# If A and B are ideals of a ring, show that the product of A and B, AB...

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?

• Please use MathJax. Also what have you tried?
– jgon
Commented Nov 29, 2019 at 22:28
• In the definition of $AB,$ you should set $a_i\in A.$ It's a typo. Commented Nov 29, 2019 at 22:37

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 $$6$$.