Let $R$ be a topological ring and $S$ a subring of $R$. Also let $X$ be a subset of $R$ such that the ideal generated by $X$ is all of $R$.
If $X \subset \bar{S}$, can we prove that $S$ is dense in $R$?
2026-03-25 11:08:38.1774436918
criterion for dense subrings
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Since the ideal generated by $X$ is the smallest ideal containing $X$, it follows that $\bar{S} = R$ whenever $\bar{S}$ is an ideal. This would imply $S$ is necessarily dense in $R$.
What isn't as clear to me however is that $\bar{S}$ must be an ideal a priori. Even if $S$ is a subring it does not imply $S$ is an ideal. Why then must $\bar{S}$ be an ideal?