# Thread: Principal Ideal

1. ## Principal Ideal

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

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

2. ## Re: Principal Ideal

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

3. ## Re: Principal Ideal

I think that we need to assume that $c \neq 0$ and that $cA \neq (0)$. With those assumptions, I think I have a proof if $R$ 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.

4. ## Re: Principal Ideal

Yeah, I'm not sure that this is true either. Unless I'm being stupid $(2,x)$ is not principal in $\mathbb{Z}_6[x]$ yet $3(2,x)=(3x)$. Am I being stupid?

5. ## 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 $R$ is an integral domain" - I missed the sentence. I can definitely do it with this new assumption.

Sorry for the trouble!