<s>[INST] Irreducible factorisation of polynomial over quotient field Let $F=\mathbb{Z}_3[x]/<x^2+1>$.
Factor $x^4+2$ into irreducibles in $F[x]$.
I know that $F$ is a field since $x^2+1$ is irreducible.
The usual way to find out that a polynomial is irreducible is that it has no roots. But how to do it if the element of F is now of the form $g(x)+<x^2+1>$ where $g(x)\in\mathbb{Z}_3[x]$ and the zero element of $F$ is $<x^2+1>$. Any help is appreciated :)
 [/INST] we find that $F = \mathbb{Z}_3[x] / (x^2+1)$ (which is the field with $9$ elements) has a cyclic multiplicative group with $8$ elements, whose generator is $x+1$ 
(proof : $(x+1)^2 = 2x$, $(2x)^2 = 2$, $2^2 = 1$, hence $(x+1)^8 = 1$ but $(x+1)^4 \ne 1$)
hence in $F$ the equation $y^4 = 1$ has $4$ solutions for $y$ : $$(x+1)^2 = 2x, \qquad(x+1)^4 = 2, \qquad (x+1)^6 = x, \qquad (x+1)^8 = 1$$ and that's cool because it means that in $F[y]$, the polynomial $y^4+2$ has $4$ distinct roots, hence :
$$y^4+2 = (y-2x)(y-2)(y-x)(y-1) = (y+x)(y+1)(y+2x)(y+2)$$
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