mapping onto Zn to determine irreducibility over Q 
If $h:\mathbb  Z \rightarrow \mathbb Z_n$ is a natural homomorphism, let $\bar{h}:\mathbb Z[x]\rightarrow\mathbb Z_n[x]$ be defined by $\bar{h}(a_0+a_1x+...+a_nx^n)=h(a_0)+h(a_1)x+...+h(a_n)x^n$. $\bar{h}$ is a homomorphism.
Prove if $\bar{h}(a(x))$ is irreucdible in $\mathbb Z_n[x]$ and $a(x)$ is monic then $a(x)$ is irreducible in $\mathbb Z[x]$.

I've seen another question mention this question but it hasn't been answered to my knowledge. I believe we could suppose $a(x)$ is reducible o=in $\mathbb Z[x]$, thus $a(x)=b(x)c(x)$ where deg$b(x),c(x)<$ deg$a(x)$. I'm not really sure what to do with this. I should probably use the definition of $\bar{h}$ and $h$ but I'm not sure how.
 A: Suppose then that $a(x)$ is reducible in $\Bbb Z[x]$; then we may write
$a(x) = p(x)q(x), \tag 1$
with $p(x), q(x) \in \Bbb Z[x]$ and $\deg p(x), \deg q(x) \ge 1$.  If
$\displaystyle a(x) = \sum_0^{\deg a} a_i x^i \in \Bbb Z[x], \tag 2$
$\displaystyle p(x) = \sum_0^{\deg p} p_i x^i \in \Bbb Z[x], \tag 3$
and
$\displaystyle q(x) = \sum_0^{\deg q} q_i x^i \in \Bbb Z[x], \tag 4$
then it follows that
$\deg a = \deg p + \deg q \tag 5$
and also that
$p_{\deg p} \; q_{\deg q} = a_{\deg a} = 1, \tag 6$
since $a(x)$ is monic and the coefficient of of the leading term of $a(x)$ is the product of the coefficients of the leading terms of its factors $p(x)$ and $q(x)$.  We also have
$\bar h(a(x)) = \bar h(p(x)q(x)) = \bar h(p(x)) \bar h(q(x)); \tag 7$
from which it follows that
$h(a_{\deg a}) = h(p_{\deg p}) h(q_{\deg q}) \in \Bbb Z_n[x].  \tag 8$
Since $h: \Bbb Z \to \Bbb Z_n$ is the natural homomorphism, we have
$h(1_{\Bbb Z}) = 1_{\Bbb Z_n}; \tag 9$
thus
$h(p_{\deg p}) h(q_{\deg q}) = h(a_{\deg a}) = h(1_{\Bbb Z}) = 1_{\Bbb Z_n}, \tag{10}$
which in turn implies
$h(p_{\deg p}), h(q_{\deg q}) \ne 0 \tag{11}$
in $\Bbb Z_n$; from this we infer that, since $h(p_{\deg p}), h(q_{\deg q})$ are respectively the leading coefficients of $h(p(x)), h(q(x)) \in \Bbb Z_n[x]$, that $\deg (\bar h(p(x)), \deg (\bar h(q(x)) \ge 1$, and hence that neither $\bar(h(p(x))$ nor $\bar h(q(x))$ is constant in $\Bbb Z_n[x]$.  Thus $\bar h(a(x))$ is reducible in $\Bbb Z_n[x]$, contrary to hypothesis.  We thus conclude that (1) cannot bind in $\Bbb Z[x]$, that is, that $a(x)$ is irreducible.
