# 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 $$J \cap K = (J + K)(J \cap K) \subset J K \subset J \cap K$$.

Ben Lynn blynn@cs.stanford.edu 💡