Operations on Ideals: Sum, Product, Intersection

The comaximal product–intersection identity is the direct algebraic engine behind the Chinese Remainder Theorem for general rings: when $I_1, \dots, I_n$ are pairwise comaximal, the map $R \to R/I_1 \times \cdots \times R/I_n$ is surjective with kernel $I_1 \cap \cdots \cap I_n = I_1 I_2 \cdots I_n$, reconstructing the classical CRT for integers as the special case $R = \mathbb{Z}$ and $I_k = (n_k)$ with pairwise coprime $n_k$. This appears throughout computer algebra systems for fast modular arithmetic, in RSA-style cryptography for computing modulo composite numbers via prime factors, and in error-correcting codes. In algebraic geometry, sums, products, and intersections of ideals correspond geometrically to intersections, unions (with multiplicity), and unions of the associated vanishing sets, via the Nullstellensatz dictionary between ideals in $k[x_1,\dots,x_n]$ and algebraic varieties; the distinction between $IJ$ and $I \cap J$ tracks the difference between a scheme-theoretic intersection with multiplicities and a reduced set-theoretic intersection. In commutative algebra more broadly, factorization of ideals into products of prime ideals in Dedekind domains — the modern payoff of Dedekind's original motivation — relies on exactly these operations, and questions about when $IJ = I\cap J$ recur in the theory of coprime factorizations used in module theory, representation theory (idempotent decompositions of rings via comaximal ideals), and numerical algebraic geometry for solving polynomial systems by localizing at comaximal components.

Ask the tutor: How does the two-ideal comaximal theorem extend by induction to prove the full Chinese Remainder Theorem for $n$ pairwise comaximal ideals?

Think of an ideal as encoding a "divisibility condition": in $\mathbb{Z}$, the ideal $(n)$ consists of all multiples of $n$, so membership in $(n)$ means "divisible by $n$." Under this dictionary, the three operations translate into familiar number-theoretic operations, but the translation requires care. The sum $I + J$ is the smallest ideal containing both $I$ and $J$ — it is generated by throwing all the generators of $I$ and $J$ together. In $\mathbb{Z}$, $(m) + (n) = (\gcd(m,n))$: an element is a sum of a multiple of $m$ and a multiple of $n$ exactly when it is a multiple of $\gcd(m,n)$, by Bézout's identity. So $I + J = R$ (comaximality) corresponds exactly to $\gcd(m,n) = 1$. The intersection $I \cap J$ asks for elements satisfying both divisibility conditions simultaneously — the common multiples. In $\mathbb{Z}$, $(m) \cap (n) = (\operatorname{lcm}(m,n))$. The product $IJ$ is subtler: it is generated by products of elements from each ideal, not simply their intersection. In $\mathbb{Z}$, $(m)(n) = (mn)$. Since $\operatorname{lcm}(m,n) \mid mn$ always, with equality iff $\gcd(m,n) = 1$, we recover in miniature the general fact that $IJ \subseteq I \cap J$, with equality precisely under comaximality.

Example 1 (Two moduli in $\mathbb{Z}$, non-coprime — illustrating strict containment). Let $R = \mathbb{Z}$, $I = (12)$, $J = (18)$. Sum. We compute $I + J = (\gcd(12,18)) = (6)$. To verify directly: $12 \cdot(-1) + 18\cdot 1 = 6 \in I+J$, so $(6) \subseteq I+J$. Conversely every element of $I+J$ has the form $12m + 18n = 6(2m+3n)$, a multiple of $6$, so $I+J \subseteq (6)$. Hence $I+J=(6) \ne \mathbb{Z}$: not comaximal. Intersection. $I \cap J$ consists of common multiples of $12$ and $18$, i.e. $I\cap J = (\operatorname{lcm}(12,18)) = (36)$. Product. $IJ = (12)(18) = (216)$, since the product ideal of two principal ideals $(m),(n)$ in $\mathbb{Z}$ is $(mn)$ — every generator is $12i\cdot 18j = 216 ij$, and conversely $216 = 12\cdot 18 \in IJ$ generates all multiples of $216$ as sums of such products (already $216\cdot k = (12k)\cdot 18 \in IJ$ for each $k$). Comparison. Indeed $IJ = (216) \subsetneq (36) = I \cap J$, consistent with Central results part (c), and strict because $\gcd(12,18) = 6 \neq 1$, so comaximality fails and the theorem does not force equality. Concretely $36 \in I\cap J$ but $36 \notin IJ = (216)$ since $216 \nmid 36$.

Example 2 (Comaximal ideals in a polynomial ring — verifying $IJ = I\cap J$ from scratch). Let $R = \mathbb{Q}[x]$, and let

Step 1: show $I,J$ are comaximal. We seek $a(x) \in I$, $b(x)\in J$ with $a+b=1$. Try $a(x) = c(x-1)$, $b(x) = d(x-2)$ for constants $c,d$ to be found (degree considerations: since $(x-1)$ and $(x-2)$ each have degree-1 generators, look for a linear combination with constant coefficients). We want

for all $x$, i.e. $(c+d)x - (c+2d) = 1$. Matching coefficients: $c+d = 0$ and $-(c+2d) = 1$. From the first, $d = -c$; substituting into the second: $-(c - 2c) = 1 \Rightarrow c = 1$, so $d = -1$. Check:

So $a = x-1 \in I$, $b = -(x-2) \in J$, and $a+b=1$; by the Corollary, $I+J = R$, i.e. $I,J$ are comaximal. Step 2: compute $I\cap J$ directly. An element $f(x) \in I \cap J$ satisfies $f(1)=0$ and $f(2)=0$ (since $(x-1) \mid f$ and $(x-2)\mid f$ in $\mathbb{Q}[x]$, a UFD, using that $x-1,x-2$ are non-associate irreducibles). Hence $(x-1)(x-2) \mid f$, so $I \cap J \subseteq \big((x-1)(x-2)\big)$. Conversely $(x-1)(x-2) \in I$ (it's a multiple of $x-1$) and also $\in J$ (multiple of $x-2$), so $\big((x-1)(x-2)\big) \subseteq I \cap J$. Thus

Step 3: compute $IJ$ directly. Every generator $a(x)(x-1) \cdot b(x)(x-2)$ (for $a,b \in \mathbb{Q}[x]$) is a multiple of $(x-1)(x-2)$, so $IJ \subseteq \big((x-1)(x-2)\big)$; and taking $a=b=1$ shows $(x-1)(x-2) \in IJ$ itself, so equality: $IJ = \big((x-1)(x-2)\big)$. Step 4: conclusion. $IJ = I \cap J = (x^2-3x+2)$ — exactly matching part (d) of the theorem, since $I,J$ were shown comaximal in Step 1. This also matches the general principle: the polynomial $(x-1)(x-2)$ has as roots precisely $1$ and $2$, mirroring how $\operatorname{lcm}$ and product coincide for coprime integers.