Given ideals *I* and *J* of ring *R*, we can perform the following operations to obtain new ideals:

- is an ideal of
*R*; - is an ideal of
*R*; - is an ideal of
*R*. Thus,*IJ*is the set of all finite sums of*xy*where*x*,*y*belong to*I*,*J*respectively.

Verifying that these sets are ideals is a rather routine process so we’ll leave it to the reader. Instead, we’ll spend some time explaining what these operations mean. The intersection *I* ∩ *J* needs no explanation: that’s the biggest ideal of *R* contained in both *I* and *J*. The sum *I*+*J* is the “dual” : it’s the smallest ideal of *R* which contains both *I* and *J*. Finally, *IJ* is the smallest ideal which contains *xy* for all *x* in *I* and *y* in *J*.

Note : . Indeed, if , then since I is an ideal; likewise, since J is an ideal. Hence .

The case of *R* = **Z** illustrates this quite nicely. Note that all ideals of **Z** can be written as *n***Z** for some non-negative integer *n*. Also, if and only if *b* | *a*. Thus:

*m***Z**+*n***Z**is the smallest ideal containing*m***Z**and*n***Z**, i.e.*d***Z**where*d*is the largest value dividing*m*and*n*; thus*m***Z**+*n***Z**= gcd(*m*,*n*)**Z**;*m***Z**∩*n***Z**= lcm(*m*,*n*)**Z**;*m***Z**·*n***Z**=*mn***Z**.

*Generalisation.* One can generalise the intersection and sum to arbitrary collections of ideals. Thus if {*I _{λ}*} is a collection of ideals, we can define ∩

_{λ}*I*as the set of

_{λ}*x*which occurs in

*all*

*I*and ∑

_{λ }_{λ }

*I*as the set of all finite sums of elements from

_{λ}*(we can only take finite sums since we can’t sum an infinite number of elements here without the concept of convergence).*

*I*_{λ}Although we can’t take an infinite product of *I _{λ}*, we can take any finite product of

*I*

_{1},

*I*

_{2}, …,

*I*inductively: first take

_{n}*I*

_{1}

*I*

_{2}, then (

*I*

_{1}

*I*

_{2})

*I*

_{3}, then ((

*I*

_{1}

*I*

_{2})

*I*

_{3})

*I*

_{4}… It turns out taking product of ideals is associative (

*proof: exercise*), so we can drop the brackets and write

*I*

_{1}

*I*

_{2}…

*I*.

_{n}Another way of looking at the *n*-term product *I* = *I*_{1}*I*_{2}…*I _{n}* is that

*I*is the smallest ideal which contains all products of the form

, for .

*Another example*. Consider the ring **R**[*x*, *y*] of polynomials in *x*, *y* with real coefficients. Define the ideals and , i.e. *I* (resp. *J*) is the set of polynomials divisible by *x* (resp. *y*). Then:

*I*+*J*= set of all polynomials in*x*,*y*without constant term;- = set of multiples of
*xy*; - = set of multiples of
*xy*.

Exercise. Let R and S be rings and take the ring product R × S. Prove that if I and J are ideals of R and S respectively, then I × J is an ideal of R × S. Note that the quotient is:

Prove also that conversely, any ideal of R × S must be of the form I × J for some ideals I and J of R and S respectively.

The ideals {0} × S and R × {0} give quotients which are isomorphic to R and S respectively. Hence, even though R and S are not subrings of R × S, they’re its quotient rings.