证法一:因为\(\left(f_i(x),g_j(x)\right)=1,\forall i,j=1,2\),所以可设\(f_1(x),f_2(x),g_1(x),g_2(x)\)的分解式为(标准分解式适当乘以一些0次项)
\begin{equation*}
\begin{array}{c}
f_1(x)=ap_1^{a_1}(x)p_2^{a_2}(x)\cdots p_s^{a_s}(x),\quad
f_2(x)=bp_1^{b_1}(x)p_2^{b_2}(x)\cdots p_s^{b_s}(x),\\
g_1(x)=cq_1^{c_1}(x)q_2^{c_2}(x)\cdots q_t^{c_t}(x),\quad
g_2(x)=dq_1^{d_1}(x)q_2^{d_2}(x)\cdots q_t^{d_t}(x),
\end{array}
\end{equation*}
其中\(p_1(x),p_2(x),\cdots ,p_s(x),q_1(x),q_2(x),\cdots ,q_t(x)\)是\(\mathbb{F}\)上两两互素的首一不可约多项式,\(a_i,b_i,c_j,d_j\geq 0\)且\(a_i+b_i>0,c_j+d_j>0,(i=1,\ldots,s,j=1,\ldots ,t)\),则
\begin{equation*}
\left(f_1(x),f_2(x)\right)=p_1^{k_1}(x)p_2^{k_2}(x)\cdots p_s^{k_s}(x),\left(g_1(x),g_2(x)\right)=q_1^{l_1}(x)q_2^{l_2}(x)\cdots q_t^{l_t}(x),
\end{equation*}
其中\(k_i=\min\{a_i,b_i\},l_j=\min\{c_j,d_j\},\forall i=1,\ldots,s,j=1,\ldots ,t\)。注意到
\begin{equation*}
\begin{array}{c}
f_1(x)g_1(x)=acp_1^{a_1}(x)p_2^{a_2}(x)\cdots p_s^{a_s}(x)q_1^{c_1}(x)q_2^{c_2}(x)\cdots q_t^{c_t}(x),\\
f_2(x)g_2(x)=bdp_1^{b_1}(x)p_2^{b_2}(x)\cdots p_s^{b_s}(x)q_1^{d_1}(x)q_2^{d_2}(x)\cdots q_t^{d_t}(x),
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccl}\Rightarrow\left(f_1(x)g_1(x),f_2(x)g_2(x)\right)&=&p_1^{k_1}(x)p_2^{k_2}(x)\cdots p_s^{k_s}(x)q_1^{l_1}(x)q_2^{l_2}(x)\cdots q_t^{l_t}(x)\\&=&\left(f_1(x),f_2(x)\right)\left(g_1(x),g_2(x)\right)\end{array}
\end{equation*}
证法二:设
\begin{equation*}
\left(f_1(x),f_2(x)\right)=d_1(x),\left(g_1(x),g_2(x)\right)=d_2(x),
\end{equation*}
则存在\(h_1(x),h_2(x),h_3(x),h_4(x)\in\mathbb{F}[x]\),使得
\begin{equation*}
f_1(x)=d_1(x)h_1(x),f_2(x)=d_1(x)h_2(x),g_1(x)=d_2(x)h_3(x),g_2(x)=d_2(x)h_4(x),
\end{equation*}
其中\(\left(h_1(x),h_2(x)\right)=1,\left(h_3(x),h_4(x)\right)=1\)。由条件\(\left(f_i(x),g_j(x)\right)=1,\forall i,j=1,2\)知
\begin{equation*}
\left(h_i(x),h_j(x)\right)=1,\forall i=1,2,j=3,4,
\end{equation*}
根据
推论 5.3.12 ,得
\(\left(h_1(x)h_3(x),h_2(x)h_4(x)\right)=1\)。因此
\begin{equation*}
\begin{array}{ccl}\left(f_1(x)g_1(x),f_2(x)g_2(x)\right)&=&\left(d_1(x)h_1(x)d_2(x)h_3(x),d_1(x)h_2(x)d_2(x)h_4(x)\right)\\&=&d_1(x)d_2(x)\left(h_1(x)h_3(x),h_2(x)h_4(x)\right)\\
&=&d_1(x)d_2(x),\end{array}
\end{equation*}
即\(\left(f_1(x)g_1(x),f_2(x)g_2(x)\right)=\left(f_1(x),f_2(x)\right)\left(g_1(x),g_2(x)\right)\)。