By Randall R. Holmes

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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. 1 that f (x) is irreducible over Q.

1 p−2 p−1 Now p divides each binomial coefficient pi with 0 < i < p and p2 p p = p−1 , so f (x + 1) is irreducible over Q (and hence over Z) by Eisenstein’s criterion. This establishes the claim. 10 – Exercises 10–1 Prove that the polynomial f (x) = 3x4 + 8x3 + 8x2 − 2x − 3 is irreducible over Q. Hint: Consider f (x − 1). 10–2 Prove that the polynomial f (x) = 3x2 − 7x − 5 is irreducible over Q. 1 Definition Let F be a field. A vector space over F is a triple (V, +, · ), where (V, +) is an abelian group and · is a function F × V → V (written (a, v) → av) satisfying the following for all a, b ∈ F and all v, w ∈ V : (i) a(v + w) = av + aw, (ii) (a + b)v = av + bv, 60 (iii) a(bv) = (ab)v, (iv) 1v = v.

This shows that ϕ−1 (S ) is closed under multiplication. Therefore, it is a subring of R. 4 Kernel and image Let ϕ : R → R be a homomorphism of rings. The kernel of ϕ and the image of ϕ retain their same meanings from group theory: • ker ϕ = ϕ−1 ({0}) = {a ∈ R | ϕ(a) = 0}, the kernel of ϕ, • im ϕ = ϕ(R) = {ϕ(a) | a ∈ R}, the image of ϕ. 3(iv) with S = {0}. In fact, the kernel of ϕ is even an ideal of R as we will see in the next section. 3(iii) with S = R. 5 Kernel same thing as ideal Let R be a ring.

### Abstract Algebra II by Randall R. Holmes

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