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!
Two-sided ideal generated by subset of not commutative ring
2
$\begingroup$
abstract-algebra
ring-theory
-
3Yes. This isn'$t$ too hard to prove. – 2012-06-09
1 Answers
3
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.