Vtyjkyuk

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 $a= langle a_1,a_2,....a_nrangle$

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 $r_i neq 0$.

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 $a_i in I$.