Hi:
Let a, b be elements of a commutative ring R, P ideal in R. I am frustrated by the fact that I cannot prove the following: if (ab) P then (a)(b) P. I think this should be true. If someone could give me a hand I would greatly appreciated it.
Thank you. It seems as if Zc =def {nc: n an integer}. But then Zc is not a subring of R. Can I still make sums and products involving Zc? May be I could define sums and products of subsets A, B of R by A+B = {a+b: a belongs to A, b belongs to B}, AB = set of all finite sums of the form a_1b_1+...+a_nb_n. But then A(B+C) = AB+AC not necessarily valid.
Please understand all I want now is to know about Zc and the way to operate with it. As for the proof, I found a longer one, that is in essence that given by you if I could operate freely with Zc.
is a subset (not subring) of because for all so , for all i.e. (note that we cannot write because might
not have 1 and so might not be in ) so you need to show that is the smallest ideal of which contains . that is proved very easily: is clearly an additive group of and if then
also if is any ideal of which contains then and and thus that means