Let $f(x)=x^3+2x^2+1 \in \Bbb{F}_3[x]$. Let $F=\Bbb{F}_3[x]/(f(x))$ and $\alpha=x+f(x) \Bbb{F}_3 \in F$, so that $f(\alpha)=0$ Prove that f(x) is irreducible.
Gauss' lemma: suppose $f \in \Bbb{Z}[x]$ is monic of degree>0. Then f is irreducible in $\Bbb{Z}[x]$ if and only if it is irreducible when viewed as an element of $\Bbb{Q}[x]$.
Theorem:(reduction mod p). Suppose $f \in \Bbb{Z}[x]$ is a monic polynomial of degree >0. Set $f_p \in \Bbb{Z}_{mod p}[x]$ to be the reduction mod p of f. If $f_p \in \Bbb{Z}_{mod p}[x]$ is irreducible for some prime p, then f is irreducible in $\Bbb{Z}[x]$
Can I directly argue with the second theorem since $f_3$ is irreducible by that theorem?
Since $f$ has degree $3$, $f$ is reducible in $\Bbb F_3[x]$ if and only if $f$ has a root in $\Bbb F_3$. A direct computation shows that any of $0,1,2\in\Bbb F_3$ is a root of $f$, hence $f$ is irreducible in $\Bbb F_3[x]$.