# Question about quotients of reducible elements and modules

First, I apologize if this is a stupid question. We started doing modules in my class a few days ago and I'm totally lost with the basics, I think.

Suppose we have a ring $K$ and $M$ a $K$-module. Let $k \in K$ be a reducible element. Let $l \in K$ be another reducible element and say $l = p^r$ where $p$ is some irreducible element in $K$.

What does $K/(k)$ look like? And $K/(l)$? What about $K/(p_{1}^{a_{1}} \ldots p_{n}^{a_{n}} )$ where $p_i$ is irreducible.

I know how to work with quotients involving irreducible elements. There are numerous theorems to think about and apply, but when working with reducible ones, I just can't seem to wrap my head around them. How can I think about these in terms of sums of cyclic modules?

Also, I apologize again if my terminology is wrong or my question doesn't make sense. I just started learning this all recently so I'm not quite sure how to phrase what I mean yet.

Now, if $K$ is not a UFD, this problem would be a little more difficult to explain.
For example, $\mathbb{Z}[\sqrt{-5}]$ is not a UFD since $2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})$. But, we have $\mathbb{Z}[\sqrt{-5}]\cong \mathbb{Z}[X]/(X^2+5)$. Reducing by $2\mathbb{Z}[X]$ gives $\mathbb{Z}[X]/(2,X^2+5)=(\mathbb{Z}/2\mathbb{Z})[X]/(X^2-1)=\mathbb{Z}/2\mathbb{Z}[X]/(X-1)^2$.
So, let's assume $K$ is a UFD. Then for any (nonunit and nonzero) $k\in K$ we have that $k$ decomposes into a product of irreducibles. If $k=p_1^{\alpha_1}\cdots p_s^{\alpha_s}$ is the canonical decomposition of $k$ (up to multiplication by units) we have $(k)=(p_1^{\alpha_1})\cdots(p_s^{\alpha_s})$. Also, as $\gcd(p_i,p_j)=1$ for all $i\neq j$. we have $(p_i^{\alpha_i})+(p_j^{\alpha_j})=(1)$. Thus, by the Chinese Remainder Theorem for commutative rings, we have $K/(k)\cong K/(p_i^{\alpha_i})\times \cdots \times K/(p_s^{\alpha_s})$. Here, if $\alpha_i=1$ we have that this is just $K/(p_i)$ and, as $p_i$ generates a prime ideal, $K/(p_i)$ is an integral domain. If $p_i$ is also maximal, then $K/(p_i)$ is a field.
As for $K/(p^r)$, we can work a bit with this. Let's say that $(p)$ is a maximal ideal so that $(p)(p)\cdots (p)=(p^r)$ is the product of the maximal ideals. Then for any prime ideal $I\supset (p^r)$, we have $I\supset(p)$ by properties of prime ideals. If $Q$ is a prime ideal in $K/(p^r)$, then there is a corresponding prime ideal $Q^c\subset K$ with $Q^c\supset (p)$ and therefore $Q^c=(p)$ as $(p)$ is assumed maximal. Therefore, there is only one prime ideal of $K/(p^r)$ and $K/(p^r)$ is a local ring. In fact, in any PID an irreducible element $p$ will generate a maximal ideal $(p)$. A more general observation is that $(p)$ is maximal amongst all proper principal ideals of $K$.