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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 $r_i \neq 0$.

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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 $a_i \in I$.

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