## Decide whether some given polynomials are irreducible over ZZ

Let $p(x) \in \mathbb{Z}[x]$ be a monic polynomial and let $q \in \mathbb{Z}$ be prime.

1. Show that if $\overline{p}(x)$ is irreducible in $\mathbb{Z}/(q)[x]$, then $p(x)$ is irreducible in $\mathbb{Q}[x]$.
2. Show via a counterexample that the converse to part (a) is false.
3. Use this to decide whether the following are irreducible over $\mathbb{Q}$: $a(x) = x^3 - 12x^2 + 44x - 52$, $b(x) = x^4 + x^3 + x^2 + x + 1$.

Suppose $p(x)$ is irreducible mod $q$. If $p(x) = a(x)b(x)$ is a factorization of $p(x)$ over $\mathbb{Q}$, then by Gauss’ lemma it is reducible over $\mathbb{Z}$. (That is, we may assume that $a(x), b(x) \in \mathbb{Z}[x]$. Then $\overline{p}(x) = \overline{a}(x) \overline{b}(x)$ factors in $\mathbb{Z}/(q)[x]$. Now without loss of generality, $\overline{a}(x)$ is a unit. That is, $\overline{a}(x)$ has degree 1. If the degree of $a(x)$ is greater than 1, then the leading term of $a(x)$ is divisible by $q$. In particular, $q$ divides the leading term of $p(x)$, a contradiction since $p(x)$ is monic. So $a(x)$ has degree 1, and hence is a unit in $\mathbb{Q}[x]$. Thus $p(x) \in \mathbb{Q}[x]$ is irreducible.

Note that $x^3 + 2x^2 + 8x + 2$ is irreducible over $\mathbb{Q}$ by Eisenstein’s criterion. However, mod 2 this polynomial is congruent to $x^3$, which is clearly reducible. So the converse does not hold in general.

We claim that $a(x)$ is irreducible mod 3. To see this, note that $a(x) \equiv x^3 + 2x + 2$ mod 3. By Fermat’s little theorem, $t^3 \equiv t$ mod 3 for all $t$, so that under the evaluation homomorphism we have $a(t) \equiv 2$ mod 3 for all $t$. In particular $a(x)$ has no roots mod 3, hence no linear factors. Any factorization of $a(x)$ must include a linear factor; thus $a(x)$ is irreducible mod 3. Hence $a(x)$ is irreducible in $\mathbb{Q}[x]$.

We claim that $b(x)$ is irreducible mod 2. Note that $b(x)$ has no linear factors, since (as we can easily see) $b(x)$ has no roots in $\mathbb{Z}/(2)$. Thus if $b(x)$ were to be reducible, it would factor as a product of two irreducible quadratics. Note that there are precisely four quadratic polynomials mod 2: $x^2 = (x)(x)$, $x^2+1 = (x+1)^2$, $x^2+x = x(x+1)$, and $x^2+x+1$. We claim that the last of these is irreducible. Indeed if $x^2+x+1$ is reducible mod 2, then it must have a linear factor and hence a root. However, we see this is not the case. Thus $x^2+x+1$ is the only irreducible quadratic polynomial in $\mathbb{Z}/(2)[x]$, and certainly $(x^2+x+1)^2 = x^4+x^2+1 \neq b(x)$ mod 2. So $b(x)$ is irreducible mod 2, and thus must be irreducible over $\mathbb{Q}$.