What is an irreducible element in $\Bbb{Z}_6[x]$?

Question

What is an irreducible element in $\Bbb{Z}_6[x]$?

This was a problem on our final and no one knew how to solve it. Does anyone have a method for solving this?

Answer 1

Claim. Assume that $\Bbb Z_m=\Bbb Z/m\Bbb Z$. The polynomial $f(x)\in\Bbb Z_6[x]$ is irreducible if and only if exactly one of these is true:
$(a)$ $f(x)$ is irreducible over $\Bbb Z_2$ and $f(x)\equiv \pm1\pmod{3}$ or
$(b)$ $f(x)$ is irreducible over $\Bbb Z_3$ and $f(x)\equiv 1\pmod{2}$.

Suppose that $f(x)\in \Bbb Z_6[x]$ is irreducible. Then, $f(x)$ is either irreducible or invertible over $\Bbb{Z}_2$, and over $\Bbb{Z}_3$. If $f(x)$ is invertible in both $\Bbb Z_2$ and $\Bbb Z_3$, then $f(x)$ is one of the constant polynomials $\pm1$, which are invertible (and not irreducible). This is a contradiction, so $f(x)$ must either be irreducible over $\Bbb Z_2$ or over $\Bbb Z_3$.

If $f(x)$ is invertible in neither $\Bbb Z_2$ nor $\Bbb Z_3$, then we can solve for $$u(x)\equiv 1\pmod{2}\ \ \ \wedge\ \ \ u(x)\equiv f(x)\pmod 3$$ and $$v(x)\equiv f(x)\pmod{2}\ \ \ \wedge \ \ \ v(x)\equiv 1\pmod{3}$$ for $u(x),v(x)\in\Bbb{Z}_6[x]$. These $u(x)$ and $v(x)$ are non-invertible over $\Bbb Z_6$ (since $u(x)$ is non-invertible modulo $3$, and $v(x)$ is non-invertible modulo $2$). However, $f(x)=u(x)\cdot v(x)$ in $\Bbb Z_6[x]$, which is a contradiction. Therefore, either $(a)$ or $(b)$ holds, but not both.

Conversely, suppose that $(a)$ holds. If $f(x)=p(x)\cdot q(x)$ for some $p(x),q(x)\in \Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $2$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $2$. Wlog, $p=f$ and $q=1$. As $f(x)\equiv \pm 1\pmod{3}$, $p(x)$ and $q(x)$ modulo $3$ are constants $\pm 1$. Thus, $q(x)\equiv 1\pmod{2}$ and $q(x)\equiv -1\pmod{3}$. Therefore, $q(x)= \pm 1$ in $\Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.

Finally, suppose that $(b)$ holds. If $f(x)=p(x)\cdot q(x)$ for some $p(x),q(x)\in \Bbb Z_6[x]$, then reduce $p(x)$ and $q(x)$ modulo $2$ and $3$ respectively. As $f(x)$ is irreducible mod $3$, $(p,q)=(f,1)$ or $(p,q)=(1,f)$ modulo $3$, up to sign swapping. Wlog, $p=f$ and $q=1$. As $f(x)\equiv 1\pmod{2}$, $p(x)$ and $q(x)$ modulo $2$ equal the constant $1$. Thus, $q(x)\equiv 1\pmod{2}$. Therefore, $q(x)= 1$ in $\Bbb Z_6[x]$. That is, $q(x)$ is constant, and so $f(x)$ is irreducible.

Examples. The polynomial $f(x)=3x+1$ fits $(a)$, so it is irreducible in $\Bbb{Z}_6[x]$. The polynomial $f(x)=2x+1$ fits $(b)$, so it is irreducible in $\Bbb{Z}_6[x]$.