0

I would like a hint on how to start this problem:

If $\sigma _m $ takes polynomials in $\mathbb{Z}$ and maps them to $\mathbb{Z}_m$ I want to show that if $\sigma _m (f(x))$ and $f(x)$ have the same degree and $\sigma _m (f(x))$ does not factor in $\mathbb{Z}_m[x]$ then $f(x)$ is irreducible in $\mathbb{Q}[x]$.

I have already shown that $\sigma _m$ is a ring homomorphism.

Mathematical Dummy
  • 326
  • 1
  • 6
  • Assume it does factor in $\mathbb Z[x]$, then ... – Torsten Schoeneberg Nov 04 '22 at 00:32
  • 2
    Assume that $f=gh$ in $\mathbb{Z}[x]$, then because $\sigma_m$ is a ring homomorphism, we have $\bar{f}=\bar{g}\bar{h}$, now argue that $\deg(f)=\deg(\bar{f})$, similarly for $g$. This would give you a factorization of $f$ in $\mathbb{Z}/m\mathbb{Z}$, a contradiction. – Takamoto Yuji Nov 04 '22 at 06:12
  • 2
    i think it is worth mentioning that for your polynomial irreducibility in $\mathbb{Z}[x]$ is equivalent to irreducibility in $\mathbb{Q}[x]$ due to an important lemma known as Gauss Lemma – Aleksei Kubanov Nov 04 '22 at 07:32
  • https://math.stackexchange.com/questions/2606220 – Sil Nov 20 '22 at 06:45

0 Answers0