Kernel of Ring Homomorphism is Subring
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Theorem
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then the kernel of $\phi$ is a subring of $R_1$.
Proof
From Ring Homomorphism of Addition is Group Homomorphism and Kernel of Group Homomorphism is Subgroup:
- $\struct {\map \ker \phi, +_1} \le \struct {R_1, +_1}$
where $\le$ denotes subgroup.
Let $x, y \in \map \ker \phi$.
\(\ds \map \phi {x \circ_1 y}\) | \(=\) | \(\ds \map \phi x \circ_2 \map \phi y\) | Morphism Property | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_{R_2} \circ_2 0_{R_2}\) | Definition of Kernel | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_{R_2}\) |
Thus $x \circ_1 y \in \map \ker \phi$.
Thus the conditions for Subring Test are fulfilled, and $\map \ker \phi$ is a subring of $R_1$.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 57.3$ Ring homomorphisms: $\text{(ii)}$