Let f: R -> S be a homomorphism of rings and let K = {r element of R | f(r) = 0s

Question

Let f: R -> S be a homomorphism of rings and let K = {r element of R | f(r) = 0s}. Prove that K is a subring of R.

Please show all steps and describe how you proceeded, clearly. Thank you :)

Explanation / Answer

not only kernel is a subring but also an ideal of R. now, as your requirement is concerned, we show that kernel K = { r in R/ f(r)=0s} is a subring of R. by the properties of homomorphism, we know that f(0) = 0 that means 0 must be in the kernel K. so , K is a non empty subset of R. now, suppose a, b are any two elements of K. then by definition, f(a) = 0, f(b) = 0 consider f(a-b) = f(a) - f(b) ( since f is a homomorphism) =0 - 0 = 0 f(a-b) = 0 ==> a- b must be in the kernel thus, we have shown that whenever a, b are in K, a- b is also in K. condition (1) is satisfied. f(a.b) = f(a)*f(b) since f is a homomorphism. = 0* 0 = 0 f(ab) = 0 ==> ab is in K. i.e. whenever a, b are in K, we get a.b is also in K. condition (2) is satisfied. so, by the necessary and sufficient condition for a non empty subset to be a subring, K is a subring of R.

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