# Abstract Algebra II by Randall R. Holmes

By Randall R. Holmes

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Additional resources for Abstract Algebra II

Example text

4 Eisenstein’s criterion Let f (x) = a0 + a1 x + a2 x2 + · · · + an xn be a polynomial over Z and let p be a prime number. 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.

6 Homomorphism is injective iff kernel is trivial Here is a reminder of a useful criterion from group theory for checking injectivity of a homomorphism. Let ϕ : R → R be a homomorphism of rings. 1 {0}. Theorem. The homomorphism ϕ is injective if and only if ker ϕ = Proof. Assume that ϕ is injective. Let r ∈ ker ϕ. Then ϕ(r) = 0. 3. So ϕ(r) = ϕ(0) and injectivity of ϕ gives r = 0. This shows that ker ϕ ⊆ {0}. Since a kernel is a subgroup, the other inclusion is immediate. Now assume that ker ϕ = {0}.

A similar proof shows that it has the right absorption property as well, so it is an ideal of R. Let I be an ideal of R. An element r of R is in the kernel of π if and only if I = π(r) = r + I, which occurs if and only if r ∈ I. Therefore, I = ker π. 6 Homomorphism is injective iff kernel is trivial Here is a reminder of a useful criterion from group theory for checking injectivity of a homomorphism. Let ϕ : R → R be a homomorphism of rings. 1 {0}. Theorem. The homomorphism ϕ is injective if and only if ker ϕ = Proof.