Kernel of Composition of Ring Homomorphisms

Theorem

Let $f : A \to B$ and $g : B \to C$ be ring homomorphisms.

Let $g \circ f$ be their composition.

Then the kernel of $g \circ f$ is the preimage under $f$ of the kernel of $g$:

$\map \ker {g \circ f} = f^{-1} \sqbrk {\ker g}$

Proof

By definition, the kernel of a ring homomorphism is its kernel when considered as a group homomorphism between the additive groups.

The result follows from Kernel of Composition of Group Homomorphisms.

$\blacksquare$