Proof

Kernels are ideals

Let be a ring homomorphism between rings $R$ and $S$ . We claim that the kernel $K$ of $f$ is an ideal.

Indeed, it is clearly a subgroup of the ring $R$ when viewed as just an additive group ^[1] because $f$ is a group homomorphism between the underlying additive groups, and kernels of group homomorphisms are subgroups (indeed, normal subgroups). (Proof.)

We just need to show, then, that $K$ is closed under multiplication by elements of the ring $R$ . But this is easy: if $k \in K$ and $r \in R$ , then $f (k r) = f (k) f (r) = 0 \times r = 0$ , so $k r$ is in $K$ if $k$ is.

Ideals are kernels

refer to the quotient group, and therefore introduce the quotient ring

^{^︎}
That is, after removing the multiplicative structure from the ring.

Ideals are the same thing as kernels of ring homomorphisms