Irreducibility over $\Bbb Q$

Question

I see solutions to problems on this site and in quiz solutions where the question is regarding the ırreducibility of a polynomial over $\Bbb Q$ and the answer is obtained by testing irreducibility over $\Bbb Z_2[x]$ or $\Bbb Z_3[x]$. I don't understand how testing over the latter fields yields anything for $\Bbb Q$.

Another question is regarding how a polynomial can be reducible over $\Bbb Z$ and $\Bbb R$, but not $\Bbb Q$. If it's reducible over $\Bbb Z$, shouldn't it be reducible over $\Bbb Q$ as well?

Answer 1

As to your first question, if $f(x)=g(x)h(x)$ (with all polynomials in $\Bbb Z[x]$), then also $f(x)\equiv g(x)h(x) \pmod{p}$ for any prime $p$. In other words, if you can factor $f$ over $\Bbb Z$, then you necessarily can always factor $f$ over $\Bbb Z_p$. Taking the contrapositive, if for any $p$ you can't factor $f$ over $\Bbb Z_p$, then $f$ must be irreducible over $\Bbb Z$ as well.

You are correct that a polynomial that is reducible over $\Bbb Z$ is necessarily also reducible over $\Bbb Q$. However, $x^2-2$ is irreducible over $\Bbb Q$ but reducible over $\Bbb R$ (as $f(x) = (x- \sqrt 2)(x + \sqrt 2).$)

Answer 2

The polynomial $f=3x^2-6$ is reducible over $\mathbb Z$ but not over $\mathbb Q$. Over $\mathbb Z$ it factors as $3(x^2-2)$, with both $3$ and $x^2-2$ being irreducible. This factorization does not cause $f$ to be reducible over $\mathbb Q$ since $3$ is a unit in $\mathbb Q[x]$. However, if we require that the gcd of the coefficients of $f$ (called the content of $f$) is $1$, then it is indeed the case that if $f$ is reducible over $\mathbb Z$ then it is reducible over $\mathbb Q$. This is because if $f=gh$ over $\mathbb Z$, then both $g$ and $h$ must be nonconstant (because the content of $f$ is $1$) and so $g$ and $h$ are not units in $\mathbb Q[x]$.

Continue to assume that the content of $f$ is $1$. The same reasoning, with $\mathbb Z_p$ in place of $\mathbb Q$, shows that if $f$ is irreducible over $\mathbb Z_p$ then it must be irreducible over $\mathbb Z$. Now Gauss's lemma states that if $f\in\mathbb Z[x]$ is irreducible over $\mathbb Z$, then it is also irreducible over $\mathbb Q$. Thus irreducibility over $\mathbb Z_p$ can be used to show irreducibility over $\mathbb Q$.