设
\begin{equation*}
\left(f_1(\lambda),f_2(\lambda)\right)=d_1(\lambda),\left(g_1(\lambda),g_2(\lambda)\right)=d_2(\lambda),
\end{equation*}
则存在多项式\(h_1(\lambda),h_2(\lambda),k_1(\lambda),k_2(\lambda)\),使得
\begin{equation*}
f_1(\lambda)=d_1(\lambda)h_1(\lambda),f_2(\lambda)=d_1(\lambda)h_2(\lambda),
\end{equation*}
\begin{equation*}
g_1(\lambda)=d_2(\lambda)k_1(\lambda),g_2(\lambda)=d_2(\lambda)k_2(\lambda),
\end{equation*}
其中\(\left(h_1(\lambda),h_2(\lambda)\right)=1,\left(k_1(\lambda),k_2(\lambda)\right)=1\)。由条件\(\left(f_i(\lambda),g_j(\lambda)\right)=1,\ (i,j=1,2)\)知 \(\left(h_i(\lambda),k_j(\lambda)\right)=1,(i,j=1,2)\), 则\(\left(h_1(\lambda)k_1(\lambda),h_2(\lambda)k_2(\lambda)\right)=1\)。因此
\begin{equation*}
\begin{array}{cl}
&\left(f_1(\lambda)g_1(\lambda),f_2(\lambda)g_2(\lambda)\right)\\
=&\left(d_1(\lambda)h_1(\lambda)d_2(\lambda)k_1(\lambda),d_1(\lambda)h_2(\lambda)d_2(\lambda)k_2(\lambda)\right)\\
=&d_1(\lambda)d_2(\lambda)\left(h_1(\lambda)k_1(\lambda),h_2(\lambda)k_2(\lambda)\right)\\
=&d_1(x)d_2(x),\end{array}
\end{equation*}
结论成立。