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 \(J K \subset J\cap K\).


  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)


  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 💡