Let $f$ be a monic polynomial in $\mathbb{Z}[x]$. Request to verify my thinking about a question on irreducibility of $f$.

Question

Suppose there are distinct primes $p,q$ such that $\bar{f}=gh\in F_p[x]$，$\bar{f}=uv\in F_q[x]$, where $\bar{f}$ is the image of $f$ under natural projection from $\mathbb{Z}[x]$ to $F_p[x], F_q[x]$ respectively. If $g,h,u,v$ are irreducible in their respective rings, and $\{deg(g), deg(h)\}$ as a set is not the same as $\{deg(u),deg(v)\}$, then $f$ is irreducible over $\mathbb{Q}$.

What I know: $f$ is monic, so is primitive, so is irreducible over $\mathbb{Q}$ iff irreducible over $\mathbb{Z}$. If $f$ is reducible over $\mathbb{Z}$, then take all irreducible factors of $f$ and take their image in $F_p[x]$, $F_q[x]$. These factors are all non-constant and monic since $f$ is monic, so their degrees don't decrease under projection. Therefore there are exactly 2 such factors (if there are more than 2, sending them to $F_p[x]$ could only produce more, which violates assumption). Write $f=f_1f_2$ where $f_i$ is monic irreducible in $\mathbb{Z}[x]$.

Can I continue to say that wlog, $\bar{f_1}=g, \bar{f_2}=h$ in $F_p[x]$, $\bar{f_1}=u, \bar{f_2}=v$ in $F_q[x]$ (since both are UFDs), and since they are all monic, $\{deg(g), deg(h)\}=\{deg(u), deg(v)\}$, which is contradiction?

Answer 1

A few comments:

• It is correct to conclude that when $\{\deg g,\deg h\}\neq\{\deg u,\deg v\}$, then $f$ is irreducible over $\Bbb{Z}$.
• You need to be a bit more careful identifying the reductions of the factors. The obvious alternative is when the two factors have the same degree, when the possibility of misguessing which reduction belongs to which modular factor will be present. I do believe that you had noticed this possibility when, say a quartic, splits into a product of two quadratics modulo two distinct primes.
• The irreducibility conclusion would still be valid when there are more than two modular factors, if the degrees of those factors are incompatible. For example, if modulo one prime you get factors of degrees $1+1+4$ and modulo another prime the factors have degrees $3+3$, there is no way those could come from reductions of factors in $\Bbb{Z}[x]$.
• Even if compatibility is not obviously broken, such pieces of data may be helpful in proving irreducibility. Say that you have factors of degrees $1+1+3$ modulo $p$ and factors of degrees $1+4$ modulo $q$. The only possible factorization in $\Bbb{Z}[x]$ compatible with both is $1+4$. But this implies the presencs of a rational zero (integral actually), a possibility you may be able eliminate using the rational root test.
• Even in the presence of compatible sequences of degrees you can make progress by applying the Chinese Remainder Theorem, and find what the factors must look like modulo $pq$. If you are lucky, those alternatives can be eliminated. When $g$ and $h$ (or $u$ and $v$) share the same degrees, the built in uncertainty means that the number of cases to be considered increases.