Product of Primary Ideals is the Intersection of Prime Ideals

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.

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...

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.

• Division by a principal ideal works well in the sense that if $I$ is an ideal and $I \subset (a)$ then $I = (a) \mathfrak{b}$ where $\mathfrak{b} = \{ \frac{u}{a}, u \in I\}$ is an ideal. Thus you can write your intersection of primary ideals as a product of (primary) ideals. – reuns Nov 19 '17 at 21:42
• @reuns Okay. I'm not sure how that is helpful, nor do I understand how $\frac{u}{a}$ could make any sense, especially if $a$ isn't a unit. – user193319 Nov 19 '17 at 21:49
• @reuns $Frac(R)$ hasn't been introduced yet. – user193319 Nov 19 '17 at 22:01
• There is somewhat of a typo in the statement of (c). For instance, $2$ and $-2$ are distinct primes of $\mathbb{Z}$, but $(2)(-2) = (4)$ and $(2) \cap (-2) = (2) = (-2)$. Probably what they mean to say is that the $p_i$ are non-associate primes. – André 3000 Nov 19 '17 at 22:01
• PID assume integral domain. $(a) = \{ av, v \in R\}$. For $a \ne 0$ the multiplication by $a$ is injective so recovering $v$ from $av$ is of course allowed. In other words, the fraction field is introduced automatically. Showing $I \subset \mathfrak{a} \implies I = \mathfrak{a} \mathfrak{b}$ is the main step to factorize ideals in products of prime ideals. – reuns Nov 19 '17 at 22:14

In fact, the inclusion $IJ \subseteq I \cap J$ is always true. Given ideals $I$ and $J$, an element of $IJ$ is of the form $x = a_1 b_1 + \cdots + a_t b_t$ for some $t \in \mathbb{Z}_{\geq 0}$ and $a_k \in I$ and $b_k \in J$. Since ideals are closed under arbitrary multiplication, then each term of the above sum is in both $I$ and $J$, and since ideals are closed under addition, then the sum is in $I$ and $J$, so $x \in I \cap J$.
Now, let's consider the reverse inclusion. For ease of notation, let's just consider two primary ideals $Q_1$ and $Q_2$. By part (b) we have $Q_1 = (p_1^m)$ and $Q_2 = (p_2^n)$ for some primes $p_1, p_2$ and, as per my comment, assume $p_1$ and $p_2$ are non-associate. Given $x \in Q_1 \cap Q_2$, then $x = p_1^m a$ and $x = p_2^n b$ for some $a,b \in R$. Then $p_1 \mid p_2^nb$, so $p_1 \mid p_2^n$ or $p_1 \mid b$. If $p_1 \mid p_2^n$, then by repeated using the fact that $p_1$ is prime we find that $p_1 \mid p_2$. Then $p_2 = p_1 c$ for some $c \in R$. But since primes are irreducible, then $c$ must be a unit, which contradicts that $p_1$ and $p_2$ are non-associate. Thus $p_1 \mid b$, so $b = p_1 b_1$ for some $b_1 \in R$.
Now we have $$p_1^m a = x = p_2^n b = p_2^n p_1 b_1 \, .$$ Since $R$ is a domain, then we cancel the factor of $p_1$, which yields $p_1^{m-1} a = p_2^n b_1$. Repeating the same argument as above (or by induction, if you want to be rigorous), then $p_1 \mid b_1$ so $p_1^2 \mid b$, and so on and so forth until we get $p_1^m \mid b$. Then $b = p_1^m b_m$ for some $b_m \in R$, so $$x = p_2^n b = p_2^n p_1^m b_m \in (p_1^m)(p_2^n) = Q_1 Q_2 \, .$$
Suppose we know the result for $k$ primary ideals. Then \begin{align*} Q_1 \cap \cdots \cap Q_k \cap Q_{k+1} &= (Q_1 \cap \cdots \cap Q_k) \cap Q_{k+1} = (Q_1 \cdots Q_k) \cap Q_{k+1} \, . \end{align*} Given $x \in \bigcap_{i=1}^{k+1} Q_i = (Q_1 \cdots Q_k) \cap Q_{k+1}$, then $$x = p_1^{n_1} \cdots p_k^{n_k} a \qquad \text{and } \qquad x = p_{k+1}^{n_{k+1}} b$$ for some $a,b \in R$. You can use the same proof as above to show that $p_{k+1}^{n_{k+1}} \mid a$ which will finish the induction.