Ideals are the same thing as kernels of ring homomorphisms
Kernels are ideals
Let \(f: R \to S\) 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 noteThat is, after removing the multiplicative structure from the ring. 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 \(f(kr) = f(k)f(r) = 0 \times r = 0\), so \(kr\) is in \(K\) if \(k\) is.
Ideals are kernels
refer to the quotient group, and therefore introduce the quotient ring