# Principal Ideal

• Oct 6th 2011, 06:02 PM
roninpro
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!
• Oct 7th 2011, 12:52 PM
ModusPonens
Re: Principal Ideal
Hint: Supose there are at least 2 generators of A.
• Oct 10th 2011, 01:30 PM
Petek
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.
• Oct 10th 2011, 01:36 PM
Drexel28
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?
• Oct 10th 2011, 02:48 PM
roninpro
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!