - (a^Nf)(X_1,\cdots,X_m) = O(N+1)$, 因此由形式幂级数复合的阶有 \begin{equation*} f(ag_1,\cdots,ag_m) = a^Nf(g_1,\cdots,g_m) + O((N+1)N'). \end{equation*}

\begin{equation*} \displaystyle{\lim_{n\to\infty}} x_n = a\Longleftrightarrow \displaystyle{\lim_{n\to\infty}} d(x_n,a)=0, \end{equation*}

\begin{equation*} \sigma|_{K'}\in \operatorname{Aut}(K'/K)_n \Longleftrightarrow i\le \psi_{L/K'}(i_0); \end{equation*}

\begin{equation*} \sigma|_{K'}\in \operatorname{Aut}(K'/K)_{\lceil \psi_{L/K'}(x)\rceil} \Longleftrightarrow \lceil \psi_{L/K'}(x)\rceil\le \psi_{L/K'}(i_0), \end{equation*} 同时 \begin{equation*} \sigma_0\in \operatorname{Aut}(L/K)_{\lceil x\rceil} \Longleftrightarrow \lceil x\rceil\le i_0. \end{equation*}

容易证明 \begin{equation} \label{form1.17} K_{m'}\subset K_m \Longleftrightarrow m'|m: \end{equation}

有$L = K(\pi)$和$\ocal_L = \ocal_K[\pi]$, 即$\ocal_L$由$\ocal_K\cup\{\pi\}$生成的子环就是自身;

\begin{align*}
&\ ((X-\alpha)^{-1}f)_k=-(\sum^k_{i=0}(f)_i\alpha^{i-(k+1)})=\frac{-\sum^k_{i=0}(f)_i\alpha^i}{\alpha^{k+1}}\\
=&\ \frac{\sum^\infty_{i=k+1}(f)_i\alpha^i}{\alpha^{k+1}}=\sum^\infty_{i=0}(f)_{i+k+1}\alpha^i\in A,
\end{align*}