Question

If A and B are ideals of a ring, show that the product of A and B is an ideal.
AB= {a1b1 +.......+anbn | ain A, b in B, n a positive integer}

Answer 1

do know the definition of an ideal?

Answer 2

or how to prove something is an ideal?

Answer 3

There is an ideal test that states: A non empty subset A of a ring R is an ideal if
1. a-b is in A whenever a, b are in A
2. ra and ar are in A whenever a is in A and r is in R

Answer 4

cool.

Answer 5

let \(a,b\in AB\) so they can be written as
\[\large a=\sum_{i=1}^{k_a}a_ib_i \]
and
\[\large b=\sum_{j=1}^{k_b}a_jb_j \]
so the question is wheter a-b is still in AB, right?

Answer 6

that and if ra and ar are in AB......right?

Answer 7

ok
\[\large a-b=\sum_{i=1}^{k_a}a_ib_j-\sum_{j=1}^{k_b}a_jb_j \]

Answer 8

that is still an element isn't it?

Answer 9

this is a finite sume of elements of the form a_tb_t, so it is an element of AB

Answer 10

now let \(r\in R\) where R is the ring where A and B are ideals. then
\[\large ra=r\sum_{i=1}^{k_a}a_ib_i=\sum_{i=1}^{k_a}r(a_ib_i)=
\sum_{i=1}^{k_a}(ra_i)b_i \]

Answer 11

since A is an ideal, then each ra_i is still in A. so the last sum is a finite sum of elements of the form ab where \(a\in A\) and \(b\in B\). therefore ra is still in AB

Answer 12

u can do the same with \(ar\).

Answer 13

thank you so much!

Answer 14

u r welcome