Reduction modulo p - common divisor Let $\mathbb{Z}[X]\longrightarrow \mathbb{Z}/p\mathbb{Z}[X]=\mathbb{F}_p[X], \sum c_iX^i\mapsto\sum \overline{c_i}X^i$ be the homomorphism "reduction modulo p" with $p$ being a prime. We furthermore have, $h(X^p)=f\cdot g$ where each polynomial is monomial. We also have $deg f=n$ and $p$ does not divide $n$. My book states: "The equation $\overline{h}(X^p)=\overline f \cdot \overline g $ shows that $\overline h$ and $\overline f$ have a common divisor in $\mathbb{F_p}[X]$." I can not figure out why this holds. I might have overseen something obvious, but I don't even have an idea. EDIT: Also, $\zeta$ is a root of both $f$ and $h(X^p)$.

Interesting!

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We just need to prove: $$h(x^p)\in (\bar{h}(x)),$$ where $(\bar{h}(x))$ is an idea. In fact, as Krish's comment, $\bar{h}(x^P)=(\bar{h}(x))^p$.

This is because we can use algebraic closed field to analysis. And $h(x)=(x+\lambda_1)\cdots(x+\lambda_n), h(x^p)=(x^p+\lambda_1)\cdots(x^p+\lambda_n)$.

As fermat's small theorem : $$ \lambda^p\equiv\lambda (mod\quad p)$$ Therefore, $$(x^p+\lambda)=(x+\lambda)^p,$$ in $Z_p[x]$.So $\bar{h}(x^P)=(\bar{h}(x))^p$.