I got this problem in my textbook.
If $A$ and $B$ are ideals of a ring, show that the product of $A$ and $B$, $$AB = \{a_1 b_1 + a_2 b_2 + \ldots+a_n b_n | a_i \in A, b_i \in B, n \text{ a positive integer}\},$$ is an ideal.
Is not $AB$ a set with just one element? not to mention an ideal or a ring?
Of course, a sum of two elements of $AB$ is also in the ideal $AB.$ Note also that for $r\in R$ and $AB \ni x=\sum_{i=1}^n a_i b_i,$ where $a_i\in A,b_i\in B,$ one has $rx=\sum_{i=1}^n ra_i b_i$ and since $ra_i\in A$ is a element of the ideal $A$, the statement follows.
You asked "is AB not a set with just one element?". The answer is no, because $AB$ consists of all elements that can be written as a finite sum of products of elements in $A$ and $B$.
A concrete example: take the ring $R = \Bbb{Z}$ and ideals $A = (2)$, $B = (3)$. In other words, $A$ is the set that consists of all multiples of $2$, and $B$ consists of all multiples of $3$.
Note that $6 \in AB$, since $6 = 2\cdot 3$ where $2 \in A$ and $3 \in B$. But also $24 \in AB$ since $24$ can be written as $24 = 2\cdot6 + 4\cdot 3$, where $2,4\in A$ and $6,3 \in B$. You could show that in this case, $AB = (6)$, or the set of multiples of $6$.
