\(\mathscr{Df1}:\varepsilon\)是\(n\)次单位根，即\(\varepsilon^n=1\)，则存在最小的正整数\(k\)使得\(\varepsilon^k=1\)（由带余除法，\(k\mid n\)），则称\(k\)为\(\varepsilon\)的阶，记作\(\operatorname{ord}(\varepsilon)=k\)。特别地，若\(\operatorname{ord}(\varepsilon)=n\)，则称\(\varepsilon\)为\(n\)次本原单位根（有\(\varphi(n)\)个）。
\(\mathscr{Df2}:\varepsilon_n=e^{\frac{2\pi i}{n}},\Phi_n(x)\triangleq\prod\limits_{1\le k\le n,\gcd(n,k)=1}(x-\varepsilon_n^k)=\prod\limits_{\operatorname{ord}(\varepsilon)=n}(x-\varepsilon)\)称为\(n\)次分圆多项式。（第二个等号是因为当且仅当\(n,k\)互素时，\(e^{\frac{2ik\pi}{n}}\)为本原单位根）
性质：
- \(\Phi_n(x)\)为首一的复系数多项式；
- 对于素数\(p\)，\(\Phi_{p^k}(x)=\frac{x^{p^k}-1}{x^{p^{k-1}}-1}\)。
\(\mathscr{Thm1}:x^n-1=\prod\limits_{d\mid n}\Phi_d(x)\)
\(\mathscr{Pf}:x^n-1=\prod\limits_{\varepsilon^n=1}(x-\varepsilon)=\prod\limits_{d\mid n}\prod\limits_{\operatorname{ord}(\varepsilon)=d}(x-\varepsilon)=\prod\limits_{d\mid n}\Phi_d(x)\quad\Box\)
\(\mathscr{Thm2}:\forall n\ge2,\Phi_n(0)=1\)
\(\mathscr{Pf1}:\Phi_n(0)=\prod\limits_{\operatorname{ord}(\varepsilon)=n}-\varepsilon=(-1)^{\varphi(n)}\cdot\prod\limits_{1\le k\le \frac{n}{2},\gcd(k,n)=1}(\varepsilon_n^k\cdot\varepsilon_n^{n-k})=(-1)^{\varphi(n)}=1(n\ge3);\Phi_2(0)=0+1=1\quad\Box\)
\(\mathscr{Pf2}:\)对\(n\)归纳。\(n=2\)时\(\Phi_2(0)=0+1=1\)。设\(<n\)成立，\(n\)时，\(\Phi_n(0)=\frac{0^n-1}{\prod\limits_{d\mid n,d\ne n}\Phi_d(0)}=\frac{-1}{0-1}=1\quad\Box\)
\(\mathscr{Thm3}:\Phi_n(x)\in\mathbb{Z}[X]\)
\(\mathscr{Pf}:\)对\(n\)归纳。\(n=1\)时\(\Phi_1(x)=x-1\)，下设\(<n\)成立，\(n\)时，记\(\prod\limits_{d\mid n,d\ne n}\Phi_d(x)=g(x)\)，则\(\Phi_n(x)\cdot g(x)=x^n-1\)，由归纳假设，\(g(x)\)为首一的整系数多项式。只需证明下面的引理就证明了原命题：
\(\mathscr{Lm}:f\in\mathbb{Z}[X],g\in\mathbb{C}[X],\)二者皆首一，且\(fg\in\mathbb{Z}[X]\)，则\(g\in\mathbb{Z}[X]\)
\(\mathscr{Pf}:令g(x)=\sum\limits_{i=0}^ng_i x^i(g_i\in\mathbb{C},g_n=1),f(x)=\sum\limits_{i=0}^mf_ix^i(f_i\in\mathbb{Z},f_m=1)，则由题，考虑次数为m+n-1的项知g_{n-1}\in\mathbb{Z},之后同理归纳即证。\quad\Box\)
\(\mathscr{Thm4}：\Phi_n(x)在\mathbb{Z}[X]中不可约。\)
