Is <S> equal to the intersection of all ideals in R that contain S?

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Homework Statement


The following are equivalent for S\subseteqR, S\neq\oslash, and R is a commutative ring with unity(multiplicative identity):

1. <S> is the ideal generated by S.
2. <S> = \bigcap(I Ideal in R, S\subseteqI) = J
3. <S> = {\sumrisi: is any integer from 1 to n, ri\inR \foralli and si\inS \foralli} = K

Homework Equations


The Attempt at a Solution


It's been some time since I worked on this and at the time I understood everything I was working on but now when I look at it, I'm thoroughly confused. Where I got stuck is showing 2 \Rightarrow 3. I've got, assume <S> = J. Choose a \inK. Let I be an ideal of R that contains S. Because each ri\inR, si\inS, each risi\inI by IO closure. Then a \in I by closure under addition. Thus a \in J and K\subseteqJ.

I'm having trouble with starting to show that J\subseteqK.
 
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S \subseteq I \Rightarrow \forall s \in S, (s) \subseteq I
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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