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Let $R$ be a non-commutative ring, and $S\subset R$ some subset. Let $I_S$ be the smallest two-sided ideal of $R$ such that $I_S\supseteq S$. Is it true that (if $R$ is unital ring) $I_S$ consist only of elements of the form: $$\sum_{s\in S}\left(\sum_{k=1}^{N_s}x_{k}sy_{k}\right),$$ or (if $R$ isn't unit ring) $I_S$ consist only of elements of the form: $$\sum_{s\in S}\left(m_{s}s+xs+sy+\sum_{k=1}^{N_s}x_{k}sy_{k}\right),$$ where $N_s\in\mathbb{N}$, $m_{s}\in\mathbb{Z}$ and $x, y, x_{k}, y_{k}\in R$. Thanks a lot!

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Yes. This isn't too hard to prove. –  Qiaochu Yuan Jun 9 '12 at 19:03
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Both statements are correct. The simplest way to prove this is to first note that $$I_S = \bigcap I$$ where the intersection runs over all ideals $I$ of $R$ that contains $S$.

Next, prove that each of the elements in question is contained in any ideal $I$ that contains $S$. This will prove that the set of such elements is contained in the intersection.

Then prove that the set of all such elements is itself an ideal that contains $S$. This will prove that the intersection is contained in the set of all such elements.

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