By Randall R. Holmes

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**Example text**

1 Theorem (Eisenstein’s criterion). If the following are satisfied, then f (x) is irreducible over Q: (i) p2 a0 , (ii) p | ai for 0 ≤ i < n, (iii) p an . Proof. Assume that the three conditions are satisfied. Suppose that f (x) is not irreducible over Q. By (ii) and (iii), f (x) is nonconstant and hence a nonzero nonunit. 1, we conclude that f (x) = g(x)h(x) with g(x) and h(x) nonconstant polynomials over Z. Let σ : Z → Zp be the reduction modulo p homomorphism. By (ii), σf (x) = σ(an )xn and by (iii), σ(an ) is nonzero and hence a unit in the field Zp .

We conclude that f (x) cannot be factored over Z as a product of two polynomials having degrees strictly less than the degree of f (x). 1, f (x) has no such factorization over Q either, using that f (x) is nonconstant since σf (x) is irreducible and hence nonconstant. 5. • We claim that the polynomial f (x) = x5 + 8x4 + 3x2 + 4x + 7 is irreducible over Q. Taking p = 2 in the theorem, we see that it is enough to show that (σf )(x) = x5 + x2 + 1 is irreducible in Z2 [x]. 57 First, neither 0 nor 1 is a zero of (σf )(x), so this polynomial has no linear factor.

2 Theorem. Let f (x) be a nonconstant polynomial over Z[x]. (i) If f (x) factors over Q as f (x) = g(x)h(x), then it factors over Z as f (x) = g1 (x)h1 (x) with deg g1 (x) = deg g(x) and deg h1 (x) = deg h(x). (ii) If f (x) is irreducible over Z, then it is irreducible over Q. (iii) If f (x) is primitive and irreducible over Q, then it is irreducible over Z. Proof. (i) Let f (x) = g(x)h(x) be a factorization of f (x) with g(x), h(x) ∈ Q[x]. We may (and do) assume that the coefficients of g(x) all have the same denominator b ∈ Z.