## Operations on Ideals

We introduce some notation:

### Ideal Generation

Let $R$ be a ring and let $X \subset R$. Then recall $\langle X \rangle$ denotes the ideal generated by $X$, that is, $\cap\{J | X\subset J \triangleleft R\}$. Note the empty set generates the zero ideal.

Set $\mathcal{L}(R) = \{J|J\triangleleft R\}$. This is a poset with respect to $\subset$. Moreover, $\mathcal{L}(A)$ is a complete lattice: for any $S \subset \mathcal{L}(A)$, we have

$\array { glb S &=& \cap\{J|J\in S\} \\ lub S &=& \langle \cup\{J|J\in S\} \rangle }$

### 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 $J,K\triangleleft R$ then $J+K=\langle J\cup K\rangle$.

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

${\sum_{i\in I}J_i } = { \langle { \cup_{i\in I} J_i } \rangle }$

Thus we see finding least upper bounds in $\mathcal{L}(R)$ is equivalent to taking sums of families of ideals.

### Products of Ideals and Sets

Now suppose $J \triangleleft R, X\subset A$. Define

$J X = { \{ { \sum_{i=1}^n a_i x_i | n\ge 1, a_i\in J, x_i \in X } \} }$

We have $J X = \langle a x | a\in J, x \in X\rangle \triangleleft R$.

Define $J x = J \{x\}$. Then we have $J x = \{a x | a \in J\}$. Note $R x = \langle x \rangle$, the principal ideal generated by $x$.

If $J,K,L \triangleleft R$, then $(J K)L = J (K L)$. More generally, if $J_1,...,J_k \triangleleft R$ then

$J_1 ... J_k = \langle x_1 ... x_k | x_i \in J_i \rangle = { \{ { \sum_{i=1}^n x_{i 1}...x_{i k} | n\ge 0 , x_{i j} \in J_j } \} }$

### Powers of Ideals

Powers of $J\triangleleft R$ are defined as follows:

$\array { J^0 &=& R \\ J^1 &=& J \\ J^k &=& J J ... J \\ &=& \langle x_1 ... x_k | x_1,...,x_k \in J\rangle }$

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

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

Example:

1. Let $R=\mathbb{Z}$, $J =\langle m \rangle, K=\langle n\rangle$ where $m,n \in \mathbb{Z}$. Then we have $J\cap K = \langle lcm(m,n)\rangle, J+K=\langle gcd(m,n)\rangle$. Hence the lattice of ideals of $\mathbb{Z}$ can be identified with the lattice

$(\mathbb{Z}_{\ge 0}, |)$

We also have $J K = \langle m n \rangle$, thus $J K = J\cap K$ if and only if $m, n$ are coprime. 2. Let $A =F[x_1,...,x_n]$ for a field $F$. Let $J=\langle x_1,...,x_n\rangle$, so $J$ consists of the polynomials with zero constant term. Then for $m\ge 1$, $J^m$ is precisely the set of the polynomials where each term has degree at least $m$.

The following are easy to verify:

1. $J(K+L)=J K + J L$

2. $J\cap(K+L)\supset (J\cap K)+(J\cap L)$

3. $J\supset K \implies J\cap(K+L) = (J\cap K) +(J\cap L)$ (modular law)

4. $(J+K)(J\cap K) \subset J K$

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

(TODO: Hasse diagram)

Example:

1. In $\mathbb{Z}$, 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 $J \cap K = (J + K)(J \cap K) \subset J K \subset J \cap K$.