if is a (commutative) ring with identity, we have so

in we have 5/1 = 10/2 = 0/2 = 0. now suppose 2/1 = 0. so there exists such that 2s =0, i.e. which is impossible.

3) Define . Is S multiplicatively closed?

Prove that as an element in and that as an element in .

because and thus similarly since we have now since for any there exists such that

4) With S as above find for each of following modules;

and

we have for the last part use this fact that

thst's right.

And (3) I would say yes because ?