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.

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- May 18th 2010, 03:15 PMENRIQUESTEFANINI[SOLVED] Operations with ideals in a ring.
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. - May 18th 2010, 04:30 PMNonCommAlg
- May 18th 2010, 05:16 PMENRIQUESTEFANINI
I'm terribly sorry. I considered a general commutative ring so, not necessarily with unit.

- May 18th 2010, 05:59 PMNonCommAlg
- May 19th 2010, 07:55 AMENRIQUESTEFANINI
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. - May 19th 2010, 08:50 AMNonCommAlg
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 - May 19th 2010, 09:00 AMNonCommAlg
the identity holds for any subsets of provided that : to see this, suppose that then and

- May 19th 2010, 10:16 AMENRIQUESTEFANINI
Hi:

Your reply has been most helpful and clarifying. Thanks a lot.

Enrique.