By Randall R. Holmes

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Iii) Every prime element of R is irreducible. (iv) If every ideal of R is principal, then every irreducible element of R is prime. Proof. 2. (i) Assume that r is irreducible. First, (r) is nonzero, since r is nonzero, and it is proper since r is not a unit. Let s be an element of R such that (r) ⊆ (s) and (s) is proper. Then s | r so that r = st for some t ∈ R. Now s is not a unit, since (s) is proper. Therefore t is a unit, since r is irreducible. Thus, r ∼ s, implying that (r) = (s). We conclude that (r) is nonzero and maximal among the proper principal ideals of R.

If f (x+c) is irreducible over Z, then f (x) is irreducible over Z and is therefore irreducible over Q if it is nonconstant. Proof. Assume that f (x + c) is irreducible over Z. The map ϕx+c : Z[x] → Z[x] given by ϕx+c (f (x)) = f (x + c) is an isomorphism (the proof that the evaluation map is a homomorphism easily generalizes to show that this map is a homomorphism as well, and this map is bijective since ϕx−c is an inverse). Since f (x + c) is irreducible over Z (which means after all that it is an irreducible element of the ring Z[x]), and since it corresponds to f (x) under the isomorphism, we conclude that f (x) is also irreducible over Z.

Showing that a given polynomial is irreducible is often difficult. For irreducibility over a field one can sometimes use the following observation about a proper factorization, together with a proof by contradiction. 1 Theorem. Assume that R is a field. If f (x) is nonzero and it has a proper factorization f (x) = g(x)h(x), then deg g(x), deg h(x) < deg f (x). Proof. Suppose that f (x) is nonzero and that it has the indicated proper factorization. 2) it follows that deg g(x), deg h(x) ≤ deg f (x).

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