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