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I'm supposed to show in a part of an exercise that if we have a ring $R$ that is a principal ideal domain, then for any ideal $I$ in $R$, $R/I$ will also be a PID.

So $I=(i)$ for some $i \in R$, and $R/I$ is $\{rI |r\in R\}$, and we know:

An ideal $J$ in $R/I$ will be an ideal $J$ in $R$ containing $I$.

So if $I \subset J$ and $I=(i)$, then $J$ should be $(j)$, where $j|i$. but wherefrom do I get the necessity that $J=(j)$, so that $J$ is also principal? and there will be several $j$ that divide $i$, but should it not be only one?

And then I shall use this to show a 1-1 correspondence between ideals of $\mathbb{Z}$ and $\mathbb{N}$.

Can someone support with with the proofs?

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    $\begingroup$ The statement is false, as $\mathbb{Z}$ is a PID, but $\mathbb{Z}/(n)$ is not a PID when $n$ is not a prime (or a negative of a prime). Perhaps you mean principal ring? Also, I don't understand in what sense $\mathbb{N}$, the set of natural numbers, has ideals - it isn't a ring. $\endgroup$ Commented Dec 9, 2011 at 21:02
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    $\begingroup$ What do you mean by "ideals of $\mathbb{N}$?" Do you mean, ideals of the multiplicative semigroup? How do you expect to use a result about rings to prove something about $\mathbb{N}$, which is not a ring? $\endgroup$ Commented Dec 9, 2011 at 21:04
  • $\begingroup$ I got a set of extra-exercises for a course where it will decide about my pass or fail, and the professor wrote down the exercise for me within two hours, so I assume that he made a mistake and meant ideals of the multiplicative semigroup. (In fact, my performance in the course is so bad because the professor is quite careless and not very sorted all the time, which makes it difficult to study from lecture notes.) $\endgroup$ Commented Dec 9, 2011 at 21:25
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    $\begingroup$ @Marie.P. I will only note that your post above shows quite a bit of carelessness: you use $J$ to denote both an ideal of $R/I$ and an ideal of $R$. You wonder why $J\leq R$ is principal, even though you assume that $R$ is a principal ideal domain. You want to apply a result about rings to an object that is not a ring... Nothing wrong with being confused when one is starting, as presumably you are, but it might be more productive to concentrate on what you are doing wrong $\endgroup$ Commented Dec 9, 2011 at 21:37

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The claimed result is false: For $R/I$ to be a principal ideal domain, it must be a domain. But $R/I$ is a domain if and only if $I$ is a prime ideal. In particular, $\mathbb{Z}/6\mathbb{Z}$ is a quotient of a PID that is not a domain, hence not a PID.

What is true is that a quotient of a PID will be a principal ideal ring: every ideal of $R/I$ is principal. Your argument is confused because you are using the same letter to denote the ideal of $R/I$ and the ideal it corresponds to inside $R$. Better to use different letters.

Indeed, let $K$ be an ideal of $R/I$. By the Correspondence Theorem $K$ corresponds to an ideal $J$ of $R$ that contains $I$. Since $R$ is assumed to be a PID, then $J=(j)$ for some $j\in R$. The claim is that $K = (j+I)(R/I)$: let $k+I\in K$. Then $k+I \in J+I$, so there exists $a\in J$ such that $k+I = a+I$, which means $a-k\in I$; since $I\subseteq J$, we conclude that $a-(a-k) = k\in J$. Therefore, $k=jx$ for some $x\in R$, so $k+I = jx+I = (j+I)(x+I)\in (j+I)(R/I)$. Thus, $K\subseteq (j+I)(R/I)$. And since $j+I\in K$ and $K$ is an ideal, then $(j+I)(R/I)\subseteq K$, giving equality.

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  • $\begingroup$ What if $I$ will be a prime ideal? $\endgroup$ Commented Nov 21, 2019 at 15:40
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    $\begingroup$ @TheStudent: As the second paragraph says, the quotient of a PID is a principal ideal ring. If in addition the ideal is prime, then the quotient is a domain. A principal ideal ring that is a domain is a PID. $\endgroup$ Commented Nov 21, 2019 at 15:50

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