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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!
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Re: Principal Ideal
Hint: Supose there are at least 2 generators of A.
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Re: Principal Ideal
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.
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Re: Principal Ideal
Yeah, I'm not sure that this is true either. Unless I'm being stupid
is not principal in
yet
. Am I being stupid?
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Re: Principal Ideal
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!