Is $\Bbb Z / n \Bbb Z$ a PID?

Question

I’d like to when $\Bbb Z / n \Bbb Z$ is a PID. I don’t know if depends of the value of $n$, it is true for all $n$ or $\Bbb Z / n \Bbb Z$ is never a PID. No idea. In case it’s true I’d like to see the proof (without group theory) .

Thanks!

Answer 1

Recall that a Principal Ideal Ring is a ring $R$ in which every ideal is principal.

Theorem. Let $R$ be a principal ideal ring, and let $I$ be an ideal of $R$. Then $R/I$ is also a principal ideal ring.

Proof. Let $J$ be an ideal of $R/I$, and let $K=\{r\in R\mid r+I\in J\}$ be the pullback of $J$ to $R$. Since $R$ is a principal ideal ring, we know that there exists $a\in R$ such that $K=(a)$. I claim that $(a+I) = J$.

Since $a\in K$, then $a+I\in J$, so $(a+I)\subseteq J$. Conversely, let $x\in J$. Then $x=r+I$ for some $r\in R$, and hence $r\in K$. Therefore, $r\in (a)$, so we can write $r$ as (note that I’m not assuming $R$ is commutative or has a unity): $$r = na + xa + ay + \sum_{i=1}^m x_iay_i$$ where $n\in\mathbb{Z}$, $m\in\mathbb{N}$, $x,y,x_i,y_i\in R$. Then $$\begin{align*} x &= r+I\\ &= (na + xa + ay + \sum_{i=1}^mx_iay_i) + I\\ &= n(a+I) + (x+I)(a+I) + (a+I)(y+I)\\ &\qquad + \sum_{i=1}^m (x_i+I)(a+I)(y_i+I) \end{align*}$$ which lies in $(a+I)$. Thus, $J\subseteq (a+I)$, proving the desired equality.~$\Box$

Thus, since $\mathbb{Z}$ is a PID, hence a principal ideal ring, it follows that for any $n$, $\mathbb{Z}/n\mathbb{Z}$ is always a principal ideal ring.

In particular, $\mathbb{Z}/n\mathbb{Z}$ is a principal ideal domain if and only if it is a domain. And this happens if and only if $n\mathbb{Z}$ is a prime ideal of $\mathbb{Z}$, which happens if and only if $n$ is a prime number.