Problem
Let R be a commutative ring with unity and let $a_1,a_2,…….,a_n$ belong to R. Then prove that $I=\langle a_1,a_2,….a_n \rangle$={$r_1a_1+…+r_na_n$} is an ideal and if J is any ideal that contains $a_1a_2,…a_n$,then $I \subseteq J$.
Attempt
1) $ra,ar \in I$ where $a= \langle a_1,a_2,….a_n\rangle$
2) It is a subring.
Hence $I$ is an ideal .
Doubt
I am not sure how to prove second part.
$I=\langle a_1,a_2,….a_n \rangle$={$r_1a_1+…+r_na_n$} where $r_i \neq 0$.
Best Answer
Since
$a_i \in J, \; 1 \le i \le n, \tag 1$
then
$r_i a_i \in J, \; r_i \in R, \; 1 \le i \le n; \tag 2$
therefore,
$\displaystyle \sum_1^n r_i a_i \in J, \; \forall r_i \in R, \; 1 \le i \le n; \tag 3$
but the sums
$\displaystyle \sum_1^n r_i a_i \tag 4$
are precisely the elements of $I$; therefore,
$I \subseteq J. \tag 5$
To show $I$ is an subring merely note that if
$\displaystyle \sum_1^n r_i a_i, \; \sum_1^n s_i a_i \in I, \tag 6$
then
$\displaystyle \sum_1^n r_i a_i - \sum_1^n s_i a_i = \sum_1^n (r_i - s_i) a_i \in I; \tag 7$
as for products,
$\left (\displaystyle \sum_1^n r_i a_i \right ) \left(\displaystyle \sum_1^n s_i a_i \right ) = \displaystyle \sum_{i,j = 1}^n r_i s_j a_i a_j \in I, \tag 8$
since each $a_i \in I$.