Commutative Algebra - Operations on Ideals

We introduce some notation:

Ideal Generation

Let $R$ be a ring and let $X \subset R$ . Then recall $⟨ X ⟩$ denotes the ideal generated by $X$ , that is, $\cap {J ∣ X \subset J ◃ R}$ . Note the empty set generates the zero ideal.

Set $ℒ (R) = {J ∣ J ◃ R}$ . This is a poset with respect to $\subset$ . Moreover, $ℒ (A)$ is a complete lattice: for any $S \subset ℒ (A)$ , we have $\begin{matrix} glb S & = & \cap {J ∣ J \in S} \\ lub S & = & ⟨ \cup {J ∣ J \in S} ⟩ \end{matrix}$

Sums of Sets

If $X, Y \subset R$ then define $X + Y = {x + y ∣ x \in X, y \in Y}$ . Define $X + y = X + {y}$ . This is consistent with our notation for cosets. Also note if $J, K ◃ R$ then $J + K = ⟨ J \cup K ⟩$ .

More generally, if ${J_{i} ∣ i \in I}$ is a family of ideals in $R$ , then define $\sum_{i \in I} J_{i} = {\sum_{i \in I} x_{i} ∣ x_{i} \in J_{i}}$ where only finitely many $x_{i}$ are nonzero. Then we have $\sum_{i \in I} J_{i} = ⟨ \cup_{i \in I} J_{i} ⟩$ Thus we see finding least upper bounds in $ℒ (R)$ is equivalent to taking sums of families of ideals.

Products of Ideals and Sets

Now suppose $J ◃ R, X \subset A$ . Define $J X = {\sum_{i = 1}^{n} a_{i} x_{i} ∣ n \geq 1, a_{i} \in J, x_{i} \in X}$

We have $J X = ⟨ a x ∣ a \in J, x \in X ⟩ ◃ R$ .

Define $J x = J {x}$ . Then we have $J x = {a x ∣ a \in J}$ . Note $R x = ⟨ x ⟩$ , the principal ideal generated by $x$ .

If $J, K, L ◃ R$ , then $(J K) L = J (K L)$ . More generally, if $J_{1}, . . ., J_{k} ◃ R$ then $J_{1} . . . J_{k} = ⟨ x_{1} . . . x_{k} ∣ x_{i} \in J_{i} ⟩ = {\sum_{i = 1}^{n} x_{i 1} . . . x_{i k} ∣ n \geq 0, x_{i j} \in J_{j}}$

Powers of Ideals

Powers of $J ◃ R$ are defined as follows: $\begin{matrix} J^{0} & = & R \\ J^{1} & = & J \\ J^{k} & = & J J . . . J \\ = & ⟨ x_{1} . . . x_{k} ∣ x_{1}, . . ., x_{k} \in J ⟩ \end{matrix}$

Hence $J^{k} = {0}$ if and only if all products of $k$ elements of $J$ are zero.

Note if $J, K ◃ R$ then $J K \subset J \cap K$ .

Example

Let $R = ℤ$ , $J = ⟨ m ⟩, K = ⟨ n ⟩$ where $m, n \in ℤ$ . Then we have $J \cap K = ⟨ lcm (m, n) ⟩, J + K = ⟨ \gcd (m, n) ⟩$ . Hence the lattice of ideals of $ℤ$ can be identified with the lattice $(ℤ_{\geq 0}, ∣)$ We also have $J K = ⟨ m n ⟩$ , thus $J K = J \cap K$ if and only if $m, n$ are coprime.
Let $A = F [x_{1}, . . ., x_{n}]$ for a field $F$ . Let $J = ⟨ x_{1}, . . ., x_{n} ⟩$ , so $J$ consists of the polynomials with zero constant term. Then for $m \geq 1$ , $J^{m}$ is precisely the set of the polynomials where each term has degree at least $m$ .

The following are easy to verify:

$J (K + L) = J K + J L$
$J \cap (K + L) \supset (J \cap K) + (J \cap L)$
$J \supset K \Rightarrow J \cap (K + L) = (J \cap K) + (J \cap L)$ (modular law)
$(J + K) (J \cap K) \subset J K$

We have $(J + K) (J \cap K) \subset J K \subset J \cap K \subset J, K \subset J + K$

[TODO: Hasse diagram]

Example:

In $ℤ$ , since $lcm (a, \gcd (b, c)) = \gcd (lcm (a, b), lcm (a, c))$ for nonnegative integers $a, b, c$ we have $J \cap (K + L) = (J \cap K) + (J \cap L)$ . Also, as $a b = \gcd (a, b) lcm (a, b)$ we have $(J + K) (J \cap K) = J K$ .

We say ideals $J, K$ of a ring $R$ are coprime or maximal if $J + K = R$ . Equivalently, for some $x \in J, y \in K$ we have $x + y = 1$ . Note that if $J, K$ are coprime then $J \cap K = J K$ , since we have $J \cap K = (J + K) (J \cap K) \subset J K \subset J \cap K$ .