# Thread: Prove that if R is a PID then every ideal in S is also principal.

1. ## Prove that if R is a PID then every ideal in S is also principal.

(a) Suppose that  $\displaystyle \phi : R \rightarrow S$ is a surjective ring homomorphism. Prove that if $\displaystyle R$ is a principal ideal domain then every ideal in $\displaystyle S$ is also principal.

(b) Give an example to show that S need not be an integral domain.

2. For part (a), let $\displaystyle J$ be an ideal in $\displaystyle S$. What can you say about $\displaystyle \phi^{-1}(J)$?

3. Originally Posted by ojones
For part (a), let $\displaystyle J$ be an ideal in $\displaystyle S$. What can you say about $\displaystyle \phi^{-1}(J)$?
$\displaystyle \phi^{-1}(J)=I\subseteq R$

4. You can say a lot more than that.

5. $\displaystyle \phi^{-1}(J)=I\subseteq R$ is principal

6. Correct. Therefore?

7. $\displaystyle \phi^{-1}(J)=I\subseteq R$ is principal; therefore, all ideals $\displaystyle J \subseteq S$ are principal.

8. Nice try. $\displaystyle I$ principal means $\displaystyle I=aR$ for some $\displaystyle a\in R$. Can you guess the form of $\displaystyle J$?

9. Originally Posted by ojones
Nice try. $\displaystyle I$ principal means $\displaystyle I=aR$ for some $\displaystyle a\in R$. Can you guess the form of $\displaystyle J$?
$\displaystyle J=\phi(I)=\phi(aR)=\phi(a)*\phi(R)=\phi(a)*S$

10. There's no *. Conclusion?

For part (b), if $\displaystyle S$ fails to be an integral domain, then what?

11. Originally Posted by ojones
There's no *. Conclusion?
You mean $\displaystyle J=\phi(I)=\phi(aR)=\phi(a)\phi(R)=\phi(a)S$?
Conclusion: J is principal.

Originally Posted by ojones
For part (b), if $\displaystyle S$ fails to be an integral domain, then what?
Then $\displaystyle S$ has at least one nonzero zero-divisor. So if $\displaystyle b,c\in S$ then if $\displaystyle bc=0$, $\displaystyle b$ and $\displaystyle c$ can both be nonzero.

12. Originally Posted by JJMC89
You mean $\displaystyle J=\phi(I)=\phi(aR)=\phi(a)\phi(R)=\phi(a)S$?

Well yes, but this isn't the best way to argue. You want to show that $\displaystyle J=\phi(a)S$ or equivalently that if $\displaystyle b\in J$ then $\displaystyle b=\phi(a)s$ for some $\displaystyle s\in S$.

Conclusion: J is principal..
Right.

Then $\displaystyle S$ has at least one nonzero zero-divisor. So if $\displaystyle b,c\in S$ then if $\displaystyle bc=0$, $\displaystyle b$ and $\displaystyle c$ can both be nonzero.
OK, but what's an obvious example of a ring which is a PID, one that's not, and an obvious homorphism between the two? Hint: $\displaystyle \mathbb{Z}$ is involved.

13. $\displaystyle \mathbb{Z}$ is a PID
$\displaystyle n\mathbb{Z}, (n>1)$ is not an integral domain.
$\displaystyle \phi(z)=nz$

14. You need to be careful here. Does your definition of an integral domain require the ring to have an identity element? If so, your example is fine. I was thinking of $\displaystyle S=\mathbb{Z}/n\mathbb{Z}$ for $\displaystyle n$ not prime.

15. Yes, my definition requires unity.

Thanks for your help! This thread is one of the most helpful responses I've gotten (for any problem).

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