Let R S be two rings and f R rightarrow S be a ring homomorp

Let R. S be two rings and f: R rightarrow S be a ring homomorphism. Let J be an ideal of S and I = {r Element R| f(r)Element J) (I is an ideal which was shown before). Recall that the kernel of f is ker(f) = {r Element R| f(r) = 0_s}. Show that I ker(f)

Solution

Let f, R,S , I and J be as given.

Claim: I contains Ker(f).

Now I = {r in R | f(r) belongs to J}.

As J is an ideal, J is an additive subgroup of S .

So 0 belongs to J.

If r is in Ker (f), f(r) =0 which belongs to J, as noted above.

Hence , by definition, r is in I.

Thus Ker(f) is contained in I.