Prove that ideals are kernels. Solution Let f:R --> S be any homomorphism of rin

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Prove that ideals are kernels.

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Let f:R --> S be any homomorphism of rings and let K = ker f. We know 0 is in K, so K is non-empty. Let a, b be in K, then f(a)=f(b)=0, so f(a-b)=f(a)-f(b)=0. For any r in R, f(ra)=f(r)f(a)=f(r)*0=0. Similary, f(ar)=f(a)f(r)=0. Thus, a-b, ra and ar are also in K, hence K is an ideal. Thus, every kernel is an ideal. Now let I be an ideal of R. Form the quotient ring R/I, and consider the map f:R -->R/I given by f(r)=r+I for r in R. It is easy to see that f is a ring homomorphism with kernel I. Thus, kernels are Ideals

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Prove that ideals are kernels. Solution Let f:R --> S be any homomorphism of rin

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