Conditions for a general formula for intersections of finitely generated ideals in UFDs

Question

What are some conditions to have a general formula for computing intersections of finitely generated ideals in a UFD? $\def\lcm{\mathrm{lcm}}$ $\def\N{\mathbb{N}}$

I would like such a formula as a practical computational tool to compute intersections, particularly by hand, for example when verifying a primary decomposition.

In particular, I'm interested in conditions under which the following formula (for finite intersections) holds in general: $$ (a_1, \ldots, a_n) \cap (b_1, \ldots, b_m) = (\lcm(a_i, b_j))_{i=1,j=1}^{i = n, j= m} \tag{*} $$

I'm especially interested in polynomial rings over a field, since that is the case I have to deal with most often.

What I've found so far

• The formula (*) is always valid in a UFD for principal ideals; that is, $(a) \cap (b) = (\lcm(a, b))$. This can be readily seen from unique factorization.

• If $R$ is a UFD and a Prüfer domain, then by condition (12) in this answer (distributivity of intersection over addition), the formula (*) holds: $$(a_1, \ldots, a_n) \cap (b_1, \ldots, b_m) = ((a_1) + \cdots + (a_n)) \cap ((b_1) + \cdots + (b_m)) = \sum_{i, j} (a_i) \cap (b_j) = \sum_{i, j} (\lcm(a_i, b_j)) = (\lcm(a_i, b_j))_{i,j} $$

But polynomial rings over a field, which is the case I'm most interested in, are almost never Prüfer: indeed, only in the single variable case, and then the formula (*) trivially holds because the ring is a PID.

• In a polynomial ring over a field, $k[\{X_i\}_{i \in I}]$ if we restrict to ideals generated by (a finite number of) monomials, then (*) holds.

The proof I've come up with for this needs a bit of machinery, so, for clarity, I've written it separately as an answer below.

What I'm looking for

I'm interested in any of the following:

1. Are the above cases correct?
2. Are there broader classes of UFDs where (*) always holds?
3. Are there some conditions on the generators of the ideals that make (*) hold in a broad class of UFDs?
4. Are there some other formulae for (finite) intersections of finitely generated ideals that hold for a broad class of UFDs?

Answer 1

Monomial case

Consider the polynomial ring $R = k[\{X_i\}_{i \in I}]$ over a field $k$ for an arbitrary set of free variables $V = \{X_i\}_{i \in I}$. Let $a_1, \ldots, a_n$ and $b_1, \ldots, b_m$ be monomials in $R$. We want to show that (*) holds. $\def\lcm{\mathrm{lcm}}$ $\def\N{\mathbb{N}}$

Definitions

I define a monomial index to be a finitely supported function $m: I \to \N$, i.e., an assignment of a positive exponent to a finite number of variables. Let $M$ denote the set of all monomial indices. Define a partial order on $M$ defined pointwise, i.e. $m \le m'$ iff $m(i) \le m'(i)$ for all $i \in I$. Notice that $M$ is a lattice whose join and meet are the pointwise maximum and minimum which correspond to the lcm and gcd of the monomials respectively. Define functions $\kappa$ and $\mu$ that assign to each monomial $s = u\prod_{i \in I}X_i^{e_i}$ (where $u \in k$ and only finitely many $e_i$ are nonzero) its coefficient $\kappa(s) = u$ and its monomial index $\mu(s) = (i \mapsto e_i)$ respectively. For convenience, set $\mu(0) = (i \mapsto \infty)$. Note that for two monomials $s$ and $s'$, we have $s' \in (s)$, i.e. $s \mid s'$, iff $\mu(s') \ge \mu(s)$. Let $X^m$ denote the monomial of index $m$ with coefficient 1, i.e. $X^m = \prod_{i \in I} X^{m(i)}$. For any $f \in R$, there is a unique way to write $f$ as a sum of distinct monomials; let $f_m$ denote the $m$-th monomial of $f$ for any $m \in M$.

Proof

First, we show that: $$(a_1, \ldots, a_n) = \{f \in R \mid \forall m \in M \quad \mu(f_m) \ge \mu(a_1)\ \lor\ \cdots\ \lor\ \mu(f_m) \ge \mu(a_n)\}$$ (In words: the ideal generated by monomials consists of those polynomials whose monomials are each a multiple of at least one of the generators.) Indeed, if $f$ is in the right hand side, then we have $f_m \in \bigcup_{i =1}^n (a_i) \subseteq (a_i)_{i = 1}^n$ and so $f = \sum_{m \in M} f_m \in (a_i)_{i = 1}^n$. Conversely, if $f$ is in the left hand side, since $(a_1, \ldots, a_n) = (a_1) + \cdots + (a_n)$, we can write $f = \sum_{i =1}^n g_i$ where for each $i$ we have $g_i \in (a_i)$. Now we can decompose each $g_i$ into a sum of monomials, each of which must still be in $(a_i)$; write $g_i = \sum_{m \in M} g_{i, m}$, where $\kappa(g_{i, m})$ can only be nonzero if $m \ge \mu(a_i)$ by what we've said. Then, by adding together monomials of equal index, the decomposition of $f$ into monomials is $f = \sum_{i = 1}^n g_i = \sum_{i = 1}^n \sum_{m \in M} g_{i, m} = \sum_{m \in M} \left(\sum_{i = 1}^n \kappa(g_{i, m})\right) X^m$, where if the $m$-th monomial is nonzero then some for some $i$ we have that $\kappa(g_{i, m})$ is nonzero, thus $m \ge \mu(a_i)$. Ergo $f$ is in the right hand side.

Now (*) follows readily from the distributivity of logical and over logical or: $$ (a_1, \ldots, a_n) \cap (b_1, \ldots, b_\ell) \\ = \left\{f \in R \Bigm\vert \forall m \in M \quad \bigvee_{i = 1}^n (\mu(f_m) \ge \mu(a_i)) \right\} \cap \left\{f \in R \Bigm\vert \forall m \in M \quad \bigvee_{j = 1}^\ell (\mu(f_m) \ge \mu(b_j)) \right\} \\ = \left\{f \in R \Bigm\vert \forall m \in M \quad \left(\bigvee_{i = 1}^n (\mu(f_m) \ge \mu(a_i))\right) \land \left(\bigvee_{j = 1}^\ell (\mu(f_m) \ge \mu(b_j))\right) \right\} \\ = \left\{f \in R \Bigm\vert \forall m \in M \quad \bigvee_{i,j}(\mu(f_m) \ge \mu(a_i)\ \land\ \mu(f_m) \ge \mu(b_j)) \right\} \\ = \left\{f \in R \Bigm\vert \forall m \in M \quad \bigvee_{i,j}(\mu(f_m) \ge \sup\{\mu(a_i), \mu(b_j)\}) \right\} \\ = \left\{f \in R \Bigm\vert \forall m \in M \quad \bigvee_{i,j}(\mu(f_m) \ge \mu(\lcm(a_i, b_j)) \right\} \\ = (\lcm(a_i, b_j))_{i,j} $$