The Evaluation Homomorphism

Abstract Algebra · Axiom Academy

EXAMPLE The Evaluation Homomorphism Proving φ: ℝ[x] → ℝ defined by φ(p) = p(2) is a ring homomorphism and identifying its kernel φ evaluates any polynomial in ℝ[x] at x = 2 Excellent work! You've proved that the evaluation map is a ring homomorphism and identified its kernel: Evaluation is a Homomorphism: The map φ(p) = p(2) preserves both addition and multiplication, making it a ring homomorphism from ℝ[x] to ℝ Preserves Addition: φ(p + q) = (p + q)(2) = p(2) + q(2) = φ(p) + φ(q) for all polynomials p, q Preserves Multiplication: φ(p · q) = (p · q)(2) = p(2) · q(2) = φ(p) · φ(q) for all polynomials p, q The Kernel: ker(φ) = p ∈ ℝ[x] : p(2) = 0 consists of all polynomials with 2 as a root Kernel Structure: By the Factor Theorem, ker(φ) = ⟨x - 2⟩, the principal ideal generated by (x - 2) This example generalizes: evaluation at any point a ∈ ℝ gives a homomorphism with kernel ⟨x - a⟩. This connection between evaluation, roots, and ideals is fundamental in ring theory!

This is the written version of the interactive lesson above. See the full Abstract Algebra course.