3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

I'm not sure about sums of cyclic modules but, perhaps I can help with the ring quotients. Also, in the discussion that follows I assume all rings are commutative with unit. This is probably not what you are looking for but I hope it helps.

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 $K$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.