Thread: [SOLVED] A,B,C right ideals. (A+B)C = AC+BC?

1. [SOLVED] A,B,C right ideals. (A+B)C = AC+BC?

Hi:
In a textbook I find: if A, B, C are right ideals in a ring R, then A(B+C) = AB+AC. This I have prove. But I could also prove (A+B)C = AC+BC (1)
where again A, B, C are right ideals in R. Can (1) be true? Any hint will be welcome.

2. Originally Posted by ENRIQUESTEFANINI
Hi:
In a textbook I find: if A, B, C are right ideals in a ring R, then A(B+C) = AB+AC. This I have prove. But I could also prove (A+B)C = AC+BC (1)
where again A, B, C are right ideals in R. Can (1) be true? Any hint will be welcome.
yes it is true even for left ideals. since the product of (left, right or two-sided) ideals $\displaystyle I,J$ is defined to be $\displaystyle IJ= \{\sum_{i=1}^n x_iy_i: \ x_i \in I, y_i \in J, \ n \in \mathbb{N} \},$ we only need to prove that
$\displaystyle (a+b)c_1 \in AC+BC$ and $\displaystyle ac_1 + b c_2 \in (A+B)C,$ for all $\displaystyle a \in A, \ b \in B, \ c_1,c_2 \in C.$ these are very easy to prove:

$\displaystyle (a+b)c_1=ac_1 + bc_1 \in AC + BC$ and $\displaystyle ac_1 + bc_2 =(a+0)c_1 + (0+b)c_2 \in (A+B)C.$

3. Ideal products.

Thanks a lot, NonCommAlg.