Let $R$ be an integral domain and $p,q \in R$. Is it true that $(pq)R = pR \cap qR$?

Question

Let $R$ be an integral domain, i.e., a commutative ring with 1 that has no-zero divisors other than 0. Let $p,q \in R$. under what conditions can we ensure that $(pq)R = pR \cap qR$, where $(pq)R$ is the principal ideal generated by the product $pq$.

One inclusion is true:

If $x \in (pq)R$, then $x = p(qr) = q(pr)$ for some $r \in R$. Hence, $x = p(qr) \in pR$ and $x = q(pr) \in qR$ so that $x \in pR \cap qR$.

For the reverse inclusion:

If $x \in pR \cap qR$, then $x = pr$ and $x = qr'$ for some $r,r' \in R$. Hence, $pr = qr'$. Is it possible to ensure (additional assumptions?) that $x \in (pq)R$ in this case?

Answer 1

Suppose $R$ is a commutative ring and $p$ is not a zero divisor. Then $pR \cap qR \subseteq pqR$ if and only if, for all $r \in R$, $q \mid pr$ implies $q \mid r$.

Proof:

• Suppose $pR \cap qR \subseteq pqR$. If $q \mid pr$, then $pr$ is in $pR \cap qR$, so $pr \in pqR$ and $pr = pqs$ for some $s \in R$. Then $r = qs$ (since $p$ is not a zero divisor), so $q \mid r$.

• Suppose $q \mid pr$ implies $q \mid r$ for all $r$. If $x \in pR \cap qR$, then $x = pr = qr'$ for some $r,r'\in R$, so $q \mid pr$ which implies $q \mid r$. So letting $r = py$ we have $x = pr = pqy \in pqR$.

Here are two different sufficient conditions:

1. If $p$ and $q$ are coprime, that is, $1 = pa+qb$ for some $a,b \in R$, then it holds. For if $x = pr = qr'$, then $x = (pa+qb)x = pq(ar' + br) \in pqR$.

2. Recall that an element $q \in R$ is prime if for all $a,b \in R$, $q \mid ab$ implies $q \mid a$ or $q \mid b$. If $q$ is a prime that doesn't divide $p$, then it's true: If $q \mid pr$, then $q \mid r$ since $q$ is prime and $q \not\mid p$. (Note that it's not enough here for $q$ to be irreducible: in $R = \mathbb{Z}[\sqrt{-5}]$, we have the irreducibles $p=2$ and $q=1+\sqrt{-5}$, but $6 = 3p = (1-\sqrt{5})q$ is in $pR \cap qR$ but not $pqR$.)

Answer 2

The most natural set of conditions I can think of which would impose this equality, is to have R be a GCD domain, and have p and q be relatively prime.

Since p and q are relatively prime, their GCD is one. This means their LCM is their product, so $pR \cap qR = (pq)R$.

An alternative condition you could use, is to have R be a field, in which case the ideal generated by any non-zero element is R, so when both p and q are non-zero, then the intersection of there ideals is R, and so is the ideal generated by their product. If either p or q is zero, then the intersection of their ideals and the ideal generated by their products are both $\{0\}$.