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Thread: Principal Ideal

  1. #1
    Senior Member roninpro's Avatar
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    Principal Ideal

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

    Suppose $\displaystyle A$ is an ideal in a commutative ring $\displaystyle R$ with identity. Show that if $\displaystyle cA$ is principal for some element $\displaystyle c$, then $\displaystyle A$ 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 $\displaystyle x$ so that $\displaystyle A$ is generated by it. I started with an example, taking $\displaystyle R=\mathbb{Z}$ and the ideal to be $\displaystyle A=2\mathbb{Z}$. I multiplied it by 10: $\displaystyle 10(2\mathbb{Z})$, which turns out to be $\displaystyle 20\mathbb{Z}$. Now, in order to recover the generator for $\displaystyle A$, 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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  2. #2
    Member ModusPonens's Avatar
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    Re: Principal Ideal

    Hint: Supose there are at least 2 generators of A.
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  3. #3
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    Re: Principal Ideal

    I think that we need to assume that $\displaystyle c \neq 0$ and that $\displaystyle cA \neq (0)$. With those assumptions, I think I have a proof if $\displaystyle R$ 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.
    Last edited by Petek; Oct 10th 2011 at 02:26 PM. Reason: Correct tex tags
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Principal Ideal

    Yeah, I'm not sure that this is true either. Unless I'm being stupid $\displaystyle (2,x)$ is not principal in $\displaystyle \mathbb{Z}_6[x]$ yet $\displaystyle 3(2,x)=(3x)$. Am I being stupid?
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  5. #5
    Senior Member roninpro's Avatar
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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 $\displaystyle R$ is an integral domain" - I missed the sentence. I can definitely do it with this new assumption.

    Sorry for the trouble!
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