# Irreducible monic polynomial in $\mathbb{Q}[x]$

I am trying to show that the following is true:

Suppose $f(x)$, monic in $\mathbb{Z}[x]$, factors modulo 3 into the product of two irreducible polynomials of degree 2, and factors modulo 2 into the product of an irreducible polynomial of degree 3 and a polynomial of degree 1. Show that $f(x)$ is irreducible in $\mathbb{Q}[x]$.

I've tried writing down the possible factors of the given degrees mod 2 and 3 and then trying to see if those lead to coefficients of $f(x)$ such that the Eisenstein criterion can be applied to show irreducibility, but this hasn't really seemed to work. Not quite sure how else to approach this, so I'd appreciate any help. Thanks!

If $f(x)$ is monic in $\mathbb{Z}[X]$ and reducible in $\mathbb{Q}[X]$ then by Gauss' lemma it is also reducible in $\mathbb{Z}[X]$. Write $f(X) = g(X)h(X)$ with $g(X)$, $h(X)$ in $\mathbb{Z}[X]$ with degree of $g(X)$ either 1 or 2. Now looking at the equation mod 2 in $\mathbb{F}_2[X]$ we have that $f(X)$ factors into $g(X)h(X) = g'(X)h'(X)$ with degree of $g'$ equal to 1 and degree of $h'$ equal to 3. But $g'$ is irreducible and must therefore divide either $g$ or $h$. But in both cases if degree of $g$ is 2, this would lead to another factorization of $f$ into more than 2 irreducible factors. Since $\mathbb{F}_2[X]$ is a UFD this is impossible. Thus degree of $g$ is 1. But then the factorization of $f$ in $\mathbb{F}_3[X]$ would have a linear factor, contradicting the given condition (again by UFD property of $\mathbb{F}_3[X]$). Therefore $f(X)$ must be irreducible in $\mathbb{Q}[X]$.
• I should have also added that being monic is used when I assumed that the degrees of $g(X)$ and $h(X)$ remain the same in $\mathbb{F}_2[x]$ and $\mathbb{F}_3[X]$, because otherwise you could have leading coefficients multiples of 2 or 3. Given a polynomial $f(X)$ in $\mathbb{Z}[X]$ of degree 4 satisfying the hypotheses, you could have had the polynomial $6f(X)^2 + f(X) = f(X)(6f(X) + 1)$ which satisfies all hypotheses modulo 2 and 3 but is reducible over $\mathbb{Q}[X]$. – Tob Ernack Apr 28 '17 at 12:41