Let $R$ be a ring, $A,B$ subsets of $R$, and $a,b,a_i,b_i\in R$. Definte the multiplication $$AB:=\{\sum_{i=1}^n a_ib_i|a_i\in A,b_i\in B,n\ge1\}$$
In other words, $AB$ is the smallest subset of $(R,+)$ which contains $\{ab|a\in A,b\in B\}$ and which is closed under addiction.
when $A=\{a\}$
$$aB:=\{a\}B=\{\sum_{i=1}^sab_i|b_i\in B,s\ge1\}$$
similarly, when
$B=\{b\}$
$$Ab:=A\{b\}=\{\sum_{i=1}^sa_ib|a_i\in A,s\ge1\}$$
Then my professor
claimed that
"if $0\in B$, then
$$\{a_1,…,a_s\}B=a_1B+…+a_sB$$
$$B\{a_1,…,a_t\}=Ba_1+…+Ba_t$$
what i don't understand is that why we need $0\in B$ and how to prove the above claim. This looks very trivial but I could not figure it out. Anyone can give me some hints? Thanks.
my thought is:
$$\{a_1,…,a_s\}B=\sum_{i_1=1}^{n_1}a_1b_{i_1}+…+\sum_{i_s=1}^{n_s}a_sb_{i_s}=a_1B+…+a_sB$$
What's wrong with my reasoning as i did not use the fact that $0\in B$?
Best Answer
We use $0\in B$ to get $a_iB\subseteq a_1B+\dots +a_nB$, so that $a_ib\in a_1B+\dots +a_nB$ for every $b\in B$.
The other inclusion follows immediately from the definition of $\{a_1,\dots, a_n\} B$.
To get a counterexample, just consider $R=\Bbb Z,\ B=\{1\},\ a_i>0$ and $n\ge 2$. Then $a_i\in\{a_1,\dots,a_n\}B$ but every element of $a_1B+a_2B+\dots+a_nB$ is at least $a_1+a_2+\dots+a_n\,>a_i$.