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Show that the reduction of $f(x)=x\in \mathbb{Z}/6\mathbb{Z}[x]$ modulo $(2)$ and $(3)$ is an irreducible polynomial.

I'm unsure of how to solve this one. It's clear that if $\bar{x}= a(x)b(x)$, then $a(x)$ and $b(x)$ need to have constant coefficients whose product is divisible by $2$ (or $3$). But I don't see how to use this. I'd appreciate any help.

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    Are you looking for irreducibility of $\overline{x}$ in $(Z/6Z)/ (2)$ and in $(Z/6Z)/ (3)$? – 2017-02-13
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    @pepa.dvorak Yes – 2017-02-13

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We have $\mathbb{Z}/6\mathbb{Z}[x]/(2)\cong \mathbb{F}_2[x]$, and $x$ is irreducible in $\mathbb{F}_2[x]$, because $x=a(x)b(x)$ gives $1={\rm deg}(a)+{\rm deg(b)}$, and hence either ${\rm deg}(a)=0,{\rm deg}(b)=1$, or ${\rm deg}(b)=0,{\rm deg}(a)=1$. Here we use that $\mathbb{F}_2$ is a field. The same method works modulo $(3)$.

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    Thank you, but how do we know $\deg(a) +\deg(b) =1$. Could't some coefficients cancel after being added modulo $(2)$? – 2017-02-13
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    No, if $ab=0$ in a field, then $a=0$ or $b=0$. So we have $deg(fg)=deg(f)+deg(g)$ over fields. For $\mathbb{Z}/6$ this is not true. For example, $x=(2x+3)(3x+2)$ in this case. – 2017-02-13