Show that the set generated by $A$ is $\{r_1a_1+r_2a_1+ \dots +r_ka_k \mid k \in \Bbb N, r_i \in R, a_i \in A, i \le k\}.$

Let $R$ be a commutative ring and $A$ a subset of $R$. Show that the set generated by $A$ is $\{r_1a_1+r_2a_1+ \dots +r_ka_k \mid k \in \Bbb N, r_i \in R, a_i \in A, i \le k\}.$

For the first inclusion I've picked $x \in \langle A \rangle$. This element is of form $x=a_1a_2\dots a_n$ for some $n \in \Bbb N$. I don't have any ideas on how I can show that this element belongs in $\{r_1a_1+r_2a_1+ \dots +r_ka_k \mid k \in \Bbb N, r_i \in R, a_i \in A, i \le k\}$. Should I forget about showing the inclusions and use the fact that $\langle A \rangle$ is the smallest ideal that contains $A$?

Using this I have that for $x \in \langle A \rangle$, $rx \in \langle A \rangle$ and $xr \in \langle A \rangle$ for any $r \in R$. Also $\langle A \rangle$ is a subgroup of $R$ so for any $a,b \in \langle A \rangle$ we have $a+b \in \langle A \rangle$. Still I don't now how this can be used to prove the claim.

Solution 1:

It is incorrect that $x=a_1 a_2\dots a_n$. Usually the statement you want to prove is given as the definition, but in your case, I assume we define $\langle A \rangle$ to be the smallest ideal containing $A$.

If this is the case, one of the inclusions follows directly from what you wrote, i.e. the definition of an ideal. For the other inclusion, simply use the minimality of $\langle A \rangle$.