If $\bar f(x)$ irreducible in $\mathbb{Z}_2[x] \Rightarrow f(x)$ irreducible in $\mathbb{Q}[x]$? The example I'm working with is $\bar f(x) = x^4+x^3+\bar 1 \in \mathbb{Z}_2[x]$ , which I know is irreducible in $\mathbb{Z}_2[x]$. The text that I'm reading from seems to imply that my "if, then" statement in my title above is true for this particular example. So I wondered if it is true more generally for any polynomial in $\mathbb{Z}_p[x]$ for $p$, prime.
Thanks in advance.
 A: Not quite.  For example, $2x^2 + x$ is irreducible in $\mathbb{Z}_2[x]$, but not in $\mathbb{Z}[x]$.
Here is a correct statement: if $f\in \mathbb{Z}[x]$, and the leading coefficient of $f$ is not divisible by $2$, then $\overline{f}$ irreducible implies ${f}$ irreducible over $\mathbb{Q}$.
To see this, suppose that $f=gh$ is a non-trivial factorization, i.e. $g$ and $h$ have degree at least $1$.  Since their leading coefficients are not divisible by $2$, $\overline{g}$ and $\overline{h}$ also have degree at least $1$, so $\overline{f} = \overline{g}\overline{h}$ is a nontrivial factorization.
A: from the book 'Abstract algebra, David Dummit and Richard Foote, proposition 12, p309'
Proposition 12: Let $I$ be a proper ideal in the integral domain $R$ and let $p(x)$ be a nonconstant monic polynomial in $R[x]$. If the image of $p(x)$ in $(R/I)[x]$ cannot be factored in $(R/I)[x]$ into two polynomials of smaller degree, then $p(x)$ is irreducible in $R[x]$.
Remark: The converse to proposition $12$ does not hold. 
