2.2 Kernel and image of ring homomorphisms

Ring homomorphisms are crucial in understanding how rings relate to each other. They map elements between rings while preserving structure. Kernels and images help us analyze these mappings, revealing key information about the rings involved.

Kernels show which elements map to zero, forming ideals in the domain ring. Images tell us which elements in the codomain are reached. Together, they give insights into injectivity, surjectivity, and isomorphisms between rings.

Ring Homomorphisms: Kernel and Image

Kernel and image of ring homomorphisms

• Kernel maps domain elements to zero in codomain $\ker(f) = \{a \in R : f(a) = 0_S\}$ for $f: R \rightarrow S$
• Image consists of codomain elements reached by mapping $\text{im}(f) = \{f(a) : a \in R\}$ for $f: R \rightarrow S$
• Natural projection $\pi: \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$ has $\ker(\pi) = n\mathbb{Z}$ and $\text{im}(\pi) = \mathbb{Z}/n\mathbb{Z}$
• Inclusion map $i: \mathbb{Z} \rightarrow \mathbb{Q}$ yields $\ker(i) = \{0\}$ and $\text{im}(i) = \mathbb{Z}$

Kernel as ideal in domain ring

• Kernel forms subring of R
  • Subtraction closure: $a, b \in \ker(f) \implies f(a-b) = f(a) - f(b) = 0 - 0 = 0$
  • Multiplication closure: $a, b \in \ker(f) \implies f(ab) = f(a)f(b) = 0 \cdot 0 = 0$
• Absorption property satisfied
  • $r \in R$, $a \in \ker(f)$: $f(ra) = f(r)f(a) = f(r) \cdot 0 = 0$
  • Similarly, $f(ar) = f(a)f(r) = 0 \cdot f(r) = 0$
• Kernel meets all ideal criteria in R

Determining kernel and image

• Kernel identification
  1. Find elements mapping to zero in codomain
  2. Express in set notation
• Image determination
  1. Identify codomain elements reached by homomorphism
  2. Express in set notation
• $f: \mathbb{Z}[x] \rightarrow \mathbb{Z}$ with $f(p(x)) = p(0)$
  • $\ker(f) = \{p(x) \in \mathbb{Z}[x] : p(0) = 0\} = \{x \cdot q(x) : q(x) \in \mathbb{Z}[x]\}$
  • $\text{im}(f) = \mathbb{Z}$

Kernel vs injectivity and image vs surjectivity

• Injective homomorphism has trivial kernel (only zero element)
• Proof: $f(a) = f(b) \iff f(a-b) = 0 \iff a-b \in \ker(f)$
• Surjective homomorphism has image equal to codomain
• $f$ surjective $\iff \text{im}(f) = S$ for $f: R \rightarrow S$
• First Isomorphism Theorem: $R/\ker(f) \cong \text{im}(f)$ connects kernel, image, and ring structures