How Do Primary Ideals in a PID Relate to Their Product and Intersection?

[Bashyboy]

Homework Statement

Let $R$ be a principal ideal domain. (a) Every proper ideal is a product of $P_1...P_n$ of maximal ideals which are uniquely determined up to an order. (b) $P$ is a primary (not prime) ideal if and only $P=(p^n)$ for some prime $p \in R$ and $n \in \Bbb{N}$. (c) If $P_1,...,P_k$ are primary ideals such that $P_i = (p_i^{n_i})$ and the $p_i$ are distinct primes, then $P_1...P_k = P_1 \cap ... \cap P_k$. (d) Every proper ideal in $R$ can be expressed (uniquely up to order) as the intersection of a finite number of primary ideals.

Homework Equations

The Attempt at a Solution

I am currently working on part (c) and could use some help. Here is what I have so far. Since we are taking the $p_i$ to be distinct, I believe we can also take them not be associates of each other, but I'm not entirely certain. First note that $(p_1^{n_1})...(p_k^{n_k}) = (p_1^{n_1}..p_k^{n_k})$, since we are working in a commutative ring. Let $x \in (p_1^{n_1}..p_k^{n_k})$. Then $p_1^{n_1}...p_k^{n_k} \mid x$ and therefore $p_i^{n_i} \mid x$ for every $i$ and finally $x \in (p_i^{n_i})$ for every $i$, proving that $(p_1^{n_1})...(p_k^{n_k}) \subseteq \bigcap_{i=1}^k (p_i^{n_i})$.

Now we try to prove the other set inclusion. First, since $\bigcap_{i=1}^k (p_i^{n_i})$ is the intersection of ideals, it must also be an ideal, from which we can conclude $\bigcap_{i=1}^k (p_i^{n_i}) = (y)$ for some $y \in R$. This $y$ must be nonzero and not a unit, which means that there exist primes/irreducibles $c_i$ and integers $m_i \in \Bbb{N}$ such that $y = c_i^{m_1} ... c_s^{m_s}$, because $R$ is also a UFD. Since $(y) = \bigcap_{i=1}^k (p_i^{n_i})$, $c_i^{m_1} ... c_s^{m_s}$ must be in the intersection, which implies that, for every $i$, $p_i^{n_i} \mid c_i^{m_1} ... c_s^{m_s}$ and therefore $p_i \mid c_i^{m_1} ... c_s^{m_s}$. Because $p_i$ is prime, it follows that $p_i \mid c_j^{m_j}$ for some $j$, and through induction we get $p_i \mid c_j$. This means that $c_j = \ell_{ij} p_i$ for some $\ell_{ij} \in \Bbb{N}$, and since $c_j$ is also irreducible, $\ell_{ij}$ must be a unit, from which it follows $c_j$ and $p_i$ are associates. So, for every $i$, there exists a $j$ such that $p_i$ and $c_j$ are associates, so $k \le s$ (can we take, WLOG, $k=s$?) For convenience, relabel the elements so that $p_i$ and $c_i$ are associates...

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Okay. At this point, I am unsure of how to proceed. I am trying to show that $c_i^{m_1} ... c_s^{m_s} \in (p_1^{n_1})...(p_k^{n_k})$, but the path is obscure for some reason, although I believe I am close. I could use some hints.

 

[fresh_42]

I think it is a bit too complicated what you wrote. Let me try to sort it a little bit by only using what you already have. We have to show:

$P_1...P_k = P_1 \cap ... \cap P_k$

I don't think you need the six lines lines for the inclusion $P_1...P_k \subseteq P_1 \cap ... \cap P_k$
All $P_i$ are ideals, so any product $q_1\ldots q_k = q_1 \ldots q_{i-1}\cdot q_i \cdot q_{i+1}\ldots q_k \in P_i$, i.e. the inclusion holds.
For the other direction, say $Q = P_1 \cap ... \cap P_k$ which ...

since $\bigcap_{i=1}^k (p_i^{n_i})$ is the intersection of ideals, it must also be an ideal

... and is proper by definition. Now we have

(a) Every proper ideal is a product of $M_1...M_n$ of maximal ideals which are uniquely determined up to an order.

So $P_1 \ldots P_k \subseteq Q = (y) = M_1 \ldots M_n= (m_1) \ldots (m_n)=(m_1 \ldots m_n)$ with primes $m_i$ because maximal ideals are prime and $R$ is a PID. (This follows without any excursions to UFD and irreducibility.)

Since $(y) = \bigcap_{i=1}^k (p_i^{n_i})$

$p_i\,\vert \,p_i^{n_i}\,\vert \,m_1 \ldots m_n=y\,$ so w.l.o.g. $\,p_i=m_i\,$ for all $1 \leq i \leq k$. Thus $y=p_1\ldots p_k \cdot m_{k+1}\ldots m_n \cdot r$. Since all $p_i^{n_i}$ still divide $y$, we can cancel them one by one by repetition of the argument, so $p_1^{n_1}\ldots p_k^{n_k} \,\vert \, y$.