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}.

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