Let \(p\) be prime and let \(\zeta_p\) be a primitive \(p\)th root of unity. Is the homomorphism \[ \ZZ[t]/(t^p-1)\longrightarrow \ZZ[\zeta_p], \qquad t\longmapsto \zeta_p, \] an isomorphism?
Solution
No. The map is surjective but not injective.
Let \[ \Phi_p(t)=1+t+\cdots+t^{p-1}. \] Since \(\zeta_p\) is primitive, its minimal polynomial over \(\QQ\) is \(\Phi_p(t)\). Hence the evaluation map \[ \operatorname{ev}_{\zeta_p}:\ZZ[t]\longrightarrow \ZZ[\zeta_p] \] has kernel \((\Phi_p(t))\) and is surjective.
Because \[ t^p-1=(t-1)\Phi_p(t), \] we have \((t^p-1)\subseteq(\Phi_p(t))\). Therefore evaluation factors through \[ \bar{\operatorname{ev}}_{\zeta_p}:\ZZ[t]/(t^p-1)\longrightarrow\ZZ[\zeta_p]. \] Its kernel is \[ \frac{(\Phi_p(t))}{(t^p-1)}, \] which is nonzero: the class of \(\Phi_p(t)\) is not zero modulo \((t^p-1)\) because \(\deg\Phi_p=p-1<p\).
Thus \[ \ZZ[t]/(t^p-1) ot\cong\ZZ[\zeta_p] \] via this map. In fact \[ \ZZ[\zeta_p]\cong\ZZ[t]/(\Phi_p(t)). \]