Kernel of Ring Homomorphism is Ideal

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Theorem

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.


The kernel of $\phi$ is an ideal of $R_1$.


Proof

By Kernel of Ring Homomorphism is Subring, $\map \ker \phi$ is a subring of $R_1$.

Let $s \in \map \ker \phi$, so $\map \phi s = 0_{R_2}$.

Suppose $x \in R_1$. Then:

\(\ds \map \phi {x \circ_1 s}\) \(=\) \(\ds \map \phi x \circ_2 \map \phi s\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds \map \phi x \circ_2 0_{R_2}\) as $s \in \map \ker \phi$
\(\ds \) \(=\) \(\ds 0_{R_2}\) Properties of $0_{R_2}$


and similarly for $\map \phi {s \circ_1 x}$.

The result follows.

$\blacksquare$


Sources

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