证明多项式不可约

\section*{证明多项式不可约}设$p$为素数．求证：$f(x)=x^{p-1}+2x^{p-2}+\cdots+p$在有理数域上不可约．\begin{proof}证明以下两个引理：\begin{lemma}若多项式$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}$的所有系数均为正实数．则$f(x)$的所有复根$z$均满足$|z|\ge \min\limits_{1\le k\le n}\dfrac{a_{k-1}}{a_{k}}$．\end{lemma}\begin{proof}若不然，设$m=\min\limits_{1\le k\le n}\dfrac{a_{k-1}}{a_{k}}$，则存在某个$f(x)$的复根$z$满足：对任意$1\le k\le n$，都有$|z|<m$．由$f(z)=0$知：$(z-m)f(z)=0$，即$$a_{n}z^{n+1}+(a_{n-1}-ma_{n})z^{n}+\cdots+(a_{0}-ma_{1})z-ma_{0}=0$$于是$$\begin{aligned}|ma_{0}|=&|a_{n}z^{n+1}+(a_{n-1}-ma_{n})z^{n}+\cdots+(a_{0}-ma_{1})z|\\\le&a_{n}|z^{n+1}|+(a_{n-1}-ma_{n})|z^{n}|+\cdots+(a_{0}-ma_{1})|z|\\<&a_{n}m^{n+1}+(a_{n-1}-ma_{n})m^{n}+\cdots+(a_{0}-ma_{1})m\\=&ma_{0}\end{aligned}$$矛盾，于是引理1证毕．\end{proof}\newpage\begin{lemma}$f(x)$为首一的整系数多项式，若$f(x)$的所有复根模长均大于$1$，且$|f(0)|$为素数，则$f(x)$在有理数域上不可约．\end{lemma}\begin{proof}由高斯引理，只需证$f(x)$在整数域上不可约．\\若不然，设$f(x)=g(x)h(x)$，其中$g(x),h(x)$均为整系数多项式．\\由$|f(0)|=|g(0)|\cdot |h(0)|$为素数，可不妨设$|g(0)|=1$．\\设$g(x)$的所有复根为$z_{1},z_{2},\cdots,z_{k}$，由韦达定理：$$|g(0)|=|z_{1}z_{2}\cdots z_{k}|=|z_{1}|\cdots|z_{k}|>1$$矛盾，于是$f(x)$在有理数域上不可约，引理2证毕．\end{proof}回到原题，对$f(x)=x^{p-1}+2x^{p-2}+\cdot+p$，由引理1知，$f(x)$的所有复根$z$均满足$$|z|\ge\min\limits_{1\le k\le p-1}\dfrac{p+1-k}{p-k}=\dfrac{p}{p-1}>1$$又$|f(0)=p$为素数，由引理2知$f(x)$在有理数域上不可约，证毕．\end{proof}