I want to show these:
(a) Let $x_1,x_2,…, x_n$ be elements of a commutative ring R. Show that the ideal generated by $x_1,x_2,…,x_n$ is the smallest ideal contain $x_1,x_2,…,x_n$.
(b) Conclude that the ideal generated by $x_1,x_2,…,x_n$ is the intersection of all ideals that contain $x_1, x_2,. . ., x_n$
a) $ (x_1,x_2,….,x_n)={r_1x_1+r_2x_2+……+r_nx_n} $
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$0=0x_1+0x_2+…..+0x_n$ , $0\in I $
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let $a={r_1x_1+r_2x_2+……+r_nx_n} , b={r_{11}x_1+r_{22}x_2+……+r_{nn}x_n} $
$a-b={(r_1-r_{11})x_1+(r_2-r_{22})x_2+……+(r_n-r_{nn})x_n} $,then $a-b\in I$
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let $r \in R , a\in I $
$ra={rr_1x_1+rr_2x_2+……+rr_nx_n} $, then $ra\in I $ -
so it is ideal
In this part, I show it is ideal but I don't know how to show that is smallest ideal.
b) I know if it is smallest ideal then all ideal of $R$ contains $I$, but how to prove it?
Best Answer
Take any ideal $I$ containing your $n$ elements. Then take any element $y$ in the ideal $J$ generated by the elements. Using the properties of an ideal, can you show that $y\in I$ (so that $J\subset I$)? This should be straightforward once you write out what you want. (You seem to have the calculation mostly there, what is missing is reasoning.) The conclusion is that any ideal $I$ containing your elements contains $J$ as a subideal. Therefore $J$ is the smallest possible ideal.
For part (b), use part (a). Show the inclusion both ways: the intersection contains the generated ideal and vice versa. One of the directions needs part (a).