# 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

## 1 Answer

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 $$x = p_2^n b = p_2^n p_1^m b_m \in (p_1^m)(p_2^n) = Q_1 Q_2 \, .$$

We can use the above in the induction step to prove this result for any finite number of primary ideals. I'll leave the details to you.

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.