Let $f(x) = x^4 + x^2 + x + 1 \in \mathbb{Z}_{3}[x] $
Show that $f$ is irreducible over $\mathbb{Z}_{3}$, then factor $f$ over $K = \frac{\mathbb{Z}_{3}[x]}{(f(x))}$.
I already showed that f is irreducible over $\mathbb{Z}_{3}$ but I can't understand how it can be factorised over K when it irreducible over $\mathbb{Z}_{3}$.
Can someone explain the difference between reducibility in $\mathbb{Z}_{3}$ and in K ?
Thanks!
On
Hint: one of the roots of the polynomial $\;f\;$ over $\; K:=\Bbb F_3[x]/\langle f\rangle\;$ is the element $\;\alpha:=x+\langle f\rangle\in\Bbb K\;$, thus over this same field:
$$f(x)=(x-\alpha)q(x)\,,\;\;q(x)\in\Bbb K[x]\,,\,\,\deg q=3\implies q(x)=x^3+ax^2+bx+c\,,\,\,a,b,c\in\Bbb K$$
Now compare coefficients in each side, for example:
$$x^4+x^2+x+1=(x-\alpha)(x^3+ax^2+bx+c)\implies\begin{cases}a-\alpha=0\iff a=\alpha \\{}\\ b-\alpha^2=1\implies b=1+\alpha^2\\{}\\\text{and etc.}\end{cases}$$
The first two lines above are gotten from comparing the coefficients of $\;x^3\,,\,\,x^2\;$
The difference is as Arthur commented, and also like the difference of what happens over the rationals $\;\Bbb Q\;$ and the reals $\;\Bbb R\;$ with say$\;g(x)=x^2-2\;$, which is irreducible over the former but over the latter $\;g(x)=\left(x-\sqrt2\right)\left(x+\sqrt 2\right)\;$ . And of course $\;g(x)\in\Bbb Q[x]\;$
$K$ is a finite field hence a splitting field of $f$ over $\mathbb{Z}_3$. Hence the Galois group $G$ of the field extension $K/\mathbb{Z}_3$ is generated by the Frobenius automorphism $y\mapsto y^3$ hence if $\alpha=x+\langle f\rangle$ the roots of $f$ in $K$ are $\alpha$, $\alpha^3$, $\alpha^{3^2}$ and $\alpha^{3^3}$ hence $f$ factors in $K$ as $$f(y)=(y-\alpha)(y-\alpha^3)(y-\alpha^{3^2})(y-\alpha^{3^3}).$$ The comment from Arthur gives a nice example for your first question.
See for example this pdf by KConrad in case you wish to see details.