Thread: Principal Ideal

Hello everyone. I have been trying to work out the following problem:

Suppose is an ideal in a commutative ring with identity. Show that if is principal for some element , then is principal.

This seems like a very simple and innocuous statement, but I can't seem to prove it. My idea is to try to construct the element so that is generated by it. I started with an example, taking and the ideal to be . I multiplied it by 10: , which turns out to be . Now, in order to recover the generator for , I need to divide 20 by 10. But this is a serious issue for the general statement above, because I am not guaranteed any kind of division or cancellation. It is completely unclear to me how to pick up the single generator.

I would appreciate any input anybody has on this problem!

Hint: Supose there are at least 2 generators of A.

I think that we need to assume that and that . With those assumptions, I think I have a proof if is an integral domain, but not in general. I also couldn't see how to use ModusPonen's hint. Was this problem from a book? If so, it might be helpful to state which section it came from and whether any of the preceding problems or exercises supply clues.

Yeah, I'm not sure that this is true either. Unless I'm being stupid is not principal in yet . Am I being stupid?

Thanks for the replies guys. I was recently told that I was supposed to assume that the ring is an integral domain. Apparently, it was part of a block of problems that said "For problems x through y, assume that is an integral domain" - I missed the sentence. I can definitely do it with this new assumption.

Sorry for the trouble!